Chapter 5 Complex Numbers And Quadratic Equations

Given:

The equation is $4x + i(3x - y) = 3 + i(-6)$.

Here, $x$ and $y$ are real numbers.

---

To Find:

The values of $x$ and $y$.

---

Solution:

We are given the equation:

We know that two complex numbers $z_1 = a + ib$ and $z_2 = c + id$ are equal if and only if their real parts are equal and their imaginary parts are equal.

That is, $a = c$ and $b = d$.

Comparing the real parts of the given equation, we get:

Solving for $x$:

$x = \frac{3}{4}$

Comparing the imaginary parts of the given equation, we get:

Substitute the value of $x = \frac{3}{4}$ from equation (i) into equation (ii):

$3 \left( \frac{3}{4} \right) - y = -6$

$\frac{9}{4} - y = -6$

Rearranging the terms to solve for $y$:

$y = \frac{9}{4} + 6$

$y = \frac{9 + 4 \times 6}{4}$

$y = \frac{9 + 24}{4}$

$y = \frac{33}{4}$

---

Conclusion:

The values of $x$ and $y$ are $x = \frac{3}{4}$ and $y = \frac{33}{4}$.

(i)

Given Expression:

$(-5i) \left( \frac{1}{8}i \right)$

---

To Find:

Express the given expression in the form $a + bi$.

---

Solution:

We have:

$(-5i) \left( \frac{1}{8}i \right) = \left(-5 \times \frac{1}{8}\right) \times (i \times i)$

$= -\frac{5}{8} \times i^2$

Since $i^2 = -1$, we substitute this value:

$= -\frac{5}{8} \times (-1)$

$= \frac{5}{8}$

To express this in the form $a + bi$, we can write:

$= \frac{5}{8} + 0i$

Here, $a = \frac{5}{8}$ and $b = 0$.

---

(ii)

Given Expression:

$(-i) (2i) \left( -\frac{1}{8}i\right)^{3}$

---

To Find:

Express the given expression in the form $a + bi$.

---

Solution:

We have:

$(-i) (2i) \left( -\frac{1}{8}i\right)^{3}$

First, let's simplify $(-i)(2i)$:

$(-i)(2i) = -2 \times i^2 = -2(-1) = 2$

Next, let's simplify $\left( -\frac{1}{8}i\right)^{3}$:

$\left( -\frac{1}{8}i\right)^{3} = \left(-\frac{1}{8}\right)^3 \times i^3$

$= -\frac{1}{8 \times 8 \times 8} \times i^3$

$= -\frac{1}{512} \times i^3$

We know that $i^3 = i^2 \times i = (-1) \times i = -i$. Substituting this:

$= -\frac{1}{512} \times (-i)$

$= \frac{1}{512} i$

Now, multiply the simplified parts:

$(-i) (2i) \left( -\frac{1}{8}i\right)^{3} = (2) \times \left( \frac{1}{512} i \right)$

$= \frac{2}{512} i$

$= \frac{\cancel{2}^1}{\cancel{512}_{256}} i$

$= \frac{1}{256} i$

To express this in the form $a + bi$, we can write:

$0 + \frac{1}{256} i$

Here, $a = 0$ and $b = \frac{1}{256}$.

Given Expression:

$(5 - 3i)^3$

---

To Find:

Express the given expression in the form $a + bi$.

---

Solution:

We need to expand $(5 - 3i)^3$. We can use the binomial expansion formula for $(x - y)^3$:

$(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3$

Let $x = 5$ and $y = 3i$. Substituting these values into the formula:

$(5 - 3i)^3 = (5)^3 - 3(5)^2(3i) + 3(5)(3i)^2 - (3i)^3$

Now, let's calculate each term:

$(5)^3 = 125$

$3(5)^2(3i) = 3(25)(3i) = 75(3i) = 225i$

$3(5)(3i)^2 = 15 (3^2 i^2) = 15 (9 i^2)$

Since $i^2 = -1$:

$15 (9 (-1)) = 15 (-9) = -135$

$(3i)^3 = 3^3 i^3 = 27 i^3$

Since $i^3 = i^2 \cdot i = (-1)i = -i$:

$27 (-i) = -27i$

Now substitute these calculated values back into the expanded expression:

$(5 - 3i)^3 = 125 - 225i + (-135) - (-27i)$

$= 125 - 225i - 135 + 27i$

Group the real and imaginary parts:

$= (125 - 135) + (-225 + 27)i$

$= -10 + (-198)i$

$= -10 - 198i$

---

Conclusion:

The expression $(5 - 3i)^3$ in the form $a + bi$ is $-10 - 198i$.

Here, $a = -10$ and $b = -198$.

Given Expression:

$(-\sqrt{3} + \sqrt{-2}) (2\sqrt{3}- i)$

---

To Find:

Express the given expression in the form $a + bi$.

---

Solution:

First, simplify the term $\sqrt{-2}$:

$\sqrt{-2} = \sqrt{-1 \times 2} = \sqrt{-1} \times \sqrt{2} = i\sqrt{2}$

Substitute this back into the expression:

$(-\sqrt{3} + i\sqrt{2}) (2\sqrt{3}- i)$

Now, multiply the two complex numbers using the distributive property:

$= (-\sqrt{3})(2\sqrt{3}) + (-\sqrt{3})(-i) + (i\sqrt{2})(2\sqrt{3}) + (i\sqrt{2})(-i)$

Calculate each term:

$(-\sqrt{3})(2\sqrt{3}) = -2(\sqrt{3})^2 = -2(3) = -6$

$(-\sqrt{3})(-i) = i\sqrt{3}$

$(i\sqrt{2})(2\sqrt{3}) = i(2\sqrt{2 \times 3}) = i(2\sqrt{6})$

$(i\sqrt{2})(-i) = -i^2 \sqrt{2}$

Since $i^2 = -1$, the last term becomes:

$-(-1)\sqrt{2} = \sqrt{2}$

Combine the terms:

$ = -6 + i\sqrt{3} + i(2\sqrt{6}) + \sqrt{2}$

Group the real parts and the imaginary parts:

$= (-6 + \sqrt{2}) + (i\sqrt{3} + i(2\sqrt{6}))$

$= (-6 + \sqrt{2}) + i(\sqrt{3} + 2\sqrt{6})$

---

Conclusion:

The expression $(-\sqrt{3} + \sqrt{-2}) (2\sqrt{3}- i)$ in the form $a + bi$ is $(-6 + \sqrt{2}) + i(\sqrt{3} + 2\sqrt{6})$.

Here, $a = -6 + \sqrt{2}$ and $b = \sqrt{3} + 2\sqrt{6}$.

Given:

The complex number is $z = 2 - 3i$.

---

To Find:

The multiplicative inverse of $z$, denoted as $z^{-1}$ or $\frac{1}{z}$.

---

Solution:

The multiplicative inverse of a complex number $z$ is $\frac{1}{z}$.

So, we need to find $\frac{1}{2 - 3i}$.

To express the multiplicative inverse in the form $a + bi$, we multiply the numerator and the denominator by the complex conjugate of the denominator.

The complex conjugate of $z = 2 - 3i$ is $\bar{z} = 2 + 3i$.

Multiply $\frac{1}{2 - 3i}$ by $\frac{2 + 3i}{2 + 3i}$:

$z^{-1} = \frac{1}{2 - 3i} \times \frac{2 + 3i}{2 + 3i}$

$z^{-1} = \frac{2 + 3i}{(2 - 3i)(2 + 3i)}$

Using the property $(a - bi)(a + bi) = a^2 + b^2$ for the denominator:

$(2 - 3i)(2 + 3i) = (2)^2 + (3)^2 = 4 + 9 = 13$

Substitute this back into the expression for $z^{-1}$:

$z^{-1} = \frac{2 + 3i}{13}$

Separate the real and imaginary parts:

$z^{-1} = \frac{2}{13} + \frac{3}{13}i$

This is in the form $a + bi$, where $a = \frac{2}{13}$ and $b = \frac{3}{13}$.

---

Alternate Solution:

The multiplicative inverse of a non-zero complex number $z = a + bi$ can be found using the formula:

$z^{-1} = \frac{\bar{z}}{|z|^2}$

Where $\bar{z}$ is the complex conjugate of $z$ and $|z|^2$ is the square of the modulus of $z$.

Given $z = 2 - 3i$. Here, $a = 2$ and $b = -3$.

The complex conjugate is $\bar{z} = 2 - (-3i) = 2 + 3i$.

The square of the modulus is $|z|^2 = a^2 + b^2 = (2)^2 + (-3)^2 = 4 + 9 = 13$.

Now, substitute $\bar{z}$ and $|z|^2$ into the formula:

$z^{-1} = \frac{2 + 3i}{13}$

Expressing this in the form $a + bi$:

$z^{-1} = \frac{2}{13} + \frac{3}{13}i$

---

Conclusion:

The multiplicative inverse of $2 - 3i$ is $\frac{2}{13} + \frac{3}{13}i$.

(i)

Given Expression:

$\frac{5 + \sqrt{2}i}{1 - \sqrt{2}i}$

---

To Find:

Express the given expression in the form $a + bi$.

---

Solution:

To express the fraction in the form $a + bi$, we multiply the numerator and the denominator by the complex conjugate of the denominator.

The denominator is $1 - \sqrt{2}i$.

The complex conjugate of the denominator is $1 + \sqrt{2}i$.

Multiply the expression by $\frac{1 + \sqrt{2}i}{1 + \sqrt{2}i}$:

$\frac{5 + \sqrt{2}i}{1 - \sqrt{2}i} = \frac{5 + \sqrt{2}i}{1 - \sqrt{2}i} \times \frac{1 + \sqrt{2}i}{1 + \sqrt{2}i}$

$= \frac{(5 + \sqrt{2}i)(1 + \sqrt{2}i)}{(1 - \sqrt{2}i)(1 + \sqrt{2}i)}$

Expand the numerator:

$(5 + \sqrt{2}i)(1 + \sqrt{2}i) = 5(1) + 5(\sqrt{2}i) + (\sqrt{2}i)(1) + (\sqrt{2}i)(\sqrt{2}i)$

$ = 5 + 5\sqrt{2}i + \sqrt{2}i + (\sqrt{2})^2 i^2$

$ = 5 + (5\sqrt{2} + \sqrt{2})i + 2i^2$

$ = 5 + 6\sqrt{2}i + 2(-1)$ (Since $i^2 = -1$)

$ = 5 + 6\sqrt{2}i - 2$

$ = 3 + 6\sqrt{2}i$

Expand the denominator using the property $(a - bi)(a + bi) = a^2 + b^2$:

$(1 - \sqrt{2}i)(1 + \sqrt{2}i) = (1)^2 + (\sqrt{2})^2$

$ = 1 + 2 = 3$

Substitute the expanded numerator and denominator back into the fraction:

$\frac{3 + 6\sqrt{2}i}{3}$

Separate the real and imaginary parts:

$= \frac{3}{3} + \frac{6\sqrt{2}}{3}i$

$ = 1 + 2\sqrt{2}i$

---

Conclusion (i):

The expression $\frac{5 + \sqrt{2}i}{1 - \sqrt{2}i}$ in the form $a + bi$ is $1 + 2\sqrt{2}i$.

Here, $a = 1$ and $b = 2\sqrt{2}$.

---

(ii)

Given Expression:

$i^{-35}$

---

To Find:

Express the given expression in the form $a + bi$.

---

Solution:

We have:

$i^{-35} = \frac{1}{i^{35}}$

To simplify $i^{35}$, we find the remainder when 35 is divided by 4.

$35 = 4 \times 8 + 3$

So, the remainder is 3.

$i^{35} = i^3$

We know that $i^3 = i^2 \cdot i = (-1)i = -i$.

Substitute this back into the expression:

$i^{-35} = \frac{1}{-i}$

To express this in the form $a + bi$, multiply the numerator and denominator by $i$:

$\frac{1}{-i} \times \frac{i}{i} = \frac{i}{-i^2}$

Since $i^2 = -1$:

$= \frac{i}{-(-1)} = \frac{i}{1} = i$

To write this in the form $a + bi$:

$0 + 1i$

---

Conclusion (ii):

The expression $i^{-35}$ in the form $a + bi$ is $0 + i$.

Here, $a = 0$ and $b = 1$.

Given Expression:

$(5i) \left(-\frac{3}{5} i \right)$

---

To Find:

Express the given expression in the form $a + bi$.

---

Solution:

We need to multiply the given complex numbers:

$(5i) \left(-\frac{3}{5} i \right) = \left( 5 \times -\frac{3}{5} \right) \times (i \times i)$

First, multiply the coefficients:

$5 \times \left(-\frac{3}{5}\right) = -\frac{5 \times 3}{5} = -\frac{\cancel{15}^3}{\cancel{5}_1} = -3$

Next, multiply the imaginary units:

$i \times i = i^2$

We know that $i^2 = -1$.

Substitute the values back into the expression:

$= (-3) \times i^2$

$= (-3) \times (-1)$

$= 3$

To express this in the form $a + bi$, we can write:

$3 + 0i$

---

Conclusion:

The expression $(5i) \left(-\frac{3}{5} i \right)$ in the form $a + bi$ is $3 + 0i$.

Here, $a = 3$ and $b = 0$.

Given Expression:

$i^9 + i^{19}$

---

To Find:

Express the given expression in the form $a + bi$.

---

Solution:

We need to simplify the powers of $i$. We know the cycle of powers of $i$: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$.

To simplify $i^n$, we can find the remainder when $n$ is divided by 4.

Simplify $i^9$:

Divide 9 by 4:

$9 = 4 \times 2 + 1$

The remainder is 1. Therefore:

$i^9 = i^1 = i$

Simplify $i^{19}$:

Divide 19 by 4:

$19 = 4 \times 4 + 3$

The remainder is 3. Therefore:

$i^{19} = i^3$

We know $i^3 = -i$. So:

$i^{19} = -i$

Now, substitute these simplified values back into the original expression:

$i^9 + i^{19} = i + (-i)$

$= i - i$

$= 0$

To express this in the form $a + bi$, we can write:

$0 + 0i$

---

Conclusion:

The expression $i^9 + i^{19}$ in the form $a + bi$ is $0 + 0i$.

Here, $a = 0$ and $b = 0$.

Given Expression:

$i^{-39}$

---

To Find:

Express the given expression in the form $a + bi$.

---

Solution:

We have the expression $i^{-39}$. Using the property of exponents $x^{-n} = \frac{1}{x^n}$, we can write:

$i^{-39} = \frac{1}{i^{39}}$

To simplify $i^{39}$, we find the remainder when 39 is divided by 4. The cycle of powers of $i$ repeats every 4 ($i^1=i, i^2=-1, i^3=-i, i^4=1$).

Divide 39 by 4:

$39 = 4 \times 9 + 3$

The remainder is 3. Therefore:

$i^{39} = i^3$

We know that $i^3 = i^2 \cdot i = (-1)i = -i$.

So, $i^{39} = -i$.

Substitute this value back into the expression for $i^{-39}$:

$i^{-39} = \frac{1}{-i}$

To express this in the form $a + bi$, we need to eliminate $i$ from the denominator. Multiply the numerator and the denominator by $i$:

$\frac{1}{-i} \times \frac{i}{i} = \frac{i}{-i^2}$

Since $i^2 = -1$, substitute this value:

$= \frac{i}{-(-1)}$

$= \frac{i}{1}$

$= i$

To write this in the standard form $a + bi$:

$0 + 1i$

---

Conclusion:

The expression $i^{-39}$ in the form $a + bi$ is $0 + i$.

Here, $a = 0$ and $b = 1$.

Given Expression:

$3(7 + i7) + i (7 + i7)$

---

To Find:

Express the given expression in the form $a + bi$.

---

Solution:

We need to simplify the given expression $3(7 + i7) + i (7 + i7)$.

First, distribute the terms:

$3(7 + i7) = 3 \times 7 + 3 \times (i7) = 21 + 21i$

$i(7 + i7) = i \times 7 + i \times (i7) = 7i + 7i^2$

Now substitute $i^2 = -1$ in the second part:

$7i + 7i^2 = 7i + 7(-1) = 7i - 7$

Now add the results of the two distributions:

$3(7 + i7) + i (7 + i7) = (21 + 21i) + (7i - 7)$

Group the real parts and the imaginary parts:

$= (21 - 7) + (21i + 7i)$

Combine the terms:

$= 14 + (21 + 7)i$

$= 14 + 28i$

---

Conclusion:

The expression $3(7 + i7) + i (7 + i7)$ in the form $a + bi$ is $14 + 28i$.

Here, $a = 14$ and $b = 28$.

Given Expression:

$(1 - i) - (-1 + i6)$

---

To Find:

Express the given expression in the form $a + bi$.

---

Solution:

We need to simplify the expression $(1 - i) - (-1 + i6)$.

First, remove the parentheses. Remember to distribute the negative sign to both terms in the second parenthesis:

$(1 - i) - (-1 + i6) = 1 - i - (-1) - (+i6)$

$= 1 - i + 1 - 6i$

Now, group the real parts and the imaginary parts:

$= (1 + 1) + (-i - 6i)$

Combine the real terms and the imaginary terms:

$= 2 + (-1 - 6)i$

$= 2 + (-7)i$

$= 2 - 7i$

---

Conclusion:

The expression $(1 - i) - (-1 + i6)$ in the form $a + bi$ is $2 - 7i$.

