Chapter 4 Complex Numbers And Quadratic Equations

Given:

The equation $4x + i(3x – y) = 3 + i(– 6)$, where $x, y \in \mathbb{R}$.

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To Find:

The values of $x$ and $y$.

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Solution:

We are given the equation:

$4x + i(3x – y) = 3 + i(– 6)$

For two complex numbers to be equal, their real parts must be equal and their imaginary parts must be equal.

Equating the real parts:

Equating the imaginary parts:

From equation (i), we can find the value of $x$:

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

Now, substitute the value of $x$ into equation (ii):

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

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

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

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

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

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

Thus, the values of $x$ and $y$ are $\mathbf{x = \frac{3}{4}}$ and $\mathbf{y = \frac{33}{4}}$.

To Express:

The given expressions in the form of $a + bi$.

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Solution (i):

We need to express $(-5i) \left( \frac{1}{8}i \right)$ in the form $a + bi$.

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

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

Since $i^2 = -1$, we have:

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

$= \frac{5}{8}$

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

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

Thus, $(-5i) \left( \frac{1}{8}i \right)$ in the form $a + bi$ is $\mathbf{\frac{5}{8} + 0i}$.

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Solution (ii):

We need to express $(-i) (2i) \left( -\frac{1}{8}i\right)^{3}$ in the form $a + bi$.

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

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

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

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

Since $i^3 = i^2 \times i = (-1) \times i = -i$, we have:

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

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

Now, substitute this back into the original expression:

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

$= (-1 \times 2 \times \frac{1}{512}) \times (i \times i \times i)$

$= -\frac{2}{512} \times i^3$

$= -\frac{1}{256} \times i^3$

Again, using $i^3 = -i$:

$= -\frac{1}{256} \times (-i)$

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

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

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

Thus, $(-i) (2i) \left( -\frac{1}{8}i\right)^{3}$ in the form $a + bi$ is $\mathbf{0 + \frac{1}{256}i}$.

To Express:

$(5 - 3i)^3$ in the form $a + bi$.

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Solution:

We need to express $(5 - 3i)^3$ in the form $a + bi$.

We can use the binomial expansion formula $(x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3$.

Here, $x = 5$ and $y = 3i$.

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

Calculate each term:

$(5)^3 = 125$

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

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

Since $i^2 = -1$, we have $15(9i^2) = 15(9)(-1) = 15(-9) = -135$

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

Since $i^3 = i^2 \times i = (-1) \times i = -i$, we have $27 i^3 = 27(-i) = -27i$

Now, substitute these values back into the expansion:

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

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

Group the real and imaginary terms:

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

$(5 - 3i)^3 = -10 + (-198)i$

$(5 - 3i)^3 = -10 - 198i$

The expression $(5 - 3i)^3$ in the form $a + bi$ is $\mathbf{-10 - 198i}$. Here, $a = -10$ and $b = -198$.

To Express:

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

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Solution:

We are given the expression $(-\sqrt{3} + \sqrt{-2}) (2\sqrt{3}- i)$.

First, simplify the term $\sqrt{-2}$. Since $\sqrt{-2} = \sqrt{2 \times -1} = \sqrt{2}\sqrt{-1} = \sqrt{2}i$, the expression becomes:

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

Now, multiply the two complex numbers using the distributive property (FOIL):

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

Calculate each term:

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

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

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

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

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

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

Now, combine the terms:

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

Group the real and imaginary parts:

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

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

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

Given:

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

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To Find:

The multiplicative inverse of $2 - 3i$.

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Solution:

Let the complex number be $z = 2 - 3i$.

The multiplicative inverse of a complex number $z = a + bi$ is given by the formula:

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

where $\overline{z}$ is the conjugate of $z$ and $|z|^2$ is the square of the modulus of $z$.

For $z = 2 - 3i$, the real part is $a = 2$ and the imaginary part is $b = -3$.