Here, $a = 2$ and $b = -7$.

Given Expression:

$\left( \frac{1}{5} +i\frac{2}{5}\right) - \left( 4+i\frac{5}{2}\right)$

---

To Find:

Express the given expression in the form $a + bi$.

---

Solution:

We need to simplify the expression $\left( \frac{1}{5} +i\frac{2}{5}\right) - \left( 4+i\frac{5}{2}\right)$.

First, remove the parentheses. Distribute the negative sign to the terms in the second parenthesis:

$= \frac{1}{5} + i\frac{2}{5} - 4 - i\frac{5}{2}$

Group the real parts and the imaginary parts:

$= \left( \frac{1}{5} - 4 \right) + \left( i\frac{2}{5} - i\frac{5}{2} \right)$

Combine the real parts. Find a common denominator (5):

$\frac{1}{5} - 4 = \frac{1}{5} - \frac{4 \times 5}{5} = \frac{1}{5} - \frac{20}{5} = \frac{1 - 20}{5} = -\frac{19}{5}$

Combine the imaginary parts. Factor out $i$ and find a common denominator (10):

$i\left( \frac{2}{5} - \frac{5}{2} \right) = i\left( \frac{2 \times 2}{5 \times 2} - \frac{5 \times 5}{2 \times 5} \right)$

$ = i\left( \frac{4}{10} - \frac{25}{10} \right)$

$ = i\left( \frac{4 - 25}{10} \right)$

$ = i\left( -\frac{21}{10} \right) = -\frac{21}{10}i$

Combine the simplified real and imaginary parts:

$ = -\frac{19}{5} - \frac{21}{10}i$

---

Conclusion:

The expression $\left( \frac{1}{5} +i\frac{2}{5}\right) - \left( 4+i\frac{5}{2}\right)$ in the form $a + bi$ is $-\frac{19}{5} - \frac{21}{10}i$.

Here, $a = -\frac{19}{5}$ and $b = -\frac{21}{10}$.

Given Expression:

$\left[ \left( \frac{1}{3} +i\frac{7}{3}\right)+\left( 4+i\frac{1}{3}\right) \right]-\left( -\frac{4}{3}+i \right)$

---

To Find:

Express the given expression in the form $a + bi$.

---

Solution:

First, simplify the expression inside the square brackets:

$\left( \frac{1}{3} +i\frac{7}{3}\right)+\left( 4+i\frac{1}{3}\right)$

Group the real parts and the imaginary parts:

$= \left( \frac{1}{3} + 4 \right) + \left( i\frac{7}{3} + i\frac{1}{3} \right)$

Combine the real parts (common denominator is 3):

$\frac{1}{3} + \frac{4 \times 3}{3} = \frac{1}{3} + \frac{12}{3} = \frac{1 + 12}{3} = \frac{13}{3}$

Combine the imaginary parts:

$i\left( \frac{7}{3} + \frac{1}{3} \right) = i\left( \frac{7 + 1}{3} \right) = i\frac{8}{3}$

So, the expression inside the square brackets simplifies to:

$\frac{13}{3} + i\frac{8}{3}$

Now, substitute this back into the original expression:

$\left( \frac{13}{3} + i\frac{8}{3} \right) - \left( -\frac{4}{3}+i \right)$

Remove the parentheses, distributing the negative sign:

$ = \frac{13}{3} + i\frac{8}{3} - \left(-\frac{4}{3}\right) - i$

$ = \frac{13}{3} + i\frac{8}{3} + \frac{4}{3} - i$

Group the real parts and the imaginary parts:

$ = \left( \frac{13}{3} + \frac{4}{3} \right) + \left( i\frac{8}{3} - i \right)$

Combine the real parts:

$\frac{13 + 4}{3} = \frac{17}{3}$

Combine the imaginary parts:

$i\left( \frac{8}{3} - 1 \right) = i\left( \frac{8}{3} - \frac{3}{3} \right) = i\left( \frac{8 - 3}{3} \right) = i\frac{5}{3}$

Combine the simplified real and imaginary parts:

$ = \frac{17}{3} + i\frac{5}{3}$

---

Conclusion:

The expression $\left[ \left( \frac{1}{3} +i\frac{7}{3}\right)+\left( 4+i\frac{1}{3}\right) \right]-\left( -\frac{4}{3}+i \right)$ in the form $a + bi$ is $\frac{17}{3} + i\frac{5}{3}$.

Here, $a = \frac{17}{3}$ and $b = \frac{5}{3}$.

Given Expression:

$(1 - i)^4$

---

To Find:

Express the given expression in the form $a + bi$.

---

Solution:

We need to compute $(1 - i)^4$. We can write this as $((1 - i)^2)^2$.

First, let's calculate $(1 - i)^2$ using the formula $(x - y)^2 = x^2 - 2xy + y^2$:

$(1 - i)^2 = (1)^2 - 2(1)(i) + (i)^2$

$= 1 - 2i + i^2$

Since $i^2 = -1$, we substitute this value:

$= 1 - 2i + (-1)$

$= 1 - 2i - 1$

$= -2i$

Now, we need to calculate $((1 - i)^2)^2$, which is $(-2i)^2$:

$(1 - i)^4 = (-2i)^2$

$= (-2)^2 \times (i)^2$

$ = 4 \times i^2$

Substitute $i^2 = -1$ again:

$= 4 \times (-1)$

$= -4$

To express this in the standard form $a + bi$, we write:

$-4 + 0i$

---

Conclusion:

The expression $(1 - i)^4$ in the form $a + bi$ is $-4 + 0i$.

Here, $a = -4$ and $b = 0$.

Given Expression:

$\left( \frac{1}{3}+3i \right)^{3}$

---

To Find:

Express the given expression in the form $a + bi$.

---

Solution:

We need to expand $\left( \frac{1}{3} + 3i \right)^{3}$. We use the binomial expansion formula for $(x + y)^3$:

$(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$

Let $x = \frac{1}{3}$ and $y = 3i$. Substitute these values into the formula:

$\left( \frac{1}{3} + 3i \right)^3 = \left(\frac{1}{3}\right)^3 + 3\left(\frac{1}{3}\right)^2(3i) + 3\left(\frac{1}{3}\right)(3i)^2 + (3i)^3$

Now, calculate each term:

$\left(\frac{1}{3}\right)^3 = \frac{1^3}{3^3} = \frac{1}{27}$

$3\left(\frac{1}{3}\right)^2(3i) = 3 \times \frac{1}{9} \times 3i = \frac{3 \times 3}{9} i = \frac{9}{9} i = i$

$3\left(\frac{1}{3}\right)(3i)^2 = 3 \times \frac{1}{3} \times (3^2 i^2) = 1 \times (9 i^2) = 9i^2$

Since $i^2 = -1$:

$9i^2 = 9(-1) = -9$

$(3i)^3 = 3^3 \times i^3 = 27 i^3$

Since $i^3 = i^2 \cdot i = (-1)i = -i$:

$27 i^3 = 27(-i) = -27i$

Now substitute these calculated values back into the expanded expression:

$\left( \frac{1}{3} + 3i \right)^3 = \frac{1}{27} + i + (-9) + (-27i)$

$= \frac{1}{27} + i - 9 - 27i$

Group the real parts and the imaginary parts:

$= \left( \frac{1}{27} - 9 \right) + (i - 27i)$

Combine the real parts (find a common denominator):

$\frac{1}{27} - 9 = \frac{1}{27} - \frac{9 \times 27}{27} = \frac{1 - 243}{27} = -\frac{242}{27}$

Combine the imaginary parts:

$(1 - 27)i = -26i$

Combine the simplified real and imaginary parts:

$ = -\frac{242}{27} - 26i$

---

Conclusion:

The expression $\left( \frac{1}{3}+3i \right)^{3}$ in the form $a + bi$ is $-\frac{242}{27} - 26i$.

Here, $a = -\frac{242}{27}$ and $b = -26$.

Given Expression:

$\left( -2-\frac{1}{3}i\right)^{3}$

---

To Find:

Express the given expression in the form $a + bi$.

---

Solution:

We need to compute $\left( -2-\frac{1}{3}i\right)^{3}$.

First, factor out $-1$ from the base:

$\left( -2-\frac{1}{3}i\right)^{3} = \left[ -1 \left( 2+\frac{1}{3}i \right) \right]^3$

$ = (-1)^3 \left( 2+\frac{1}{3}i \right)^3$

$ = - \left( 2+\frac{1}{3}i \right)^3$

Now, we use the binomial expansion formula for $(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$.

Let $x = 2$ and $y = \frac{1}{3}i$.

$\left( 2+\frac{1}{3}i \right)^3 = (2)^3 + 3(2)^2\left(\frac{1}{3}i\right) + 3(2)\left(\frac{1}{3}i\right)^2 + \left(\frac{1}{3}i\right)^3$

Calculate each term:

$(2)^3 = 8$

$3(2)^2\left(\frac{1}{3}i\right) = 3(4)\left(\frac{1}{3}i\right) = 12 \times \frac{1}{3}i = 4i$

$3(2)\left(\frac{1}{3}i\right)^2 = 6\left(\frac{1}{9}i^2\right) = \frac{6}{9}i^2 = \frac{2}{3}i^2$

Since $i^2 = -1$:

$\frac{2}{3}i^2 = \frac{2}{3}(-1) = -\frac{2}{3}$

$\left(\frac{1}{3}i\right)^3 = \left(\frac{1}{3}\right)^3 i^3 = \frac{1}{27}i^3$

Since $i^3 = -i$:

$\frac{1}{27}i^3 = \frac{1}{27}(-i) = -\frac{1}{27}i$

Substitute these values back into the expansion of $\left( 2+\frac{1}{3}i \right)^3$:

$ = 8 + 4i - \frac{2}{3} - \frac{1}{27}i$

Group the real and imaginary parts:

$= \left( 8 - \frac{2}{3} \right) + \left( 4i - \frac{1}{27}i \right)$

Combine the real parts:

$8 - \frac{2}{3} = \frac{24}{3} - \frac{2}{3} = \frac{22}{3}$

Combine the imaginary parts:

$4i - \frac{1}{27}i = \left( 4 - \frac{1}{27} \right)i = \left( \frac{108}{27} - \frac{1}{27} \right)i = \frac{107}{27}i$

So, $\left( 2+\frac{1}{3}i \right)^3 = \frac{22}{3} + \frac{107}{27}i$.

Finally, apply the negative sign from the beginning:

$\left( -2-\frac{1}{3}i\right)^{3} = - \left( \frac{22}{3} + \frac{107}{27}i \right)$

$= -\frac{22}{3} - \frac{107}{27}i$

---

Conclusion:

The expression $\left( -2-\frac{1}{3}i\right)^{3}$ in the form $a + bi$ is $-\frac{22}{3} - \frac{107}{27}i$.

Here, $a = -\frac{22}{3}$ and $b = -\frac{107}{27}$.

Given:

The complex number is $z = 4 - 3i$.

---

To Find:

The multiplicative inverse of $z$, denoted as $z^{-1}$ or $\frac{1}{z}$.

---

Solution:

The multiplicative inverse of a complex number $z$ is given by $\frac{1}{z}$.

So, we need to calculate $z^{-1} = \frac{1}{4 - 3i}$.

To express the inverse in the standard form $a + bi$, we multiply the numerator and the denominator by the complex conjugate of the denominator.

The complex conjugate of $z = 4 - 3i$ is $\bar{z} = 4 + 3i$.

Multiply $\frac{1}{4 - 3i}$ by $\frac{4 + 3i}{4 + 3i}$:

$z^{-1} = \frac{1}{4 - 3i} \times \frac{4 + 3i}{4 + 3i}$

$z^{-1} = \frac{4 + 3i}{(4 - 3i)(4 + 3i)}$

Using the property $(a - bi)(a + bi) = a^2 + b^2$ for the denominator:

$(4 - 3i)(4 + 3i) = (4)^2 + (3)^2 = 16 + 9 = 25$

Substitute this back into the expression for $z^{-1}$:

$z^{-1} = \frac{4 + 3i}{25}$

Separate the real and imaginary parts to get the form $a + bi$:

$z^{-1} = \frac{4}{25} + \frac{3}{25}i$

---

Alternate Solution:

The multiplicative inverse of a non-zero complex number $z = a + bi$ can be calculated using the formula:

$z^{-1} = \frac{\bar{z}}{|z|^2}$

Where $\bar{z}$ is the complex conjugate of $z$ and $|z|^2$ is the square of the modulus of $z$.

Given $z = 4 - 3i$. Here, $a = 4$ and $b = -3$.

The complex conjugate is $\bar{z} = 4 - (-3i) = 4 + 3i$.

The square of the modulus is $|z|^2 = a^2 + b^2 = (4)^2 + (-3)^2 = 16 + 9 = 25$.

Now, substitute $\bar{z}$ and $|z|^2$ into the formula:

$z^{-1} = \frac{4 + 3i}{25}$

Expressing this in the form $a + bi$:

$z^{-1} = \frac{4}{25} + \frac{3}{25}i$

---

Conclusion:

The multiplicative inverse of $4 - 3i$ is $\frac{4}{25} + \frac{3}{25}i$.

Given:

The complex number is $z = \sqrt{5} + 3i$.

---

To Find:

The multiplicative inverse of $z$, denoted as $z^{-1}$ or $\frac{1}{z}$.

---

Solution:

The multiplicative inverse of a complex number $z$ is given by $\frac{1}{z}$.

So, we need to calculate $z^{-1} = \frac{1}{\sqrt{5} + 3i}$.

To express the inverse in the standard form $a + bi$, we multiply the numerator and the denominator by the complex conjugate of the denominator.

The complex conjugate of $z = \sqrt{5} + 3i$ is $\bar{z} = \sqrt{5} - 3i$.

Multiply $\frac{1}{\sqrt{5} + 3i}$ by $\frac{\sqrt{5} - 3i}{\sqrt{5} - 3i}$:

$z^{-1} = \frac{1}{\sqrt{5} + 3i} \times \frac{\sqrt{5} - 3i}{\sqrt{5} - 3i}$

$z^{-1} = \frac{\sqrt{5} - 3i}{(\sqrt{5} + 3i)(\sqrt{5} - 3i)}$

Using the property $(a + bi)(a - bi) = a^2 + b^2$ for the denominator:

$(\sqrt{5} + 3i)(\sqrt{5} - 3i) = (\sqrt{5})^2 + (3)^2 = 5 + 9 = 14$

Substitute this back into the expression for $z^{-1}$:

$z^{-1} = \frac{\sqrt{5} - 3i}{14}$

Separate the real and imaginary parts to get the form $a + bi$:

$z^{-1} = \frac{\sqrt{5}}{14} - \frac{3}{14}i$

---

Alternate Solution:

The multiplicative inverse of a non-zero complex number $z = a + bi$ can be calculated using the formula:

$z^{-1} = \frac{\bar{z}}{|z|^2}$

Where $\bar{z}$ is the complex conjugate of $z$ and $|z|^2$ is the square of the modulus of $z$.

Given $z = \sqrt{5} + 3i$. Here, $a = \sqrt{5}$ and $b = 3$.

The complex conjugate is $\bar{z} = \sqrt{5} - 3i$.

The square of the modulus is $|z|^2 = a^2 + b^2 = (\sqrt{5})^2 + (3)^2 = 5 + 9 = 14$.

Now, substitute $\bar{z}$ and $|z|^2$ into the formula:

$z^{-1} = \frac{\sqrt{5} - 3i}{14}$

Expressing this in the form $a + bi$:

$z^{-1} = \frac{\sqrt{5}}{14} - \frac{3}{14}i$

---

Conclusion:

The multiplicative inverse of $\sqrt{5} + 3i$ is $\frac{\sqrt{5}}{14} - \frac{3}{14}i$.

Given:

The complex number is $z = -i$. This can also be written as $z = 0 - i$.

---

To Find:

The multiplicative inverse of $z$, denoted as $z^{-1}$ or $\frac{1}{z}$.

---

Solution:

The multiplicative inverse of a complex number $z$ is given by $\frac{1}{z}$.

So, we need to calculate $z^{-1} = \frac{1}{-i}$.

To express the inverse in the standard form $a + bi$, we multiply the numerator and the denominator by the complex conjugate of the denominator.

The complex conjugate of $z = 0 - i$ is $\bar{z} = 0 + i = i$.

Multiply $\frac{1}{-i}$ by $\frac{i}{i}$ (Note: Multiplying by $\frac{i}{i}$ is a common technique to simplify fractions with $i$ in the denominator):

$z^{-1} = \frac{1}{-i} \times \frac{i}{i}$

$z^{-1} = \frac{i}{-i^2}$

Since $i^2 = -1$, substitute this value:

$z^{-1} = \frac{i}{-(-1)}$

$z^{-1} = \frac{i}{1}$

$z^{-1} = i$

Separate the real and imaginary parts to get the form $a + bi$:

$z^{-1} = 0 + 1i$

---

Alternate Solution:

The multiplicative inverse of a non-zero complex number $z = a + bi$ can be calculated using the formula:

$z^{-1} = \frac{\bar{z}}{|z|^2}$

Where $\bar{z}$ is the complex conjugate of $z$ and $|z|^2$ is the square of the modulus of $z$.

Given $z = -i = 0 - 1i$. Here, $a = 0$ and $b = -1$.

The complex conjugate is $\bar{z} = 0 - (-1)i = 0 + 1i = i$.

The square of the modulus is $|z|^2 = a^2 + b^2 = (0)^2 + (-1)^2 = 0 + 1 = 1$.

Now, substitute $\bar{z}$ and $|z|^2$ into the formula:

$z^{-1} = \frac{i}{1}$

$z^{-1} = i$

Expressing this in the form $a + bi$:

$z^{-1} = 0 + 1i$

---

Conclusion:

The multiplicative inverse of $-i$ is $i$, which can be written as $0 + i$.

Given Expression:

Let the given expression be $Z$.

$Z = \frac{(3 + i\sqrt{5})(3 - i\sqrt{5})}{(\sqrt{3} + i\sqrt{2}) - (\sqrt{3} - i\sqrt{2})}$

---

To Find:

Express the given expression $Z$ in the form $a + bi$.

---

Solution:

We first simplify the numerator and the denominator separately.

Numerator:

The numerator is $(3 + i\sqrt{5})(3 - i\sqrt{5})$.

This is in the form $(a + ib)(a - ib)$, which equals $a^2 + b^2$.

Here, $a = 3$ and $b = \sqrt{5}$.

Numerator $= (3)^2 + (\sqrt{5})^2$

$= 9 + 5$

$ = 14$

Denominator:

The denominator is $(\sqrt{3} + i\sqrt{2}) - (\sqrt{3} - i\sqrt{2})$.

Remove the parentheses:

Denominator $= \sqrt{3} + i\sqrt{2} - \sqrt{3} - (-i\sqrt{2})$

$= \sqrt{3} + i\sqrt{2} - \sqrt{3} + i\sqrt{2}$

Group the real and imaginary parts:

$= (\sqrt{3} - \sqrt{3}) + (i\sqrt{2} + i\sqrt{2})$

$ = 0 + 2i\sqrt{2}$

$ = 2\sqrt{2}i$

Combine Numerator and Denominator:

Now substitute the simplified numerator and denominator back into the expression for $Z$:

$Z = \frac{14}{2\sqrt{2}i}$

Simplify the fraction:

$Z = \frac{\cancel{14}^7}{\cancel{2}\sqrt{2}i} = \frac{7}{\sqrt{2}i}$

To express $Z$ in the form $a + bi$, we need to eliminate $i$ from the denominator. Multiply the numerator and denominator by $i$:

$Z = \frac{7}{\sqrt{2}i} \times \frac{i}{i}$

$Z = \frac{7i}{\sqrt{2}i^2}$

Since $i^2 = -1$:

$Z = \frac{7i}{\sqrt{2}(-1)}$

$Z = -\frac{7i}{\sqrt{2}}$

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{2}$:

$Z = -\frac{7i}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$

$Z = -\frac{7\sqrt{2}i}{(\sqrt{2})^2}$

$Z = -\frac{7\sqrt{2}i}{2}$

Finally, write the expression in the form $a + bi$:

$Z = 0 - \frac{7\sqrt{2}}{2}i$

---

Conclusion:

The expression $\frac{(3 + i\sqrt{5})(3 - i\sqrt{5})}{(\sqrt{3} + i\sqrt{2}) - (\sqrt{3} - i\sqrt{2})}$ in the form $a + bi$ is $0 - \frac{7\sqrt{2}}{2}i$.

Here, $a = 0$ and $b = -\frac{7\sqrt{2}}{2}$.

Given:

The complex number is $z = 1 + i\sqrt{3}$.

To Find:

The polar form of the complex number $z$.

---

Solution:

The standard form of a complex number is $z = x + iy$. Comparing this with the given number, we have:

$x = 1$ and $y = \sqrt{3}$.

The polar form of a complex number is given by $z = r(\cos \theta + i \sin \theta)$, where $r$ is the modulus and $\theta$ is the argument.

Step 1: Calculate the modulus, $r$.

The modulus is calculated using the formula $r = |z| = \sqrt{x^2 + y^2}$.

$r = \sqrt{(1)^2 + (\sqrt{3})^2}$

$r = \sqrt{1 + 3} = \sqrt{4}$

$r = 2$

Step 2: Calculate the argument, $\theta$.

The argument $\theta$ must satisfy the equations:

$\cos \theta = \frac{x}{r} = \frac{1}{2}$

$\sin \theta = \frac{y}{r} = \frac{\sqrt{3}}{2}$

Since both $\cos \theta$ and $\sin \theta$ are positive, the angle $\theta$ lies in the first quadrant.

The principal value of $\theta$ that satisfies these conditions is $\theta = \frac{\pi}{3}$.

Step 3: Write the polar form.

Substitute the values of $r=2$ and $\theta = \frac{\pi}{3}$ into the polar form expression:

$z = 2 \left( \cos\left(\frac{\pi}{3}\right) + i \sin\left(\frac{\pi}{3}\right) \right)$

---

The polar form of the complex number $z = 1 + i\sqrt{3}$ is $2 \left( \cos\frac{\pi}{3} + i \sin\frac{\pi}{3} \right)$.

Given:

The complex number is $z = \frac{-16}{1 + i\sqrt{3}}$.

To Find:

The polar form of the complex number $z$.

---

Solution:

Step 1: Convert the complex number to the standard form $x + iy$.

To do this, we multiply the numerator and the denominator by the conjugate of the denominator, which is $1 - i\sqrt{3}$.

$z = \frac{-16}{1 + i\sqrt{3}} \times \frac{1 - i\sqrt{3}}{1 - i\sqrt{3}}$

$z = \frac{-16(1 - i\sqrt{3})}{(1)^2 + (\sqrt{3})^2}$

$z = \frac{-16(1 - i\sqrt{3})}{1 + 3} = \frac{-16(1 - i\sqrt{3})}{4}$

$z = -4(1 - i\sqrt{3})$

$z = -4 + 4i\sqrt{3}$

Now, we have $x = -4$ and $y = 4\sqrt{3}$.

Step 2: Calculate the modulus, $r$.

$r = |z| = \sqrt{x^2 + y^2} = \sqrt{(-4)^2 + (4\sqrt{3})^2}$

$r = \sqrt{16 + 16 \times 3} = \sqrt{16 + 48} = \sqrt{64}$

$r = 8$

Step 3: Calculate the argument, $\theta$.

The argument $\theta$ must satisfy:

$\cos \theta = \frac{x}{r} = \frac{-4}{8} = -\frac{1}{2}$

$\sin \theta = \frac{y}{r} = \frac{4\sqrt{3}}{8} = \frac{\sqrt{3}}{2}$

Since $\cos \theta$ is negative and $\sin \theta$ is positive, the angle $\theta$ lies in the second quadrant.

The reference angle $\alpha$ for which $\cos \alpha = \frac{1}{2}$ is $\alpha = \frac{\pi}{3}$.

For the second quadrant, the principal argument is $\theta = \pi - \alpha$.

$\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$

Step 4: Write the polar form.

Substitute the values of $r=8$ and $\theta = \frac{2\pi}{3}$ into the polar form expression:

$z = 8 \left( \cos\left(\frac{2\pi}{3}\right) + i \sin\left(\frac{2\pi}{3}\right) \right)$

---

The polar form of the complex number $\frac{-16}{1 + i\sqrt{3}}$ is $8 \left( \cos\frac{2\pi}{3} + i \sin\frac{2\pi}{3} \right)$.

Given:

The complex number is $z = -1 - i\sqrt{3}$.

To Find:

The modulus and the argument of the complex number $z$.

---

Solution:

The given complex number is in the standard form $z = x + iy$, where:

$x = -1$

$y = -\sqrt{3}$

Modulus:

The modulus of $z$, denoted as $r$ or $|z|$, is given by the formula $r = \sqrt{x^2 + y^2}$.

$r = \sqrt{(-1)^2 + (-\sqrt{3})^2}$

$r = \sqrt{1 + 3} = \sqrt{4}$

$r = 2$

So, the modulus is 2.

Argument:

Let $\theta$ be the argument of $z$. The argument must satisfy:

$\cos \theta = \frac{x}{r} = \frac{-1}{2}$

$\sin \theta = \frac{y}{r} = \frac{-\sqrt{3}}{2}$

Since both $\cos \theta$ and $\sin \theta$ are negative, the angle $\theta$ lies in the third quadrant.

Let $\alpha$ be the reference angle, such that $\tan \alpha = \left|\frac{y}{x}\right| = \left|\frac{-\sqrt{3}}{-1}\right| = \sqrt{3}$.

This gives $\alpha = \frac{\pi}{3}$.

For an angle in the third quadrant, the principal argument (in the range $(-\pi, \pi]$) is given by $\theta = -(\pi - \alpha)$.

$\theta = -\left(\pi - \frac{\pi}{3}\right) = -\frac{2\pi}{3}$

So, the argument is $-\frac{2\pi}{3}$.

---

Conclusion:

For the complex number $z = -1 - i\sqrt{3}$, the modulus is 2 and the argument is $-\frac{2\pi}{3}$.

Given:

The complex number is $z = -\sqrt{3} + i$.

To Find:

The modulus and the argument of the complex number $z$.

---

Solution:

The given complex number is in the standard form $z = x + iy$, where:

$x = -\sqrt{3}$

$y = 1$

Modulus:

The modulus is $r = \sqrt{x^2 + y^2}$.

$r = \sqrt{(-\sqrt{3})^2 + (1)^2}$

$r = \sqrt{3 + 1} = \sqrt{4}$

$r = 2$

So, the modulus is 2.

Argument:

Let $\theta$ be the argument. It must satisfy:

$\cos \theta = \frac{x}{r} = \frac{-\sqrt{3}}{2}$

$\sin \theta = \frac{y}{r} = \frac{1}{2}$

Since $\cos \theta$ is negative and $\sin \theta$ is positive, the angle $\theta$ lies in the second quadrant.

The reference angle $\alpha$ is given by $\tan \alpha = \left|\frac{y}{x}\right| = \left|\frac{1}{-\sqrt{3}}\right| = \frac{1}{\sqrt{3}}$.

This gives $\alpha = \frac{\pi}{6}$.

For an angle in the second quadrant, the principal argument is $\theta = \pi - \alpha$.

$\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$

So, the argument is $\frac{5\pi}{6}$.

---

Conclusion:

For the complex number $z = -\sqrt{3} + i$, the modulus is 2 and the argument is $\frac{5\pi}{6}$.

Given:

The complex number is $z = 1 - i$.

To Find:

The polar form of the complex number $z$.

---

Solution:

The complex number is $z = 1 - i$, so $x = 1$ and $y = -1$.

Step 1: Find the modulus $r$.

$r = \sqrt{x^2 + y^2} = \sqrt{(1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$.

Step 2: Find the argument $\theta$.

$\cos \theta = \frac{x}{r} = \frac{1}{\sqrt{2}}$

$\sin \theta = \frac{y}{r} = \frac{-1}{\sqrt{2}}$

Since $\cos \theta$ is positive and $\sin \theta$ is negative, $\theta$ is in the fourth quadrant.

The reference angle is $\alpha = \frac{\pi}{4}$. For the fourth quadrant, the principal argument is $\theta = -\alpha = -\frac{\pi}{4}$.

Step 3: Write the polar form.

The polar form is $z = r(\cos \theta + i \sin \theta)$.

$z = \sqrt{2} \left( \cos\left(-\frac{\pi}{4}\right) + i \sin\left(-\frac{\pi}{4}\right) \right)$

---

The polar form is $\sqrt{2} \left( \cos(-\frac{\pi}{4}) + i \sin(-\frac{\pi}{4}) \right)$.

Given:

The complex number is $z = -1 + i$.

To Find:

The polar form of the complex number $z$.

---

Solution:

The complex number is $z = -1 + i$, so $x = -1$ and $y = 1$.

Step 1: Find the modulus $r$.

$r = \sqrt{x^2 + y^2} = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2}$.

Step 2: Find the argument $\theta$.

$\cos \theta = \frac{x}{r} = \frac{-1}{\sqrt{2}}$

$\sin \theta = \frac{y}{r} = \frac{1}{\sqrt{2}}$

Since $\cos \theta$ is negative and $\sin \theta$ is positive, $\theta$ is in the second quadrant.

The reference angle is $\alpha = \frac{\pi}{4}$. For the second quadrant, the principal argument is $\theta = \pi - \alpha = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

Step 3: Write the polar form.

The polar form is $z = r(\cos \theta + i \sin \theta)$.

$z = \sqrt{2} \left( \cos\frac{3\pi}{4} + i \sin\frac{3\pi}{4} \right)$

---

The polar form is $\sqrt{2} \left( \cos\frac{3\pi}{4} + i \sin\frac{3\pi}{4} \right)$.

Given:

The complex number is $z = -1 - i$.

To Find:

The polar form of the complex number $z$.

---

Solution:

The complex number is $z = -1 - i$, so $x = -1$ and $y = -1$.

Step 1: Find the modulus $r$.

$r = \sqrt{x^2 + y^2} = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$.

Step 2: Find the argument $\theta$.

$\cos \theta = \frac{x}{r} = \frac{-1}{\sqrt{2}}$

$\sin \theta = \frac{y}{r} = \frac{-1}{\sqrt{2}}$

Since both $\cos \theta$ and $\sin \theta$ are negative, $\theta$ is in the third quadrant.

The reference angle is $\alpha = \frac{\pi}{4}$. For the third quadrant, the principal argument is $\theta = -(\pi - \alpha) = -(\pi - \frac{\pi}{4}) = -\frac{3\pi}{4}$.

Step 3: Write the polar form.

The polar form is $z = r(\cos \theta + i \sin \theta)$.

$z = \sqrt{2} \left( \cos\left(-\frac{3\pi}{4}\right) + i \sin\left(-\frac{3\pi}{4}\right) \right)$

---

The polar form is $\sqrt{2} \left( \cos(-\frac{3\pi}{4}) + i \sin(-\frac{3\pi}{4}) \right)$.

Given:

The complex number is $z = -3$.

To Find:

The polar form of the complex number $z$.

---

Solution:

The complex number is $z = -3 + 0i$, so $x = -3$ and $y = 0$.

Step 1: Find the modulus $r$.

$r = \sqrt{x^2 + y^2} = \sqrt{(-3)^2 + (0)^2} = \sqrt{9} = 3$.

Step 2: Find the argument $\theta$.

$\cos \theta = \frac{x}{r} = \frac{-3}{3} = -1$

$\sin \theta = \frac{y}{r} = \frac{0}{3} = 0$

The angle for which $\cos\theta = -1$ and $\sin\theta = 0$ is $\theta = \pi$.

Step 3: Write the polar form.

The polar form is $z = r(\cos \theta + i \sin \theta)$.

$z = 3(\cos\pi + i \sin\pi)$

---

The polar form is $3(\cos\pi + i \sin\pi)$.

Given:

The complex number is $z = \sqrt{3} + i$.

To Find:

The polar form of the complex number $z$.

---

Solution:

The complex number is $z = \sqrt{3} + i$, so $x = \sqrt{3}$ and $y = 1$.

Step 1: Find the modulus $r$.

$r = \sqrt{x^2 + y^2} = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2$.

Step 2: Find the argument $\theta$.

$\cos \theta = \frac{x}{r} = \frac{\sqrt{3}}{2}$

$\sin \theta = \frac{y}{r} = \frac{1}{2}$

Since both $\cos \theta$ and $\sin \theta$ are positive, $\theta$ is in the first quadrant.

The angle that satisfies these conditions is $\theta = \frac{\pi}{6}$.

Step 3: Write the polar form.

The polar form is $z = r(\cos \theta + i \sin \theta)$.

$z = 2 \left( \cos\frac{\pi}{6} + i \sin\frac{\pi}{6} \right)$

---

The polar form is $2 \left( \cos\frac{\pi}{6} + i \sin\frac{\pi}{6} \right)$.

Given:

The complex number is $z = i$.

To Find:

The polar form of the complex number $z$.

---

Solution:

The complex number is $z = 0 + 1i$, so $x = 0$ and $y = 1$.

Step 1: Find the modulus $r$.

$r = \sqrt{x^2 + y^2} = \sqrt{(0)^2 + (1)^2} = \sqrt{1} = 1$.

Step 2: Find the argument $\theta$.

$\cos \theta = \frac{x}{r} = \frac{0}{1} = 0$

$\sin \theta = \frac{y}{r} = \frac{1}{1} = 1$

The angle for which $\cos\theta = 0$ and $\sin\theta = 1$ is $\theta = \frac{\pi}{2}$.

Step 3: Write the polar form.

The polar form is $z = r(\cos \theta + i \sin \theta)$.

$z = 1 \left( \cos\frac{\pi}{2} + i \sin\frac{\pi}{2} \right)$

---

The polar form is $\cos\frac{\pi}{2} + i \sin\frac{\pi}{2}$.