The conjugate of $z$ is $\overline{z} = a - bi = 2 - (-3)i = 2 + 3i$.

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

Now, substitute these values into the formula for the multiplicative inverse:

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

We can write this in the form $a + bi$:

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

Thus, the multiplicative inverse of $2 - 3i$ is $\mathbf{\frac{2}{13} + \frac{3}{13}i}$.

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Alternate Solution:

Let the multiplicative inverse of $z = 2 - 3i$ be $w = x + yi$, where $x$ and $y$ are real numbers.

By the definition of multiplicative inverse, $z \times w = 1$.

$(2 - 3i)(x + yi) = 1$

Expand the left side:

$2(x + yi) - 3i(x + yi) = 1$

$2x + 2yi - 3xi - 3yi^2 = 1$

Since $i^2 = -1$, we have:

$2x + 2yi - 3xi - 3y(-1) = 1$

$2x + 2yi - 3xi + 3y = 1$

Group the real and imaginary terms:

$(2x + 3y) + (2y - 3x)i = 1 + 0i$

For two complex numbers to be equal, their real and imaginary parts must be equal.

Equating the real parts:

Equating the imaginary parts:

From equation (ii), we can express $y$ in terms of $x$:

$2y = 3x$

$y = \frac{3}{2}x$

Substitute this value of $y$ into equation (i):

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

$2x + \frac{9}{2}x = 1$

Multiply the entire equation by 2 to eliminate the fraction:

$4x + 9x = 2$

$13x = 2$

$x = \frac{2}{13}$

Now, substitute the value of $x$ back into the equation for $y$:

$y = \frac{3}{2}x = \frac{3}{2} \times \frac{2}{13} = \frac{3}{13}$

So, the multiplicative inverse is $w = x + yi = \frac{2}{13} + \frac{3}{13}i$.

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

To Express:

The given expressions in the form of $a + bi$.

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Solution (i):

We need to express $\frac{5 + \sqrt{2}i}{1 - \sqrt{2}i}$ in the form $a + bi$.

To do this, we multiply the numerator and the denominator by the conjugate of the denominator.

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

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

Multiply the numerators:

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

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

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

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

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

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

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

Multiply the denominators:

$(1 - \sqrt{2}i)(1 + \sqrt{2}i)$ is in the form $(a - bi)(a + bi) = a^2 + b^2$.

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

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

$= 1 + 2$

$= 3$

Now, combine the simplified numerator and denominator:

$\frac{5 + \sqrt{2}i}{1 - \sqrt{2}i} = \frac{3 + 6\sqrt{2}i}{3}$

Separate the real and imaginary parts:

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

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

Thus, $\frac{5 + \sqrt{2}i}{1 - \sqrt{2}i}$ in the form $a + bi$ is $\mathbf{1 + 2\sqrt{2}i}$.

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Solution (ii):

We need to express $i^{-35}$ in the form $a + bi$.

We can write $i^{-35}$ as $\frac{1}{i^{35}}$.

To simplify $i^{35}$, we divide the exponent by 4 and consider the remainder. The powers of $i$ cycle with a period of 4 ($i^1=i, i^2=-1, i^3=-i, i^4=1$).

Divide 35 by 4: $35 = 4 \times 8 + 3$. The remainder is 3.

Therefore, $i^{35} = i^{4 \times 8 + 3} = (i^4)^8 \times i^3 = (1)^8 \times i^3 = 1 \times i^3 = i^3$.

We know that $i^3 = -i$.

So, $i^{-35} = \frac{1}{i^{35}} = \frac{1}{-i}$.

To express $\frac{1}{-i}$ in the form $a + bi$, multiply the numerator and denominator by $i$:

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

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

Since $i^2 = -1$, we have $-i^2 = -(-1) = 1$.

$= \frac{i}{1}$

$= i$

To express this in the form $a + bi$, we write it as $0 + 1i$ or simply $0 + i$.