Given:

The quadratic equation is $x^2 + 2 = 0$.

---

To Find:

The solution(s) for $x$.

---

Solution:

We are given the equation:

$x^2 + 2 = 0$

To solve for $x$, we first isolate the $x^2$ term. Subtract 2 from both sides of the equation:

$x^2 = -2$

Now, take the square root of both sides to find the values of $x$:

$x = \pm \sqrt{-2}$

Since the square root of a negative number involves the imaginary unit $i$, where $i = \sqrt{-1}$, we can rewrite $\sqrt{-2}$ as:

$x = \pm \sqrt{(-1) \times 2}$

$x = \pm \sqrt{-1} \times \sqrt{2}$

$x = \pm i \sqrt{2}$

Thus, the solutions are $x = i\sqrt{2}$ and $x = -i\sqrt{2}$.

---

Conclusion:

The solutions to the quadratic equation $x^2 + 2 = 0$ are $x = i\sqrt{2}$ and $x = -i\sqrt{2}$.

Given:

The quadratic equation is $x^2 + x + 1 = 0$.

---

To Find:

The solution(s) for $x$.

---

Solution:

The given equation is a quadratic equation of the form $ax^2 + bx + c = 0$.

Comparing $x^2 + x + 1 = 0$ with the standard form, we have:

$a = 1$

$b = 1$

$c = 1$

We can use the quadratic formula to find the solutions for $x$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

First, calculate the discriminant, $D = b^2 - 4ac$:

$D = (1)^2 - 4(1)(1)$

$D = 1 - 4$

$D = -3$

Since the discriminant is negative ($D < 0$), the solutions will be complex.

Now substitute the values of $a$, $b$, and the discriminant $D$ into the quadratic formula:

$x = \frac{-(1) \pm \sqrt{-3}}{2(1)}$

$x = \frac{-1 \pm \sqrt{-3}}{2}$

We know that $\sqrt{-3} = \sqrt{-1 \times 3} = \sqrt{-1} \times \sqrt{3} = i\sqrt{3}$.

Substitute this back into the expression for $x$:

$x = \frac{-1 \pm i\sqrt{3}}{2}$

The two solutions are:

$x_1 = \frac{-1 + i\sqrt{3}}{2}$

$x_2 = \frac{-1 - i\sqrt{3}}{2}$

---

Conclusion:

The solutions to the quadratic equation $x^2 + x + 1 = 0$ are $x = \frac{-1 + i\sqrt{3}}{2}$ and $x = \frac{-1 - i\sqrt{3}}{2}$.

Given:

The quadratic equation is $\sqrt{5} x^2 + x + \sqrt{5} = 0$.

---

To Find:

The solution(s) for $x$.

---

Solution:

The given equation is a quadratic equation of the form $ax^2 + bx + c = 0$.

Comparing $\sqrt{5} x^2 + x + \sqrt{5} = 0$ with the standard form, we identify the coefficients:

$a = \sqrt{5}$

$b = 1$

$c = \sqrt{5}$

We use the quadratic formula to find the solutions for $x$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

First, calculate the discriminant, $D = b^2 - 4ac$:

$D = (1)^2 - 4(\sqrt{5})(\sqrt{5})$

$D = 1 - 4(5)$

$D = 1 - 20$

$D = -19$

Since the discriminant is negative ($D < 0$), the solutions will be complex.

Now substitute the values of $a$, $b$, and the discriminant $D$ into the quadratic formula:

$x = \frac{-(1) \pm \sqrt{-19}}{2(\sqrt{5})}$

$x = \frac{-1 \pm \sqrt{-19}}{2\sqrt{5}}$

We know that $\sqrt{-19} = \sqrt{-1 \times 19} = \sqrt{-1} \times \sqrt{19} = i\sqrt{19}$.

Substitute this back into the expression for $x$:

$x = \frac{-1 \pm i\sqrt{19}}{2\sqrt{5}}$

The two solutions are:

$x_1 = \frac{-1 + i\sqrt{19}}{2\sqrt{5}}$

$x_2 = \frac{-1 - i\sqrt{19}}{2\sqrt{5}}$

---

Conclusion:

The solutions to the quadratic equation $\sqrt{5} x^2 + x + \sqrt{5} = 0$ are $x = \frac{-1 + i\sqrt{19}}{2\sqrt{5}}$ and $x = \frac{-1 - i\sqrt{19}}{2\sqrt{5}}$.

Given:

The quadratic equation is $x^2 + 3 = 0$.

---

To Find:

The solution(s) for $x$.

---

Solution:

We are given the equation:

$x^2 + 3 = 0$

To solve for $x$, we first isolate the $x^2$ term. Subtract 3 from both sides of the equation:

$x^2 = -3$

Now, take the square root of both sides to find the values of $x$:

$x = \pm \sqrt{-3}$

Since the square root of a negative number involves the imaginary unit $i$, where $i = \sqrt{-1}$, we can rewrite $\sqrt{-3}$ as:

$x = \pm \sqrt{(-1) \times 3}$

$x = \pm \sqrt{-1} \times \sqrt{3}$

$x = \pm i \sqrt{3}$

Thus, the solutions are $x = i\sqrt{3}$ and $x = -i\sqrt{3}$.

---

Conclusion:

The solutions to the quadratic equation $x^2 + 3 = 0$ are $x = i\sqrt{3}$ and $x = -i\sqrt{3}$.

Given:

The quadratic equation is $2x^2 + x + 1 = 0$.

---

To Find:

The solution(s) for $x$.

---

Solution:

The given equation is a quadratic equation of the form $ax^2 + bx + c = 0$.

Comparing $2x^2 + x + 1 = 0$ with the standard form, we identify the coefficients:

$a = 2$

$b = 1$

$c = 1$

We use the quadratic formula to find the solutions for $x$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

First, calculate the discriminant, $D = b^2 - 4ac$:

$D = (1)^2 - 4(2)(1)$

$D = 1 - 8$

$D = -7$

Since the discriminant is negative ($D < 0$), the solutions will be complex.

Now substitute the values of $a$, $b$, and the discriminant $D$ into the quadratic formula:

$x = \frac{-(1) \pm \sqrt{-7}}{2(2)}$

$x = \frac{-1 \pm \sqrt{-7}}{4}$

We know that $\sqrt{-7} = \sqrt{-1 \times 7} = \sqrt{-1} \times \sqrt{7} = i\sqrt{7}$.

Substitute this back into the expression for $x$:

$x = \frac{-1 \pm i\sqrt{7}}{4}$

The two solutions are:

$x_1 = \frac{-1 + i\sqrt{7}}{4}$

$x_2 = \frac{-1 - i\sqrt{7}}{4}$

---

Conclusion:

The solutions to the quadratic equation $2x^2 + x + 1 = 0$ are $x = \frac{-1 + i\sqrt{7}}{4}$ and $x = \frac{-1 - i\sqrt{7}}{4}$.

Given:

The quadratic equation is $x^2 + 3x + 9 = 0$.

---

To Find:

The solution(s) for $x$.

---

Solution:

The given equation is a quadratic equation of the form $ax^2 + bx + c = 0$.

Comparing $x^2 + 3x + 9 = 0$ with the standard form, we identify the coefficients:

$a = 1$

$b = 3$

$c = 9$

We use the quadratic formula to find the solutions for $x$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

First, calculate the discriminant, $D = b^2 - 4ac$:

$D = (3)^2 - 4(1)(9)$

$D = 9 - 36$

$D = -27$

Since the discriminant is negative ($D < 0$), the solutions will be complex.

Now substitute the values of $a$, $b$, and the discriminant $D$ into the quadratic formula:

$x = \frac{-(3) \pm \sqrt{-27}}{2(1)}$

$x = \frac{-3 \pm \sqrt{-27}}{2}$

We simplify $\sqrt{-27}$:

$\sqrt{-27} = \sqrt{-1 \times 27} = \sqrt{-1} \times \sqrt{27} = i \sqrt{9 \times 3} = i (3\sqrt{3}) = 3i\sqrt{3}$

Substitute this back into the expression for $x$:

$x = \frac{-3 \pm 3i\sqrt{3}}{2}$

The two solutions are:

$x_1 = \frac{-3 + 3i\sqrt{3}}{2}$

$x_2 = \frac{-3 - 3i\sqrt{3}}{2}$

---

Conclusion:

The solutions to the quadratic equation $x^2 + 3x + 9 = 0$ are $x = \frac{-3 + 3i\sqrt{3}}{2}$ and $x = \frac{-3 - 3i\sqrt{3}}{2}$.

Given:

The quadratic equation is $-x^2 + x - 2 = 0$.

---

To Find:

The solution(s) for $x$.

---

Solution:

The given equation is a quadratic equation of the form $ax^2 + bx + c = 0$.

Comparing $-x^2 + x - 2 = 0$ with the standard form, we identify the coefficients:

$a = -1$

$b = 1$

$c = -2$

We use the quadratic formula to find the solutions for $x$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

First, calculate the discriminant, $D = b^2 - 4ac$:

$D = (1)^2 - 4(-1)(-2)$

$D = 1 - 8$

$D = -7$

Since the discriminant is negative ($D < 0$), the solutions will be complex.

Now substitute the values of $a$, $b$, and the discriminant $D$ into the quadratic formula:

$x = \frac{-(1) \pm \sqrt{-7}}{2(-1)}$

$x = \frac{-1 \pm \sqrt{-7}}{-2}$

We know that $\sqrt{-7} = \sqrt{-1 \times 7} = \sqrt{-1} \times \sqrt{7} = i\sqrt{7}$.

Substitute this back into the expression for $x$:

$x = \frac{-1 \pm i\sqrt{7}}{-2}$

To simplify, we can multiply the numerator and denominator by $-1$:

$x = \frac{(-1)(-1 \pm i\sqrt{7})}{(-1)(-2)}$

$x = \frac{1 \mp i\sqrt{7}}{2}$

The two solutions are:

$x_1 = \frac{1 - i\sqrt{7}}{2}$

$x_2 = \frac{1 + i\sqrt{7}}{2}$

---

Conclusion:

The solutions to the quadratic equation $-x^2 + x - 2 = 0$ are $x = \frac{1 - i\sqrt{7}}{2}$ and $x = \frac{1 + i\sqrt{7}}{2}$.

Given:

The quadratic equation is $x^2 + 3x + 5 = 0$.

---

To Find:

The solution(s) for $x$.

---

Solution:

The given equation is a quadratic equation of the form $ax^2 + bx + c = 0$.

Comparing $x^2 + 3x + 5 = 0$ with the standard form, we identify the coefficients:

$a = 1$

$b = 3$

$c = 5$

We use the quadratic formula to find the solutions for $x$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

First, calculate the discriminant, $D = b^2 - 4ac$:

$D = (3)^2 - 4(1)(5)$

$D = 9 - 20$

$D = -11$

Since the discriminant is negative ($D < 0$), the solutions will be complex.

Now substitute the values of $a$, $b$, and the discriminant $D$ into the quadratic formula:

$x = \frac{-(3) \pm \sqrt{-11}}{2(1)}$

$x = \frac{-3 \pm \sqrt{-11}}{2}$

We know that $\sqrt{-11} = \sqrt{-1 \times 11} = \sqrt{-1} \times \sqrt{11} = i\sqrt{11}$.

Substitute this back into the expression for $x$:

$x = \frac{-3 \pm i\sqrt{11}}{2}$

The two solutions are:

$x_1 = \frac{-3 + i\sqrt{11}}{2}$

$x_2 = \frac{-3 - i\sqrt{11}}{2}$

---

Conclusion:

The solutions to the quadratic equation $x^2 + 3x + 5 = 0$ are $x = \frac{-3 + i\sqrt{11}}{2}$ and $x = \frac{-3 - i\sqrt{11}}{2}$.

Given:

The quadratic equation is $x^2 - x + 2 = 0$.

---

To Find:

The solution(s) for $x$.

---

Solution:

The given equation is a quadratic equation of the form $ax^2 + bx + c = 0$.

Comparing $x^2 - x + 2 = 0$ with the standard form, we identify the coefficients:

$a = 1$

$b = -1$

$c = 2$

We use the quadratic formula to find the solutions for $x$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

First, calculate the discriminant, $D = b^2 - 4ac$:

$D = (-1)^2 - 4(1)(2)$

$D = 1 - 8$

$D = -7$

Since the discriminant is negative ($D < 0$), the solutions will be complex.

Now substitute the values of $a$, $b$, and the discriminant $D$ into the quadratic formula:

$x = \frac{-(-1) \pm \sqrt{-7}}{2(1)}$

$x = \frac{1 \pm \sqrt{-7}}{2}$

We know that $\sqrt{-7} = \sqrt{-1 \times 7} = \sqrt{-1} \times \sqrt{7} = i\sqrt{7}$.

Substitute this back into the expression for $x$:

$x = \frac{1 \pm i\sqrt{7}}{2}$

The two solutions are:

$x_1 = \frac{1 + i\sqrt{7}}{2}$

$x_2 = \frac{1 - i\sqrt{7}}{2}$

---

Conclusion:

The solutions to the quadratic equation $x^2 - x + 2 = 0$ are $x = \frac{1 + i\sqrt{7}}{2}$ and $x = \frac{1 - i\sqrt{7}}{2}$.

Given:

The quadratic equation is $\sqrt{2} x^2 + x + \sqrt{2} = 0$.

---

To Find:

The solution(s) for $x$.

---

Solution:

The given equation is a quadratic equation of the form $ax^2 + bx + c = 0$.

Comparing $\sqrt{2} x^2 + x + \sqrt{2} = 0$ with the standard form, we identify the coefficients:

$a = \sqrt{2}$

$b = 1$

$c = \sqrt{2}$

We use the quadratic formula to find the solutions for $x$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

First, calculate the discriminant, $D = b^2 - 4ac$:

$D = (1)^2 - 4(\sqrt{2})(\sqrt{2})$

$D = 1 - 4(2)$

$D = 1 - 8$

$D = -7$

Since the discriminant is negative ($D < 0$), the solutions will be complex.

Now substitute the values of $a$, $b$, and the discriminant $D$ into the quadratic formula:

$x = \frac{-(1) \pm \sqrt{-7}}{2(\sqrt{2})}$

$x = \frac{-1 \pm \sqrt{-7}}{2\sqrt{2}}$

We know that $\sqrt{-7} = \sqrt{-1 \times 7} = \sqrt{-1} \times \sqrt{7} = i\sqrt{7}$.

Substitute this back into the expression for $x$:

$x = \frac{-1 \pm i\sqrt{7}}{2\sqrt{2}}$

The two solutions are:

$x_1 = \frac{-1 + i\sqrt{7}}{2\sqrt{2}}$

$x_2 = \frac{-1 - i\sqrt{7}}{2\sqrt{2}}$

---

Conclusion:

The solutions to the quadratic equation $\sqrt{2} x^2 + x + \sqrt{2} = 0$ are $x = \frac{-1 + i\sqrt{7}}{2\sqrt{2}}$ and $x = \frac{-1 - i\sqrt{7}}{2\sqrt{2}}$.

Given:

The quadratic equation is $\sqrt{3} x^2 - \sqrt{2} x + 3\sqrt{3} = 0$.

---

To Find:

The solution(s) for $x$.

---

Solution:

The given equation is a quadratic equation of the form $ax^2 + bx + c = 0$.

Comparing $\sqrt{3} x^2 - \sqrt{2} x + 3\sqrt{3} = 0$ with the standard form, we identify the coefficients:

$a = \sqrt{3}$

$b = -\sqrt{2}$

$c = 3\sqrt{3}$

We use the quadratic formula to find the solutions for $x$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

First, calculate the discriminant, $D = b^2 - 4ac$:

$D = (-\sqrt{2})^2 - 4(\sqrt{3})(3\sqrt{3})$

$D = 2 - 4(3 \times (\sqrt{3})^2)$

$D = 2 - 4(3 \times 3)$

$D = 2 - 4(9)$

$D = 2 - 36$

$D = -34$

Since the discriminant is negative ($D < 0$), the solutions will be complex.

Now substitute the values of $a$, $b$, and the discriminant $D$ into the quadratic formula:

$x = \frac{-(-\sqrt{2}) \pm \sqrt{-34}}{2(\sqrt{3})}$

$x = \frac{\sqrt{2} \pm \sqrt{-34}}{2\sqrt{3}}$

We know that $\sqrt{-34} = \sqrt{-1 \times 34} = \sqrt{-1} \times \sqrt{34} = i\sqrt{34}$.

Substitute this back into the expression for $x$:

$x = \frac{\sqrt{2} \pm i\sqrt{34}}{2\sqrt{3}}$

The two solutions are:

$x_1 = \frac{\sqrt{2} + i\sqrt{34}}{2\sqrt{3}}$

$x_2 = \frac{\sqrt{2} - i\sqrt{34}}{2\sqrt{3}}$

---

Conclusion:

The solutions to the quadratic equation $\sqrt{3} x^2 - \sqrt{2} x + 3\sqrt{3} = 0$ are $x = \frac{\sqrt{2} + i\sqrt{34}}{2\sqrt{3}}$ and $x = \frac{\sqrt{2} - i\sqrt{34}}{2\sqrt{3}}$.

Given:

The quadratic equation is $x^2 + x + \frac{1}{\sqrt{2}} = 0$.

---

To Find:

The solution(s) for $x$.

---

Solution:

The given equation is $x^2 + x + \frac{1}{\sqrt{2}} = 0$.

To simplify, we can multiply the entire equation by $\sqrt{2}$ to eliminate the fraction:

$\sqrt{2}(x^2 + x + \frac{1}{\sqrt{2}}) = \sqrt{2} \times 0$

$\sqrt{2}x^2 + \sqrt{2}x + 1 = 0$

This is a quadratic equation of the form $ax^2 + bx + c = 0$.

Comparing $\sqrt{2}x^2 + \sqrt{2}x + 1 = 0$ with the standard form, we identify the coefficients:

$a = \sqrt{2}$

$b = \sqrt{2}$

$c = 1$

We use the quadratic formula to find the solutions for $x$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

First, calculate the discriminant, $D = b^2 - 4ac$:

$D = (\sqrt{2})^2 - 4(\sqrt{2})(1)$

$D = 2 - 4\sqrt{2}$

Since $4\sqrt{2} > 2$, the discriminant is negative ($D < 0$), so the solutions will be complex.

Now substitute the values of $a$, $b$, and the discriminant $D$ into the quadratic formula:

$x = \frac{-(\sqrt{2}) \pm \sqrt{2 - 4\sqrt{2}}}{2(\sqrt{2})}$

$x = \frac{-\sqrt{2} \pm \sqrt{-(4\sqrt{2} - 2)}}{2\sqrt{2}}$

We know that $\sqrt{-(4\sqrt{2} - 2)} = \sqrt{-1} \times \sqrt{4\sqrt{2} - 2} = i\sqrt{4\sqrt{2} - 2}$.

Substitute this back into the expression for $x$:

$x = \frac{-\sqrt{2} \pm i\sqrt{4\sqrt{2} - 2}}{2\sqrt{2}}$

We can separate the real and imaginary parts:

$x = \frac{-\sqrt{2}}{2\sqrt{2}} \pm i \frac{\sqrt{4\sqrt{2} - 2}}{2\sqrt{2}}$

$x = -\frac{1}{2} \pm i \sqrt{\frac{4\sqrt{2} - 2}{(2\sqrt{2})^2}}$

$x = -\frac{1}{2} \pm i \sqrt{\frac{4\sqrt{2} - 2}{8}}$

$x = -\frac{1}{2} \pm i \sqrt{\frac{2(2\sqrt{2} - 1)}{8}}$

$x = -\frac{1}{2} \pm i \sqrt{\frac{2\sqrt{2} - 1}{4}}$

$x = -\frac{1}{2} \pm i \frac{\sqrt{2\sqrt{2} - 1}}{2}$

$x = \frac{-1 \pm i\sqrt{2\sqrt{2} - 1}}{2}$

The two solutions are:

$x_1 = \frac{-1 + i\sqrt{2\sqrt{2} - 1}}{2}$

$x_2 = \frac{-1 - i\sqrt{2\sqrt{2} - 1}}{2}$

---

Conclusion:

The solutions to the quadratic equation $x^2 + x + \frac{1}{\sqrt{2}} = 0$ are $x = \frac{-1 + i\sqrt{2\sqrt{2} - 1}}{2}$ and $x = \frac{-1 - i\sqrt{2\sqrt{2} - 1}}{2}$.

Given:

The quadratic equation is $x^2 + \frac{x}{\sqrt{2}} + 1 = 0$.

---

To Find:

The solution(s) for $x$.

---

Solution:

The given equation is $x^2 + \frac{x}{\sqrt{2}} + 1 = 0$.

To simplify, we can multiply the entire equation by $\sqrt{2}$ to eliminate the fraction in the coefficient of $x$:

$\sqrt{2} \left( x^2 + \frac{x}{\sqrt{2}} + 1 \right) = \sqrt{2} \times 0$

$\sqrt{2}x^2 + \sqrt{2} \cdot \frac{x}{\sqrt{2}} + \sqrt{2} \cdot 1 = 0$

$\sqrt{2}x^2 + x + \sqrt{2} = 0$

This is now a quadratic equation in the standard form $ax^2 + bx + c = 0$.

Comparing $\sqrt{2}x^2 + x + \sqrt{2} = 0$ with the standard form, we identify the coefficients:

$a = \sqrt{2}$

$b = 1$

$c = \sqrt{2}$

We use the quadratic formula to find the solutions for $x$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

First, calculate the discriminant, $D = b^2 - 4ac$:

$D = (1)^2 - 4(\sqrt{2})(\sqrt{2})$

$D = 1 - 4(2)$

$D = 1 - 8$

$D = -7$

Since the discriminant is negative ($D < 0$), the solutions will be complex.

Now substitute the values of $a$, $b$, and the discriminant $D$ into the quadratic formula:

$x = \frac{-(1) \pm \sqrt{-7}}{2(\sqrt{2})}$

$x = \frac{-1 \pm \sqrt{-7}}{2\sqrt{2}}$

We know that $\sqrt{-7} = \sqrt{-1 \times 7} = \sqrt{-1} \times \sqrt{7} = i\sqrt{7}$.

Substitute this back into the expression for $x$:

$x = \frac{-1 \pm i\sqrt{7}}{2\sqrt{2}}$

The two solutions are:

$x_1 = \frac{-1 + i\sqrt{7}}{2\sqrt{2}}$

$x_2 = \frac{-1 - i\sqrt{7}}{2\sqrt{2}}$

---

Conclusion:

The solutions to the quadratic equation $x^2 + \frac{x}{\sqrt{2}} + 1 = 0$ are $x = \frac{-1 + i\sqrt{7}}{2\sqrt{2}}$ and $x = \frac{-1 - i\sqrt{7}}{2\sqrt{2}}$.

Given:

The complex number is $Z = \frac{(3 - 2i)(2 + 3i)}{(1 + 2i)(2 - i)}$.

---

To Find:

The conjugate of the complex number $Z$, denoted as $\bar{Z}$.

---

Solution:

First, let's simplify the numerator and the denominator of the expression.

Numerator: $(3 - 2i)(2 + 3i)$

$(3 - 2i)(2 + 3i) = 3(2) + 3(3i) - 2i(2) - 2i(3i)$

$ = 6 + 9i - 4i - 6i^2$

$ = 6 + (9 - 4)i - 6(-1)$ (Since $i^2 = -1$)

$ = 6 + 5i + 6$

$ = 12 + 5i$

Denominator: $(1 + 2i)(2 - i)$

$(1 + 2i)(2 - i) = 1(2) + 1(-i) + 2i(2) + 2i(-i)$

$ = 2 - i + 4i - 2i^2$

$ = 2 + (-1 + 4)i - 2(-1)$ (Since $i^2 = -1$)

$ = 2 + 3i + 2$

$ = 4 + 3i$

Now, substitute the simplified numerator and denominator back into the expression for $Z$:

$Z = \frac{12 + 5i}{4 + 3i}$

To express $Z$ in the standard form $a + bi$, multiply the numerator and denominator by the conjugate of the denominator, which is $4 - 3i$.

$Z = \frac{12 + 5i}{4 + 3i} \times \frac{4 - 3i}{4 - 3i}$

$Z = \frac{(12 + 5i)(4 - 3i)}{(4 + 3i)(4 - 3i)}$

Simplify the new numerator:

$(12 + 5i)(4 - 3i) = 12(4) + 12(-3i) + 5i(4) + 5i(-3i)$

$ = 48 - 36i + 20i - 15i^2$

$ = 48 + (-36 + 20)i - 15(-1)$

$ = 48 - 16i + 15$

$ = 63 - 16i$

Simplify the new denominator using $(a + ib)(a - ib) = a^2 + b^2$:

$(4 + 3i)(4 - 3i) = (4)^2 + (3)^2 = 16 + 9 = 25$

Substitute these back into the expression for $Z$:

$Z = \frac{63 - 16i}{25}$

Write $Z$ in the standard form $a + bi$:

$Z = \frac{63}{25} - \frac{16}{25}i$

Now, find the conjugate of $Z$. The conjugate $\bar{Z}$ of a complex number $Z = a + bi$ is $\bar{Z} = a - bi$.

$\bar{Z} = \frac{63}{25} - \left( -\frac{16}{25}i \right)$

$\bar{Z} = \frac{63}{25} + \frac{16}{25}i$

---

Conclusion:

The conjugate of the complex number $\frac{(3 - 2i)(2 + 3i)}{(1 + 2i)(2 - i)}$ is $\frac{63}{25} + \frac{16}{25}i$.

(i)

Given:

The complex number is $z_1 = \frac{1 + i}{1 - i}$.

---

To Find:

The modulus and argument of $z_1$.

---

Solution:

First, we simplify the complex number $z_1$ into the standard form $x + iy$.

Multiply the numerator and denominator by the conjugate of the denominator $(1 + i)$:

$z_1 = \frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i}$

$z_1 = \frac{(1 + i)^2}{(1 - i)(1 + i)}$

Numerator: $(1 + i)^2 = 1^2 + 2(1)(i) + i^2 = 1 + 2i - 1 = 2i$.

Denominator: $(1 - i)(1 + i) = 1^2 + 1^2 = 1 + 1 = 2$.

So, $z_1 = \frac{2i}{2} = i$.

In the standard form $z_1 = x + iy$, we have $z_1 = 0 + 1i$.

Thus, $x = 0$ and $y = 1$.

Modulus:

The modulus $r_1 = |z_1|$ is:

$r_1 = \sqrt{x^2 + y^2} = \sqrt{0^2 + 1^2} = \sqrt{1} = 1$.

Argument:

Let $\theta_1$ be the argument. It satisfies:

The angle $\theta_1$ (principal value in $(-\pi, \pi]$) for which $\cos \theta_1 = 0$ and $\sin \theta_1 = 1$ is $\theta_1 = \frac{\pi}{2}$.

---

Conclusion (i):

For $z_1 = \frac{1 + i}{1 - i}$:

• Modulus is $1$.
• Argument is $\frac{\pi}{2}$.

---

(ii)

Given:

The complex number is $z_2 = \frac{1}{1 + i}$.

---

To Find:

The modulus and argument of $z_2$.

---

Solution:

First, we simplify the complex number $z_2$ into the standard form $x + iy$.

Multiply the numerator and denominator by the conjugate of the denominator $(1 - i)$:

$z_2 = \frac{1}{1 + i} \times \frac{1 - i}{1 - i}$

$z_2 = \frac{1 - i}{(1 + i)(1 - i)}$

Numerator: $1 - i$.

Denominator: $(1 + i)(1 - i) = 1^2 + 1^2 = 1 + 1 = 2$.

So, $z_2 = \frac{1 - i}{2} = \frac{1}{2} - \frac{1}{2}i$.

In the standard form $z_2 = x + iy$, we have $x = \frac{1}{2}$ and $y = -\frac{1}{2}$.

Modulus:

The modulus $r_2 = |z_2|$ is:

$r_2 = \sqrt{x^2 + y^2} = \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2}$

$r_2 = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{2}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$ or $\frac{\sqrt{2}}{2}$.

Argument:

Let $\theta_2$ be the argument. It satisfies:

Since $\cos \theta_2 > 0$ and $\sin \theta_2 < 0$, the angle $\theta_2$ lies in the fourth quadrant.

Let $\alpha$ be the reference angle. $\tan \alpha = \left|\frac{y}{x}\right| = \left|\frac{-1/2}{1/2}\right| = |-1| = 1$. So, $\alpha = \frac{\pi}{4}$.

For the fourth quadrant, the principal argument is $\theta_2 = -\alpha$.

$\theta_2 = -\frac{\pi}{4}$.

---

Conclusion (ii):

For $z_2 = \frac{1}{1 + i}$:

• Modulus is $\frac{1}{\sqrt{2}}$ (or $\frac{\sqrt{2}}{2}$).
• Argument is $-\frac{\pi}{4}$.

Given:

The relation between complex numbers:

where $x, y, a, b$ are real numbers, and it is assumed that $a - ib \neq 0$ (i.e., $a$ and $b$ are not both zero).

---

To Prove:

$x^2 + y^2 = 1$

---

Proof:

We are given the equation:

$x + iy = \frac{a + ib}{a - ib}$

Taking the modulus on both sides of equation (i), we get:

$|x + iy| = \left| \frac{a + ib}{a - ib} \right|$

Using the property of modulus $\left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}$, we have:

$|x + iy| = \frac{|a + ib|}{|a - ib|}$

Recall that the modulus of a complex number $p + iq$ is $|p + iq| = \sqrt{p^2 + q^2}$.

Therefore:

$|x + iy| = \sqrt{x^2 + y^2}$

$|a + ib| = \sqrt{a^2 + b^2}$

$|a - ib| = \sqrt{a^2 + (-b)^2} = \sqrt{a^2 + b^2}$

Substituting these values into the modulus equation:

$\sqrt{x^2 + y^2} = \frac{\sqrt{a^2 + b^2}}{\sqrt{a^2 + b^2}}$

Since $a$ and $b$ are not both zero, $\sqrt{a^2 + b^2} \neq 0$. Thus, the fraction on the right side simplifies to 1:

$\sqrt{x^2 + y^2} = 1$

Squaring both sides of the equation:

$(\sqrt{x^2 + y^2})^2 = 1^2$

$x^2 + y^2 = 1$

Hence, it is proved that $x^2 + y^2 = 1$.

Given:

The complex number $z = \frac{3 + 2i \sin \theta}{1 - 2i \sin \theta}$.

The condition that $z$ is purely real.

---

To Find:

The real values of $\theta$ that satisfy the condition.

---

Solution:

Let the given complex number be $z$.

$z = \frac{3 + 2i \sin \theta}{1 - 2i \sin \theta}$

To determine when $z$ is purely real, we first express it in the standard form $a + bi$. We multiply the numerator and the denominator by the conjugate of the denominator, which is $1 + 2i \sin \theta$.

$z = \frac{(3 + 2i \sin \theta)(1 + 2i \sin \theta)}{(1 - 2i \sin \theta)(1 + 2i \sin \theta)}$

Simplify the numerator:

Numerator $= 3(1) + 3(2i \sin \theta) + (2i \sin \theta)(1) + (2i \sin \theta)(2i \sin \theta)$

$ = 3 + 6i \sin \theta + 2i \sin \theta + 4i^2 \sin^2 \theta$

$ = 3 + (6 + 2)i \sin \theta + 4(-1) \sin^2 \theta$ (Since $i^2 = -1$)

$ = 3 + 8i \sin \theta - 4 \sin^2 \theta$

$ = (3 - 4 \sin^2 \theta) + i(8 \sin \theta)$

Simplify the denominator using the property $(a - ib)(a + ib) = a^2 + b^2$:

Denominator $= (1)^2 + (2 \sin \theta)^2 = 1 + 4 \sin^2 \theta$

Now substitute the simplified numerator and denominator back into the expression for $z$:

$z = \frac{(3 - 4 \sin^2 \theta) + i(8 \sin \theta)}{1 + 4 \sin^2 \theta}$

Separate the real and imaginary parts:

$z = \frac{3 - 4 \sin^2 \theta}{1 + 4 \sin^2 \theta} + i \left( \frac{8 \sin \theta}{1 + 4 \sin^2 \theta} \right)$

A complex number is purely real if its imaginary part is zero.

Set the imaginary part to zero:

$\frac{8 \sin \theta}{1 + 4 \sin^2 \theta} = 0$

The denominator $1 + 4 \sin^2 \theta$ is always greater than or equal to 1 (since $\sin^2 \theta \ge 0$), so it is never zero. Thus, the fraction is zero if and only if the numerator is zero:

$8 \sin \theta = 0$

$\sin \theta = 0$

The general solution for $\sin \theta = 0$ is given by:

$\theta = n\pi$, where $n$ is any integer ($n \in \mathbb{Z}$).

---

Conclusion:

The complex number $\frac{3 + 2i \sin \theta}{1 - 2i \sin \theta}$ is purely real when $\theta = n\pi$, where $n$ is any integer.

Given:

The complex number $z = \frac{i - 1}{\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}}$.

---

To Find:

Convert the complex number $z$ into its polar form $r(\cos \theta + i \sin \theta)$.

---

Solution:

Let the numerator be $z_1 = i - 1 = -1 + i$.

Let the denominator be $z_2 = \cos \frac{\pi}{3} + i\sin \frac{\pi}{3}$.

So, $z = \frac{z_1}{z_2}$.

We will first convert $z_1$ and $z_2$ into their polar forms.

Polar form of $z_1 = -1 + i$:

Comparing with $x + iy$, we have $x = -1$ and $y = 1$.

Modulus $r_1 = |z_1| = \sqrt{x^2 + y^2} = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2}$.

Argument $\theta_1$: Since $x < 0$ and $y > 0$, $z_1$ lies in the second quadrant.

The reference angle $\alpha = \tan^{-1}\left|\frac{y}{x}\right| = \tan^{-1}\left|\frac{1}{-1}\right| = \tan^{-1}(1) = \frac{\pi}{4}$.

For the second quadrant, the argument $\theta_1 = \pi - \alpha = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

Thus, the polar form of $z_1$ is $z_1 = \sqrt{2} \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right)$.

Polar form of $z_2 = \cos \frac{\pi}{3} + i\sin \frac{\pi}{3}$:

The denominator $z_2$ is already in polar form.

Modulus $r_2 = |z_2| = 1$.

Argument $\theta_2 = \frac{\pi}{3}$.

Polar form of $z = \frac{z_1}{z_2}$:

Using the division property for complex numbers in polar form:

$z = \frac{r_1(\cos \theta_1 + i \sin \theta_1)}{r_2(\cos \theta_2 + i \sin \theta_2)} = \frac{r_1}{r_2} \left( \cos(\theta_1 - \theta_2) + i \sin(\theta_1 - \theta_2) \right)$

The modulus of $z$ is $r = \frac{r_1}{r_2} = \frac{\sqrt{2}}{1} = \sqrt{2}$.

The argument of $z$ is $\theta = \theta_1 - \theta_2 = \frac{3\pi}{4} - \frac{\pi}{3}$.

$\theta = \frac{3 \times 3 \pi - 4 \times \pi}{12} = \frac{9\pi - 4\pi}{12} = \frac{5\pi}{12}$.

Therefore, the polar form of $z$ is:

$z = \sqrt{2} \left( \cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12} \right)$

---

Conclusion:

The polar form of the complex number $z = \frac{i - 1}{\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}}$ is $\sqrt{2} \left( \cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12} \right)$.

Given:

The expression to evaluate is $\left[ i^{18} + \left( \frac{1}{i} \right)^{25} \right]^3$.

---

To Find:

The value of the given expression.

---

Solution:

We first simplify the terms inside the bracket.

Simplify $i^{18}$:

The powers of $i$ cycle every 4 terms: $i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1$.

To find $i^{18}$, divide 18 by 4:

$18 = 4 \times 4 + 2$

The remainder is 2. Therefore:

$i^{18} = i^2 = -1$

Simplify $\left( \frac{1}{i} \right)^{25}$:

First, simplify $\frac{1}{i}$:

$\frac{1}{i} = \frac{1}{i} \times \frac{i}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i$

Now, calculate $(-i)^{25}$:

$(-i)^{25} = (-1)^{25} \times (i)^{25} = -1 \times i^{25}$

To find $i^{25}$, divide 25 by 4:

$25 = 4 \times 6 + 1$

The remainder is 1. Therefore:

$i^{25} = i^1 = i$

So, $(-i)^{25} = -1 \times i = -i$.

Substitute back into the bracket:

$i^{18} + \left( \frac{1}{i} \right)^{25} = -1 + (-i) = -1 - i$

Calculate the cube of the result:

We need to evaluate $(-1 - i)^3$.

$(-1 - i)^3 = [-(1 + i)]^3 = (-1)^3 (1 + i)^3 = -(1 + i)^3$

Expand $(1 + i)^3$ using the formula $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$:

$(1 + i)^3 = (1)^3 + 3(1)^2(i) + 3(1)(i)^2 + (i)^3$

$ = 1 + 3(1)i + 3i^2 + i^3$

$ = 1 + 3i + 3(-1) + (-i)$ (Since $i^2 = -1$ and $i^3 = -i$)

$ = 1 + 3i - 3 - i$

$ = (1 - 3) + (3i - i)$

$ = -2 + 2i$

Now, apply the negative sign:

$-(1 + i)^3 = -(-2 + 2i) = 2 - 2i$

---

Conclusion:

The value of the expression $\left[ i^{18} + \left( \frac{1}{i} \right)^{25} \right]^3$ is $2 - 2i$.

Given:

Two arbitrary complex numbers $z_1$ and $z_2$.

---

To Prove:

Re($z_1 z_2$) = Re($z_1$) Re($z_2$) – Im($z_1$) Im($z_2$)

---

Proof:

Let the two complex numbers be represented in their standard forms:

$z_1 = a + ib$

$z_2 = c + id$

Where $a, b, c, d$ are real numbers.

From these definitions, we can identify the real and imaginary parts:

Re($z_1$) = $a$

Im($z_1$) = $b$

Re($z_2$) = $c$

Im($z_2$) = $d$

Now, let's calculate the product $z_1 z_2$:

$z_1 z_2 = (a + ib)(c + id)$

$z_1 z_2 = a(c + id) + ib(c + id)$

$z_1 z_2 = ac + iad + ibc + i^2bd$

$z_1 z_2 = ac + iad + ibc - bd$ (Since $i^2 = -1$)

Group the real and imaginary terms:

$z_1 z_2 = (ac - bd) + i(ad + bc)$

The real part of the product $z_1 z_2$ is the term without $i$:

Now, let's evaluate the right-hand side (RHS) of the equation we need to prove:

RHS = Re($z_1$) Re($z_2$) – Im($z_1$) Im($z_2$)

RHS = $(a)(c) - (b)(d)$

Comparing equations (i) and (ii), we see that:

Re($z_1 z_2$) = Re($z_1$) Re($z_2$) – Im($z_1$) Im($z_2$)

Hence, the statement is proved.

Given:

The complex expression $Z = \left( \frac{1}{1 - 4i} -\frac{2}{1 + i}\right) \left( \frac{3 - 4i}{5 + i} \right)$.

---

To Find:

Reduce the given expression to the standard form $a + bi$.

---

Solution:

Let's simplify the expression step by step.

First, simplify the term inside the first parenthesis:

$P_1 = \frac{1}{1 - 4i} - \frac{2}{1 + i}$

Find a common denominator: $(1 - 4i)(1 + i)$.

$P_1 = \frac{1(1 + i) - 2(1 - 4i)}{(1 - 4i)(1 + i)}$

Simplify the numerator:

$1 + i - 2 + 8i = (1 - 2) + (i + 8i) = -1 + 9i$

Simplify the denominator:

$(1 - 4i)(1 + i) = 1(1) + 1(i) - 4i(1) - 4i(i) = 1 + i - 4i - 4i^2$

$ = 1 - 3i - 4(-1) = 1 - 3i + 4 = 5 - 3i$

So, the first part is:

$P_1 = \frac{-1 + 9i}{5 - 3i}$

The second part of the expression is:

$P_2 = \frac{3 - 4i}{5 + i}$

Now multiply the two parts $P_1$ and $P_2$:

$Z = P_1 \times P_2 = \left( \frac{-1 + 9i}{5 - 3i} \right) \left( \frac{3 - 4i}{5 + i} \right)$

$Z = \frac{(-1 + 9i)(3 - 4i)}{(5 - 3i)(5 + i)}$

Simplify the numerator of $Z$:

$(-1 + 9i)(3 - 4i) = -1(3) - 1(-4i) + 9i(3) + 9i(-4i)$

$ = -3 + 4i + 27i - 36i^2$

$ = -3 + 31i - 36(-1)$

$ = -3 + 31i + 36 = 33 + 31i$

Simplify the denominator of $Z$:

$(5 - 3i)(5 + i) = 5(5) + 5(i) - 3i(5) - 3i(i)$

$ = 25 + 5i - 15i - 3i^2$

$ = 25 - 10i - 3(-1)$

$ = 25 - 10i + 3 = 28 - 10i$

So, the expression becomes:

$Z = \frac{33 + 31i}{28 - 10i}$

To reduce this to the standard form $a + bi$, multiply the numerator and denominator by the conjugate of the denominator, which is $28 + 10i$.

$Z = \frac{33 + 31i}{28 - 10i} \times \frac{28 + 10i}{28 + 10i}$

$Z = \frac{(33 + 31i)(28 + 10i)}{(28 - 10i)(28 + 10i)}$

Simplify the numerator:

$(33 + 31i)(28 + 10i) = 33(28) + 33(10i) + 31i(28) + 31i(10i)$

$ = 924 + 330i + 868i + 310i^2$

$ = 924 + 1198i + 310(-1)$

$ = 924 + 1198i - 310 = 614 + 1198i$

Simplify the denominator using $(a - ib)(a + ib) = a^2 + b^2$:

$(28 - 10i)(28 + 10i) = (28)^2 + (10)^2 = 784 + 100 = 884$

Substitute these back into the expression for $Z$:

$Z = \frac{614 + 1198i}{884}$

Separate the real and imaginary parts and simplify the fractions:

$Z = \frac{614}{884} + i \frac{1198}{884}$

$Z = \frac{\cancel{614}^{307}}{\cancel{884}_{442}} + i \frac{\cancel{1198}^{599}}{\cancel{884}_{442}}$

$Z = \frac{307}{442} + i \frac{599}{442}$

---

Conclusion:

The standard form of the given expression is $\frac{307}{442} + i \frac{599}{442}$.

Here, $a = \frac{307}{442}$ and $b = \frac{599}{442}$.

Given:

The relation between complex numbers:

where $x, y, a, b, c, d$ are real numbers, and $c - id \neq 0$.

---

To Prove:

$(x^2 + y^2)^2 = \frac{a^2 + b^2}{c^2 + d^2}$

---

Proof:

We start with the given equation:

$x - iy = \sqrt{\frac{a - ib}{c - id}}$

Taking the modulus on both sides of equation (i):

$|x - iy| = \left| \sqrt{\frac{a - ib}{c - id}} \right|$

Using the properties of modulus, $|z^n| = |z|^n$ and $|\sqrt{z}| = \sqrt{|z|}$ (for the principal square root):

$|x - iy| = \sqrt{\left| \frac{a - ib}{c - id} \right|}$

Using the property $\left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}$:

$|x - iy| = \sqrt{\frac{|a - ib|}{|c - id|}}$

Recall that the modulus of a complex number $p + iq$ is $|p + iq| = \sqrt{p^2 + q^2}$. Also, $|p - iq| = \sqrt{p^2 + (-q)^2} = \sqrt{p^2 + q^2}$.

Therefore:

$|x - iy| = \sqrt{x^2 + (-y)^2} = \sqrt{x^2 + y^2}$

$|a - ib| = \sqrt{a^2 + (-b)^2} = \sqrt{a^2 + b^2}$

$|c - id| = \sqrt{c^2 + (-d)^2} = \sqrt{c^2 + d^2}$

Substitute these values back into the modulus equation:

$\sqrt{x^2 + y^2} = \sqrt{\frac{\sqrt{a^2 + b^2}}{\sqrt{c^2 + d^2}}}$

Squaring both sides:

$(\sqrt{x^2 + y^2})^2 = \left( \sqrt{\frac{\sqrt{a^2 + b^2}}{\sqrt{c^2 + d^2}}} \right)^2$

$x^2 + y^2 = \frac{\sqrt{a^2 + b^2}}{\sqrt{c^2 + d^2}}$

Squaring both sides again:

$(x^2 + y^2)^2 = \left( \frac{\sqrt{a^2 + b^2}}{\sqrt{c^2 + d^2}} \right)^2$

$(x^2 + y^2)^2 = \frac{(\sqrt{a^2 + b^2})^2}{(\sqrt{c^2 + d^2})^2}$

$(x^2 + y^2)^2 = \frac{a^2 + b^2}{c^2 + d^2}$

Hence, the statement is proved.

---

Alternate Proof (Using Conjugates):

Given:

Taking the conjugate of both sides:

$\overline{x - iy} = \overline{\sqrt{\frac{a - ib}{c - id}}}$

$x + iy = \sqrt{\overline{\left(\frac{a - ib}{c - id}\right)}}$

$x + iy = \sqrt{\frac{\overline{a - ib}}{\overline{c - id}}}$

Multiply equation (1) by equation (2):

$(x - iy)(x + iy) = \sqrt{\frac{a - ib}{c - id}} \times \sqrt{\frac{a + ib}{c + id}}$

$x^2 - (iy)^2 = \sqrt{\frac{(a - ib)(a + ib)}{(c - id)(c + id)}}$

$x^2 - i^2y^2 = \sqrt{\frac{a^2 - (ib)^2}{c^2 - (id)^2}}$

$x^2 - (-1)y^2 = \sqrt{\frac{a^2 - i^2b^2}{c^2 - i^2d^2}}$

$x^2 + y^2 = \sqrt{\frac{a^2 - (-1)b^2}{c^2 - (-1)d^2}}$

$x^2 + y^2 = \sqrt{\frac{a^2 + b^2}{c^2 + d^2}}$

Squaring both sides:

$(x^2 + y^2)^2 = \left( \sqrt{\frac{a^2 + b^2}{c^2 + d^2}} \right)^2$

$(x^2 + y^2)^2 = \frac{a^2 + b^2}{c^2 + d^2}$

Hence, proved.

(i)

Given:

The complex number $z_1 = \frac{1 + 7i}{(2 - i)^2}$.

---

To Find:

Convert the complex number $z_1$ into its polar form $r(\cos \theta + i \sin \theta)$.

---

Solution:

First, simplify the denominator:

$(2 - i)^2 = 2^2 - 2(2)(i) + i^2 = 4 - 4i - 1 = 3 - 4i$.

So, the complex number is:

$z_1 = \frac{1 + 7i}{3 - 4i}$

To convert $z_1$ to the standard form $x + iy$, multiply the numerator and denominator by the conjugate of the denominator, which is $3 + 4i$.

$z_1 = \frac{1 + 7i}{3 - 4i} \times \frac{3 + 4i}{3 + 4i} = \frac{(1 + 7i)(3 + 4i)}{(3 - 4i)(3 + 4i)}$

Numerator: $(1 + 7i)(3 + 4i) = 1(3) + 1(4i) + 7i(3) + 7i(4i)$

$ = 3 + 4i + 21i + 28i^2 = 3 + 25i - 28 = -25 + 25i$.

Denominator: $(3 - 4i)(3 + 4i) = 3^2 + 4^2 = 9 + 16 = 25$.

Substitute back:

$z_1 = \frac{-25 + 25i}{25} = \frac{-25}{25} + \frac{25i}{25} = -1 + i$.

Now, convert $z_1 = -1 + i$ to polar form.

Comparing with $x + iy$, we have $x = -1$ and $y = 1$.

Modulus $r$:

$r = \sqrt{x^2 + y^2} = \sqrt{(-1)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}$.

Argument $\theta$:

The complex number lies in the second quadrant since $x < 0$ and $y > 0$.

The reference angle $\alpha = \tan^{-1}\left|\frac{y}{x}\right| = \tan^{-1}\left|\frac{1}{-1}\right| = \tan^{-1}(1) = \frac{\pi}{4}$.

For the second quadrant, the principal argument is $\theta = \pi - \alpha = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

Polar Form:

The polar form is $z_1 = r(\cos \theta + i \sin \theta)$.

$z_1 = \sqrt{2} \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right)$.

---

(ii)

Given:

The complex number $z_2 = \frac{1 + 3i}{1 - 2i}$.

---

To Find:

Convert the complex number $z_2$ into its polar form $r(\cos \theta + i \sin \theta)$.

---

Solution:

To convert $z_2$ to the standard form $x + iy$, multiply the numerator and denominator by the conjugate of the denominator, which is $1 + 2i$.

$z_2 = \frac{1 + 3i}{1 - 2i} \times \frac{1 + 2i}{1 + 2i} = \frac{(1 + 3i)(1 + 2i)}{(1 - 2i)(1 + 2i)}$

Numerator: $(1 + 3i)(1 + 2i) = 1(1) + 1(2i) + 3i(1) + 3i(2i)$

$ = 1 + 2i + 3i + 6i^2 = 1 + 5i - 6 = -5 + 5i$.

Denominator: $(1 - 2i)(1 + 2i) = 1^2 + 2^2 = 1 + 4 = 5$.

Substitute back:

$z_2 = \frac{-5 + 5i}{5} = \frac{-5}{5} + \frac{5i}{5} = -1 + i$.

Now, convert $z_2 = -1 + i$ to polar form.

Comparing with $x + iy$, we have $x = -1$ and $y = 1$.

Modulus $r$:

$r = \sqrt{x^2 + y^2} = \sqrt{(-1)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}$.

Argument $\theta$:

The complex number lies in the second quadrant since $x < 0$ and $y > 0$.

The reference angle $\alpha = \tan^{-1}\left|\frac{y}{x}\right| = \tan^{-1}\left|\frac{1}{-1}\right| = \tan^{-1}(1) = \frac{\pi}{4}$.

For the second quadrant, the principal argument is $\theta = \pi - \alpha = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

Polar Form:

The polar form is $z_2 = r(\cos \theta + i \sin \theta)$.

$z_2 = \sqrt{2} \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right)$.

---

Conclusion:

(i) The polar form of $\frac{1 + 7i}{(2 - i)^2}$ is $\sqrt{2} \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right)$.

(ii) The polar form of $\frac{1 + 3i}{1 - 2i}$ is $\sqrt{2} \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right)$.

Given:

The quadratic equation is $3x^2 - 4x + \frac{20}{3} = 0$.

---

To Find:

The solution(s) for $x$.

---

Solution:

The given equation is $3x^2 - 4x + \frac{20}{3} = 0$.

To eliminate the fraction, we can multiply the entire equation by 3:

$3 \left( 3x^2 - 4x + \frac{20}{3} \right) = 3 \times 0$

$9x^2 - 12x + 20 = 0$

This is a quadratic equation of the form $ax^2 + bx + c = 0$.

Comparing $9x^2 - 12x + 20 = 0$ with the standard form, we identify the coefficients:

$a = 9$

$b = -12$

$c = 20$

We use the quadratic formula to find the solutions for $x$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

First, calculate the discriminant, $D = b^2 - 4ac$:

$D = (-12)^2 - 4(9)(20)$

$D = 144 - 720$

$D = -576$

Since the discriminant is negative ($D < 0$), the solutions will be complex.

Now substitute the values of $a$, $b$, and the discriminant $D$ into the quadratic formula:

$x = \frac{-(-12) \pm \sqrt{-576}}{2(9)}$

$x = \frac{12 \pm \sqrt{-576}}{18}$

We simplify $\sqrt{-576}$:

$\sqrt{-576} = \sqrt{-1 \times 576} = \sqrt{-1} \times \sqrt{576} = i \times 24 = 24i$

Substitute this back into the expression for $x$:

$x = \frac{12 \pm 24i}{18}$

Separate the real and imaginary parts and simplify:

$x = \frac{12}{18} \pm \frac{24}{18}i$

$x = \frac{2 \times \cancel{6}}{3 \times \cancel{6}} \pm \frac{4 \times \cancel{6}}{3 \times \cancel{6}}i$

$x = \frac{2}{3} \pm \frac{4}{3}i$

The two solutions are:

$x_1 = \frac{2}{3} + \frac{4}{3}i$

$x_2 = \frac{2}{3} - \frac{4}{3}i$

---

Conclusion:

The solutions to the quadratic equation $3x^2 - 4x + \frac{20}{3} = 0$ are $x = \frac{2}{3} + \frac{4}{3}i$ and $x = \frac{2}{3} - \frac{4}{3}i$.

Given:

The quadratic equation is $x^2 - 2x + \frac{3}{2} = 0$.

---

To Find:

The solution(s) for $x$.

---

Solution:

The given equation is $x^2 - 2x + \frac{3}{2} = 0$.

To eliminate the fraction, we can multiply the entire equation by 2:

$2 \left( x^2 - 2x + \frac{3}{2} \right) = 2 \times 0$

$2x^2 - 4x + 3 = 0$

This is a quadratic equation of the form $ax^2 + bx + c = 0$.

Comparing $2x^2 - 4x + 3 = 0$ with the standard form, we identify the coefficients:

$a = 2$

$b = -4$

$c = 3$

We use the quadratic formula to find the solutions for $x$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

First, calculate the discriminant, $D = b^2 - 4ac$:

$D = (-4)^2 - 4(2)(3)$

$D = 16 - 24$

$D = -8$

Since the discriminant is negative ($D < 0$), the solutions will be complex.

Now substitute the values of $a$, $b$, and the discriminant $D$ into the quadratic formula:

$x = \frac{-(-4) \pm \sqrt{-8}}{2(2)}$

$x = \frac{4 \pm \sqrt{-8}}{4}$

We simplify $\sqrt{-8}$:

$\sqrt{-8} = \sqrt{-1 \times 8} = \sqrt{-1} \times \sqrt{8} = i \sqrt{4 \times 2} = i (2\sqrt{2}) = 2i\sqrt{2}$

Substitute this back into the expression for $x$:

$x = \frac{4 \pm 2i\sqrt{2}}{4}$

Separate the real and imaginary parts and simplify:

$x = \frac{4}{4} \pm \frac{2i\sqrt{2}}{4}$

$x = 1 \pm i \frac{\cancel{2}\sqrt{2}}{\cancel{4}_2}$

$x = 1 \pm i \frac{\sqrt{2}}{2}$

The two solutions are:

$x_1 = 1 + i \frac{\sqrt{2}}{2}$

$x_2 = 1 - i \frac{\sqrt{2}}{2}$

---

Conclusion:

The solutions to the quadratic equation $x^2 - 2x + \frac{3}{2} = 0$ are $x = 1 + i \frac{\sqrt{2}}{2}$ and $x = 1 - i \frac{\sqrt{2}}{2}$.

Given:

The quadratic equation is $27x^2 - 10x + 1 = 0$.

---

To Find:

The solution(s) for $x$.

---

Solution:

The given equation is $27x^2 - 10x + 1 = 0$.

This is a quadratic equation of the form $ax^2 + bx + c = 0$.

Comparing $27x^2 - 10x + 1 = 0$ with the standard form, we identify the coefficients:

$a = 27$

$b = -10$

$c = 1$

We use the quadratic formula to find the solutions for $x$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

First, calculate the discriminant, $D = b^2 - 4ac$:

$D = (-10)^2 - 4(27)(1)$

$D = 100 - 108$

$D = -8$

Since the discriminant is negative ($D < 0$), the solutions will be complex.

Now substitute the values of $a$, $b$, and the discriminant $D$ into the quadratic formula:

$x = \frac{-(-10) \pm \sqrt{-8}}{2(27)}$

$x = \frac{10 \pm \sqrt{-8}}{54}$

We simplify $\sqrt{-8}$:

$\sqrt{-8} = \sqrt{-1 \times 8} = \sqrt{-1} \times \sqrt{8} = i \sqrt{4 \times 2} = i (2\sqrt{2}) = 2i\sqrt{2}$

Substitute this back into the expression for $x$:

$x = \frac{10 \pm 2i\sqrt{2}}{54}$

Factor out 2 from the numerator and simplify the fraction:

$x = \frac{2(5 \pm i\sqrt{2})}{54}$

$x = \frac{\cancel{2}(5 \pm i\sqrt{2})}{\cancel{54}_{27}}$

$x = \frac{5 \pm i\sqrt{2}}{27}$

The two solutions are:

$x_1 = \frac{5 + i\sqrt{2}}{27}$

$x_2 = \frac{5 - i\sqrt{2}}{27}$

---

Conclusion:

The solutions to the quadratic equation $27x^2 - 10x + 1 = 0$ are $x = \frac{5 + i\sqrt{2}}{27}$ and $x = \frac{5 - i\sqrt{2}}{27}$.

Given:

The quadratic equation is $21x^2 - 28x + 10 = 0$.

---

To Find:

The solution(s) for $x$.

---

Solution:

The given equation is $21x^2 - 28x + 10 = 0$.

This is a quadratic equation of the form $ax^2 + bx + c = 0$.

Comparing $21x^2 - 28x + 10 = 0$ with the standard form, we identify the coefficients:

$a = 21$

$b = -28$

$c = 10$

We use the quadratic formula to find the solutions for $x$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

First, calculate the discriminant, $D = b^2 - 4ac$:

$D = (-28)^2 - 4(21)(10)$

$D = 784 - 840$

$D = -56$

Since the discriminant is negative ($D < 0$), the solutions will be complex.

Now substitute the values of $a$, $b$, and the discriminant $D$ into the quadratic formula:

$x = \frac{-(-28) \pm \sqrt{-56}}{2(21)}$

$x = \frac{28 \pm \sqrt{-56}}{42}$

We simplify $\sqrt{-56}$:

$\sqrt{-56} = \sqrt{-1 \times 56} = \sqrt{-1} \times \sqrt{4 \times 14} = i \times 2\sqrt{14} = 2i\sqrt{14}$

Substitute this back into the expression for $x$:

$x = \frac{28 \pm 2i\sqrt{14}}{42}$

Factor out 2 from the numerator and simplify the fraction:

$x = \frac{2(14 \pm i\sqrt{14})}{42}$

$x = \frac{\cancel{2}(14 \pm i\sqrt{14})}{\cancel{42}_{21}}$

$x = \frac{14 \pm i\sqrt{14}}{21}$

We can separate the real and imaginary parts:

$x = \frac{14}{21} \pm i \frac{\sqrt{14}}{21}$

$x = \frac{2 \times \cancel{7}}{3 \times \cancel{7}} \pm i \frac{\sqrt{14}}{21}$

$x = \frac{2}{3} \pm i \frac{\sqrt{14}}{21}$

The two solutions are:

$x_1 = \frac{14 + i\sqrt{14}}{21}$ or $x_1 = \frac{2}{3} + i \frac{\sqrt{14}}{21}$

$x_2 = \frac{14 - i\sqrt{14}}{21}$ or $x_2 = \frac{2}{3} - i \frac{\sqrt{14}}{21}$

---

Conclusion:

The solutions to the quadratic equation $21x^2 - 28x + 10 = 0$ are $x = \frac{14 + i\sqrt{14}}{21}$ and $x = \frac{14 - i\sqrt{14}}{21}$ (or equivalently, $x = \frac{2}{3} \pm i \frac{\sqrt{14}}{21}$).

Given:

The complex numbers $z_1 = 2 - i$ and $z_2 = 1 + i$.

---

To Find:

The value of the modulus $\left| \frac{z_{1} + z_{2} + 1}{z_{1} - z_{2} + 1} \right|$.

---

Solution:

Let the expression be $Z = \frac{z_1 + z_2 + 1}{z_1 - z_2 + 1}$. We need to find $|Z|$.

First, calculate the numerator $N = z_1 + z_2 + 1$:

$N = (2 - i) + (1 + i) + 1$

$N = (2 + 1 + 1) + (-i + i)$

$N = 4 + 0i = 4$

Next, calculate the denominator $D = z_1 - z_2 + 1$:

$D = (2 - i) - (1 + i) + 1$

$D = 2 - i - 1 - i + 1$

$D = (2 - 1 + 1) + (-i - i)$

$D = 2 - 2i$

Now substitute the numerator and denominator back into the expression for $Z$:

$Z = \frac{4}{2 - 2i}$

We need to find the modulus of $Z$, i.e., $|Z| = \left| \frac{4}{2 - 2i} \right|$.

Using the property of modulus $\left| \frac{z_a}{z_b} \right| = \frac{|z_a|}{|z_b|}$:

$|Z| = \frac{|4|}{|2 - 2i|}$

The modulus of the numerator is $|4| = \sqrt{4^2 + 0^2} = 4$.

The modulus of the denominator is $|2 - 2i| = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8}$.

$\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$.

So, the modulus of $Z$ is:

$|Z| = \frac{4}{2\sqrt{2}}$

$|Z| = \frac{2}{\sqrt{2}}$

Rationalize the denominator:

$|Z| = \frac{2}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{2\sqrt{2}}{2} = \sqrt{2}$

---

Conclusion:

The value of $\left| \frac{z_{1} + z_{2} + 1}{z_{1} - z_{2} + 1} \right|$ is $\sqrt{2}$.

Given:

The relation between complex numbers:

where $a, b, x$ are real numbers.

---

To Prove:

$a^2 + b^2 = \frac{(x^2 + 1)^2}{(2x^2 + 1)^2}$

---

Proof:

We start with the given equation:

$a + ib = \frac{(x + i)^2}{2x^2 + 1}$

Take the modulus of both sides of equation (i):

$|a + ib| = \left| \frac{(x + i)^2}{2x^2 + 1} \right|$

Using the properties of modulus: $|z_1 / z_2| = |z_1| / |z_2|$ and $|z^n| = |z|^n$:

$|a + ib| = \frac{|(x + i)^2|}{|2x^2 + 1|}$

$|a + ib| = \frac{|x + i|^2}{|2x^2 + 1|}$

Recall that the modulus of a complex number $p + iq$ is $|p + iq| = \sqrt{p^2 + q^2}$.

Also, since $x$ is real, $2x^2 + 1$ is a positive real number. The modulus of a positive real number $k$ is $k$ itself. So, $|2x^2 + 1| = 2x^2 + 1$.

Therefore:

$|a + ib| = \sqrt{a^2 + b^2}$

$|x + i| = \sqrt{x^2 + 1^2} = \sqrt{x^2 + 1}$

Substitute these values back into the modulus equation:

$\sqrt{a^2 + b^2} = \frac{(\sqrt{x^2 + 1})^2}{2x^2 + 1}$

$\sqrt{a^2 + b^2} = \frac{x^2 + 1}{2x^2 + 1}$

Now, square both sides of the equation:

$(\sqrt{a^2 + b^2})^2 = \left( \frac{x^2 + 1}{2x^2 + 1} \right)^2$

$a^2 + b^2 = \frac{(x^2 + 1)^2}{(2x^2 + 1)^2}$

Hence, the statement is proved.

Given:

The complex numbers $z_1 = 2 - i$ and $z_2 = -2 + i$.

---

(i) To Find: $Re \left(\frac{z_{1}z_{2}}{\overline{z_{1}}}\right)$

---

Solution (i):

First, find the necessary components:

The conjugate of $z_1$ is $\overline{z_1} = \overline{2 - i} = 2 + i$.

The product $z_1 z_2$ is:

$z_1 z_2 = (2 - i)(-2 + i)$

$ = 2(-2) + 2(i) - i(-2) - i(i)$

$ = -4 + 2i + 2i - i^2$

$ = -4 + 4i - (-1)$ (Since $i^2 = -1$)

$ = -4 + 4i + 1$

$ = -3 + 4i$

Now, calculate the fraction $\frac{z_1 z_2}{\overline{z_1}}$:

$\frac{z_1 z_2}{\overline{z_1}} = \frac{-3 + 4i}{2 + i}$

To simplify, multiply the numerator and denominator by the conjugate of the denominator, which is $2 - i$:

$\frac{-3 + 4i}{2 + i} \times \frac{2 - i}{2 - i} = \frac{(-3 + 4i)(2 - i)}{(2 + i)(2 - i)}$

Numerator: $(-3)(2) + (-3)(-i) + (4i)(2) + (4i)(-i)$

$ = -6 + 3i + 8i - 4i^2$

$ = -6 + 11i - 4(-1)$

$ = -6 + 11i + 4 = -2 + 11i$

Denominator: $(2)^2 + (1)^2 = 4 + 1 = 5$.

So, the fraction is:

$\frac{z_1 z_2}{\overline{z_1}} = \frac{-2 + 11i}{5} = -\frac{2}{5} + \frac{11}{5}i$

The real part (Re) of this complex number is the term without $i$.

$Re \left(\frac{z_1 z_2}{\overline{z_1}}\right) = -\frac{2}{5}$

---

(ii) To Find: $Im \left(\frac{1}{z_{1}\overline{z_{1}}}\right)$

---

Solution (ii):

First, calculate the product $z_1 \overline{z_1}$.

We have $z_1 = 2 - i$ and $\overline{z_1} = 2 + i$.

$z_1 \overline{z_1} = (2 - i)(2 + i)$

Using the property $(a - ib)(a + ib) = a^2 + b^2$ (or $|z_1|^2$):

$z_1 \overline{z_1} = (2)^2 + (1)^2 = 4 + 1 = 5$

Now, find the reciprocal $\frac{1}{z_1 \overline{z_1}}$:

$\frac{1}{z_1 \overline{z_1}} = \frac{1}{5}$

To find the imaginary part, we express $\frac{1}{5}$ in the standard form $a + bi$:

$\frac{1}{5} = \frac{1}{5} + 0i$

The imaginary part (Im) is the coefficient of $i$.

$Im \left(\frac{1}{z_1 \overline{z_1}}\right) = 0$

---

Conclusion:

(i) $Re \left(\frac{z_{1}z_{2}}{\overline{z_{1}}}\right) = -\frac{2}{5}$

(ii) $Im \left(\frac{1}{z_{1}\overline{z_{1}}}\right) = 0$

Given:

The complex number $z = \frac{1 + 2i}{1 - 3i}$.

---

To Find:

The modulus ($r$) and argument ($\theta$) of the complex number $z$.

---

Solution:

First, we need to express the complex number $z$ in the standard form $x + iy$.

Multiply the numerator and the denominator by the conjugate of the denominator, which is $1 + 3i$:

$z = \frac{1 + 2i}{1 - 3i} \times \frac{1 + 3i}{1 + 3i}$

$z = \frac{(1 + 2i)(1 + 3i)}{(1 - 3i)(1 + 3i)}$

Simplify the numerator:

Numerator $= 1(1) + 1(3i) + 2i(1) + 2i(3i)$

$ = 1 + 3i + 2i + 6i^2$

$ = 1 + 5i + 6(-1)$ (Since $i^2 = -1$)

$ = 1 + 5i - 6$

$ = -5 + 5i$

Simplify the denominator using the property $(a - ib)(a + ib) = a^2 + b^2$:

Denominator $= (1)^2 + (3)^2 = 1 + 9 = 10$

Substitute the simplified numerator and denominator back:

$z = \frac{-5 + 5i}{10}$

Separate the real and imaginary parts:

$z = \frac{-5}{10} + \frac{5}{10}i$

$z = -\frac{1}{2} + \frac{1}{2}i$

Now we have the standard form $z = x + iy$, where $x = -\frac{1}{2}$ and $y = \frac{1}{2}$.

Modulus:

The modulus $r = |z|$ is calculated as:

$r = \sqrt{x^2 + y^2} = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2}$

$r = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{2}{4}} = \sqrt{\frac{1}{2}}$

$r = \frac{1}{\sqrt{2}}$

Argument:

Let $\theta$ be the argument. Since $x = -\frac{1}{2} < 0$ and $y = \frac{1}{2} > 0$, the complex number lies in the second quadrant.

The argument $\theta$ satisfies:

Let $\alpha$ be the acute reference angle:

$\alpha = \tan^{-1}\left|\frac{y}{x}\right| = \tan^{-1}\left|\frac{1/2}{-1/2}\right| = \tan^{-1}|-1| = \tan^{-1}(1) = \frac{\pi}{4}$.

For an angle in the second quadrant, the principal argument $\theta$ (in the range $(-\pi, \pi]$) is:

$\theta = \pi - \alpha = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

---

Conclusion:

For the complex number $z = \frac{1 + 2i}{1 - 3i}$:

• The modulus is $r = \frac{1}{\sqrt{2}}$.
• The argument (principal value) is $\theta = \frac{3\pi}{4}$.

Given:

The complex number expression $(x - iy)(3 + 5i)$ is the conjugate of the complex number $-6 - 24i$.

$x$ and $y$ are real numbers.

---

To Find:

The values of the real numbers $x$ and $y$.

---

Solution:

Let $z = -6 - 24i$.

The conjugate of $z$ is $\bar{z} = \overline{-6 - 24i} = -6 + 24i$.

According to the problem statement:

$(x - iy)(3 + 5i) = \bar{z}$

$(x - iy)(3 + 5i) = -6 + 24i$

Expand the left side of the equation:

$(x - iy)(3 + 5i) = x(3 + 5i) - iy(3 + 5i)$

$ = 3x + 5xi - 3iy - 5i^2y$

$ = 3x + 5xi - 3iy - 5(-1)y$ (Since $i^2 = -1$)

$ = 3x + 5y + i(5x - 3y)$

Now, equate this expression to the conjugate $\bar{z}$:

$(3x + 5y) + i(5x - 3y) = -6 + 24i$

For two complex numbers to be equal, their real parts and imaginary parts must be equal.

Equating the real parts:

Equating the imaginary parts:

We need to solve this system of linear equations for $x$ and $y$.

Multiply equation (1) by 3 and equation (2) by 5:

$3 \times (3x + 5y = -6) \implies 9x + 15y = -18$

$5 \times (5x - 3y = 24) \implies 25x - 15y = 120$

Add the two new equations:

$(9x + 15y) + (25x - 15y) = -18 + 120$

$34x = 102$

$x = \frac{102}{34}$

$x = 3$

Substitute the value of $x = 3$ into equation (1):

$3(3) + 5y = -6$

$9 + 5y = -6$

$5y = -6 - 9$

$5y = -15$

$y = \frac{-15}{5}$

$y = -3$

---

Conclusion:

The real numbers are $x = 3$ and $y = -3$.

Given:

The complex number expression $Z = \frac{1 + i}{1 - i} - \frac{1 - i}{1 + i}$.

---

To Find:

The modulus of the complex number $Z$, denoted as $|Z|$.

---

Solution:

First, we simplify the expression for $Z$. Find a common denominator, which is $(1 - i)(1 + i)$.

$Z = \frac{(1 + i)(1 + i)}{(1 - i)(1 + i)} - \frac{(1 - i)(1 - i)}{(1 - i)(1 + i)}$

$Z = \frac{(1 + i)^2 - (1 - i)^2}{(1 - i)(1 + i)}$

Simplify the numerator:

$(1 + i)^2 = 1^2 + 2(1)(i) + i^2 = 1 + 2i - 1 = 2i$.

$(1 - i)^2 = 1^2 - 2(1)(i) + i^2 = 1 - 2i - 1 = -2i$.

Numerator $= (1 + i)^2 - (1 - i)^2 = (2i) - (-2i) = 2i + 2i = 4i$.

Simplify the denominator using the property $(a - ib)(a + ib) = a^2 + b^2$:

Denominator $= (1 - i)(1 + i) = 1^2 + 1^2 = 1 + 1 = 2$.

Substitute the simplified numerator and denominator back into the expression for $Z$:

$Z = \frac{4i}{2} = 2i$.

Now, we need to find the modulus of $Z = 2i$.

Write $Z$ in the standard form $x + iy$: $Z = 0 + 2i$.

Here, $x = 0$ and $y = 2$.

The modulus $|Z|$ is calculated as:

$|Z| = \sqrt{x^2 + y^2} = \sqrt{(0)^2 + (2)^2}$

$|Z| = \sqrt{0 + 4} = \sqrt{4}$

$|Z| = 2$

---

Conclusion:

The modulus of the complex number $\frac{1 + i}{1 - i} - \frac{1 - i}{1 + i}$ is 2.

Given:

The relation $(x + iy)^3 = u + iv$, where $x, y, u, v$ are real numbers.

---

To Prove:

$\frac{u}{x} + \frac{v}{y} = 4 (x^2 - y^2)$

---

Proof:

We are given the equation:

$(x + iy)^3 = u + iv$

Expand the left side using the binomial expansion formula $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$:

$(x + iy)^3 = x^3 + 3(x^2)(iy) + 3(x)(iy)^2 + (iy)^3$

$ = x^3 + 3ix^2y + 3x(i^2y^2) + i^3y^3$

Using the properties $i^2 = -1$ and $i^3 = i^2 \cdot i = -i$:

$ = x^3 + 3ix^2y + 3x(-1)y^2 + (-i)y^3$

$ = x^3 + 3ix^2y - 3xy^2 - iy^3$

Group the real and imaginary parts:

$(x + iy)^3 = (x^3 - 3xy^2) + i(3x^2y - y^3)$

Now, equate this to the right side, $u + iv$:

$(x^3 - 3xy^2) + i(3x^2y - y^3) = u + iv$

Comparing the real and imaginary parts on both sides, we get:

Assuming $x \neq 0$ and $y \neq 0$, we can divide equation (1) by $x$ and equation (2) by $y$:

From (1): $\frac{u}{x} = \frac{x^3 - 3xy^2}{x} = \frac{x(x^2 - 3y^2)}{x} = x^2 - 3y^2$

From (2): $\frac{v}{y} = \frac{3x^2y - y^3}{y} = \frac{y(3x^2 - y^2)}{y} = 3x^2 - y^2$

Now, add the expressions for $\frac{u}{x}$ and $\frac{v}{y}$:

$\frac{u}{x} + \frac{v}{y} = (x^2 - 3y^2) + (3x^2 - y^2)$

$= x^2 - 3y^2 + 3x^2 - y^2$

$= (x^2 + 3x^2) + (-3y^2 - y^2)$

$= 4x^2 - 4y^2$

$= 4(x^2 - y^2)$

Thus, we have shown that $\frac{u}{x} + \frac{v}{y} = 4(x^2 - y^2)$, assuming $x \neq 0$ and $y \neq 0$.

(Note: If $x=0$, then $u=0$ and $v=-y^3$. If $y=0$, then $u=x^3$ and $v=0$. In these cases, the expression $\frac{u}{x} + \frac{v}{y}$ involves division by zero, so the identity holds for $x \neq 0, y \neq 0$.)

---

Conclusion:

Given $(x + iy)^3 = u + iv$, we have proved that $\frac{u}{x} + \frac{v}{y} = 4(x^2 - y^2)$.

Given:

$\alpha$ and $\beta$ are different complex numbers.

$|\beta| = 1$.

---

To Find:

The value of the modulus $\left| \frac{\beta - \alpha}{1 - \overline{\alpha}\beta} \right|$.

---

Solution:

Let $Z = \frac{\beta - \alpha}{1 - \overline{\alpha}\beta}$. We need to find $|Z|$.

We know that for any complex number $w$, $|w|^2 = w \overline{w}$.

Let's calculate $|Z|^2$:

$|Z|^2 = Z \overline{Z} = \left( \frac{\beta - \alpha}{1 - \overline{\alpha}\beta} \right) \overline{\left( \frac{\beta - \alpha}{1 - \overline{\alpha}\beta} \right)}$

Using the properties of conjugates $\overline{\left(\frac{w_1}{w_2}\right)} = \frac{\overline{w_1}}{\overline{w_2}}$ and $\overline{w_1 - w_2} = \overline{w_1} - \overline{w_2}$:

$|Z|^2 = \left( \frac{\beta - \alpha}{1 - \overline{\alpha}\beta} \right) \left( \frac{\overline{\beta - \alpha}}{\overline{1 - \overline{\alpha}\beta}} \right)$

$|Z|^2 = \left( \frac{\beta - \alpha}{1 - \overline{\alpha}\beta} \right) \left( \frac{\overline{\beta} - \overline{\alpha}}{\overline{1} - \overline{\overline{\alpha}\beta}} \right)$

Since $\overline{1} = 1$ and $\overline{\overline{\alpha}\beta} = \overline{\overline{\alpha}} \overline{\beta} = \alpha \overline{\beta}$:

$|Z|^2 = \frac{(\beta - \alpha)(\overline{\beta} - \overline{\alpha})}{(1 - \overline{\alpha}\beta)(1 - \alpha \overline{\beta})}$

Expand the numerator:

Numerator $= (\beta - \alpha)(\overline{\beta} - \overline{\alpha}) = \beta\overline{\beta} - \beta\overline{\alpha} - \alpha\overline{\beta} + \alpha\overline{\alpha}$

$= |\beta|^2 - \beta\overline{\alpha} - \alpha\overline{\beta} + |\alpha|^2$

Expand the denominator:

Denominator $= (1 - \overline{\alpha}\beta)(1 - \alpha \overline{\beta}) = 1 - \alpha\overline{\beta} - \overline{\alpha}\beta + (\overline{\alpha}\beta)(\alpha\overline{\beta})$

$ = 1 - \alpha\overline{\beta} - \overline{\alpha}\beta + \overline{\alpha}\alpha \beta\overline{\beta}$

$ = 1 - \alpha\overline{\beta} - \overline{\alpha}\beta + |\alpha|^2 |\beta|^2$

So,

$|Z|^2 = \frac{|\beta|^2 - \beta\overline{\alpha} - \alpha\overline{\beta} + |\alpha|^2}{1 - \alpha\overline{\beta} - \overline{\alpha}\beta + |\alpha|^2 |\beta|^2}$

We are given that $|\beta| = 1$, which implies $|\beta|^2 = 1^2 = 1$.

Substitute $|\beta|^2 = 1$ into the expression for $|Z|^2$:

$|Z|^2 = \frac{1 - \beta\overline{\alpha} - \alpha\overline{\beta} + |\alpha|^2}{1 - \alpha\overline{\beta} - \overline{\alpha}\beta + |\alpha|^2 (1)}$

$|Z|^2 = \frac{1 - \beta\overline{\alpha} - \alpha\overline{\beta} + |\alpha|^2}{1 - \alpha\overline{\beta} - \overline{\alpha}\beta + |\alpha|^2}$

The numerator and the denominator are identical.

$|Z|^2 = 1$

Since $|Z|$ is a modulus, it must be non-negative.

$|Z| = \sqrt{1} = 1$

(We need to ensure the denominator $1 - \overline{\alpha}\beta \neq 0$. If $1 - \overline{\alpha}\beta = 0$, then $\overline{\alpha}\beta = 1$. Taking moduli, $|\overline{\alpha}\beta| = 1 \implies |\overline{\alpha}||\beta| = 1$. Since $|\beta|=1$ and $|\overline{\alpha}|=|\alpha|$, we get $|\alpha|=1$. If $|\alpha|=1$, then $\overline{\alpha} = 1/\alpha$. Substituting into $\overline{\alpha}\beta=1$ gives $(1/\alpha)\beta=1$, which means $\beta=\alpha$. But we are given $\alpha \neq \beta$. Thus, the denominator is never zero.)

---

Conclusion:

The value of $\left| \frac{\beta - \alpha}{1 - \overline{\alpha}\beta} \right|$ is 1.

Given:

The equation $|1 - i|^x = 2^x$.

---

To Find:

The number of non-zero integral solutions for $x$.

---

Solution:

We are given the equation:

$|1 - i|^x = 2^x$

First, let's calculate the modulus of the complex number $1 - i$.

For a complex number $z = a + bi$, the modulus is $|z| = \sqrt{a^2 + b^2}$.

Here, $a = 1$ and $b = -1$.

$|1 - i| = \sqrt{(1)^2 + (-1)^2}$

$|1 - i| = \sqrt{1 + 1}$

$|1 - i| = \sqrt{2}$

Now substitute this value back into the given equation:

$(\sqrt{2})^x = 2^x$

We can write $\sqrt{2}$ as $2^{1/2}$. Substituting this:

$(2^{1/2})^x = 2^x$

Using the exponent rule $(a^m)^n = a^{mn}$:

$2^{x/2} = 2^x$

Since the bases are equal, we can equate the exponents:

$\frac{x}{2} = x$

To solve for $x$, subtract $x$ from both sides:

$\frac{x}{2} - x = 0$

$\frac{x - 2x}{2} = 0$

$\frac{-x}{2} = 0$

$-x = 0$

$x = 0$

The only integral solution found is $x = 0$.

However, the question asks for the number of non-zero integral solutions.

Since the only integral solution is 0, there are no non-zero integral solutions.

---

Conclusion:

There are no non-zero integral solutions to the equation $|1 - i|^x = 2^x$. The number of such solutions is 0.

Given:

The equation involving products of complex numbers:

where $a, b, c, d, e, f, g, h, A, B$ are real numbers.

---

To Prove:

$(a^2 + b^2 ) (c^2 + d^2 ) (e^2 + f^2 ) (g^2 + h^2 ) = A^2 + B^2$

---

Proof:

We start with the given equation (i):

$(a + ib) (c + id) (e + if) (g + ih) = A + iB$

Take the modulus of both sides of the equation:

$|(a + ib) (c + id) (e + if) (g + ih)| = |A + iB|$

We use the property that the modulus of a product of complex numbers is the product of their moduli: $|z_1 z_2 ... z_n| = |z_1| |z_2| ... |z_n|$.

$|a + ib| |c + id| |e + if| |g + ih| = |A + iB|$

Recall that the modulus of a complex number $p + iq$ is $|p + iq| = \sqrt{p^2 + q^2}$.

Apply this to each term:

$|a + ib| = \sqrt{a^2 + b^2}$

$|c + id| = \sqrt{c^2 + d^2}$

$|e + if| = \sqrt{e^2 + f^2}$

$|g + ih| = \sqrt{g^2 + h^2}$

$|A + iB| = \sqrt{A^2 + B^2}$

Substitute these modulus values back into the equation:

$\sqrt{a^2 + b^2} \sqrt{c^2 + d^2} \sqrt{e^2 + f^2} \sqrt{g^2 + h^2} = \sqrt{A^2 + B^2}$

Square both sides of the equation:

$\left( \sqrt{a^2 + b^2} \sqrt{c^2 + d^2} \sqrt{e^2 + f^2} \sqrt{g^2 + h^2} \right)^2 = \left( \sqrt{A^2 + B^2} \right)^2$

Using the property $(\sqrt{p}\sqrt{q}...)^2 = (\sqrt{p})^2 (\sqrt{q})^2 ... = pq...$

$(\sqrt{a^2 + b^2})^2 (\sqrt{c^2 + d^2})^2 (\sqrt{e^2 + f^2})^2 (\sqrt{g^2 + h^2})^2 = A^2 + B^2$

$(a^2 + b^2 ) (c^2 + d^2 ) (e^2 + f^2 ) (g^2 + h^2 ) = A^2 + B^2$

Hence, the statement is proved.

Given:

The equation is $\left( \frac{1 + i}{1 - i} \right)^{m} = 1$.

To Find:

The least positive integral value of $m$.

---

Solution:

First, we simplify the complex fraction inside the parenthesis: $\frac{1 + i}{1 - i}$.

To do this, we multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of $(1 - i)$ is $(1 + i)$.

$\frac{1 + i}{1 - i} = \frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i}$

$\frac{1 + i}{1 - i} = \frac{(1 + i)^2}{(1 - i)(1 + i)}$

Now, let's simplify the numerator and the denominator separately.

Numerator: $(1 + i)^2 = 1^2 + 2(1)(i) + i^2 = 1 + 2i - 1 = 2i$. (Since $i^2 = -1$)

Denominator: $(1 - i)(1 + i) = 1^2 - (i)^2 = 1 - (-1) = 1 + 1 = 2$.

Substitute these back into the fraction:

$\frac{1 + i}{1 - i} = \frac{2i}{2} = i$

Now, the given equation becomes:

$(i)^m = 1$

---

We need to find the least positive integer value of $m$ for which $i^m = 1$.

Let's examine the powers of $i$:

• $i^1 = i$
• $i^2 = -1$
• $i^3 = i^2 \cdot i = -1 \cdot i = -i$
• $i^4 = i^2 \cdot i^2 = (-1)(-1) = 1$

We see that the first time the power of $i$ equals 1 is when the exponent is 4.

The powers of $i$ repeat in a cycle of 4, so $i^m = 1$ whenever $m$ is a multiple of 4 (i.e., $m = 4k$ for any positive integer $k$).

The positive integral values of $m$ that satisfy the equation are 4, 8, 12, 16, and so on.

The least of these positive integral values is 4.

---

Therefore, the least positive integral value of $m$ is 4.