Thus, $i^{-35}$ in the form $a + bi$ is $\mathbf{0 + i}$.

We need to express the given complex number in the form $a + bi$.

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

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

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

$= -3 \times (-1)$

$= 3$

To express this in the form $a + bi$, we write the imaginary part as $0i$.

$3 = 3 + 0i$

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

To Express:

$i^9 + i^{19}$ in the form $a + bi$.

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Solution:

We need to express $i^9 + i^{19}$ in the form $a + bi$.

We use the property that powers of $i$ repeat in a cycle of 4: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$. For any integer $n$, $i^n = i^{n \pmod 4}$ if $n \pmod 4 \neq 0$, and $i^n = 1$ if $n$ is a multiple of 4.

For $i^9$, we divide the exponent 9 by 4:

$9 = 4 \times 2 + 1$

The remainder is 1, so $i^9 = i^1 = i$.

For $i^{19}$, we divide the exponent 19 by 4:

$19 = 4 \times 4 + 3$

The remainder is 3, so $i^{19} = i^3 = -i$.

Now, substitute these simplified values back into the expression:

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

$= i - i$

$= 0$

To express 0 in the form $a + bi$, we write it as:

$0 = 0 + 0i$

Thus, $i^9 + i^{19}$ in the form $a + bi$ is $\mathbf{0 + 0i}$.

To Express:

$i^{-39}$ in the form $a + bi$.

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Solution:

We need to express $i^{-39}$ in the form $a + bi$.

We can write $i^{-39}$ as $\frac{1}{i^{39}}$.

To simplify $i^{39}$, we use the property that powers of $i$ repeat in a cycle of 4. We divide the exponent by 4 and look at the remainder.

Divide 39 by 4:

$39 \div 4$ gives a quotient of 9 and a remainder of 3.

So, $39 = 4 \times 9 + 3$.

Therefore, $i^{39} = i^{4 \times 9 + 3} = (i^4)^9 \times i^3$.

Since $i^4 = 1$, we have $(i^4)^9 = 1^9 = 1$.

So, $i^{39} = 1 \times i^3 = i^3$.

We know that $i^3 = -i$.

Substituting this back into the original expression:

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

To express this in the form $a + bi$, we multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of $-i$ is $i$.

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

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

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

Since $i^2 = -1$, we have $-i^2 = -(-1) = 1$.

$= \frac{i}{1}$

$= i$

To express $i$ in the form $a + bi$, where $a$ and $b$ are real numbers, we write it as:

$i = 0 + 1i$

Thus, $i^{-39}$ in the form $a + bi$ is $\mathbf{0 + i}$.

To Express:

$3(7 + i7) + i (7 + i7)$ in the form $a + bi$.

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Solution:

We are given the expression $3(7 + i7) + i (7 + i7)$.

First, distribute the numbers outside the parentheses:

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

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

Substitute $i^2 = -1$ into the second expression:

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

Now, add the results of the two distributed terms:

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

Group the real parts and the imaginary parts:

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

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

$= 14 + 28i$

The expression is in the form $a + bi$, where $a = 14$ and $b = 28$.

Thus, $3(7 + i7) + i (7 + i7)$ in the form $a + bi$ is $\mathbf{14 + 28i}$.

To Express:

$(1 - i) - ( –1 + i6)$ in the form $a + bi$.

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Solution:

We are given the expression $(1 - i) - ( –1 + i6)$.

To subtract complex numbers, we subtract their corresponding real and imaginary parts:

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

Simplify the real part:

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

Simplify the imaginary part:

$-i - i6 = (-1 - 6)i = -7i$

Combine the real and imaginary parts:

$(1 - i) - ( –1 + i6) = 2 + (-7i)$

$= 2 - 7i$

The expression is in the form $a + bi$, where $a = 2$ and $b = -7$.

Thus, $(1 - i) - ( –1 + i6)$ in the form $a + bi$ is $\mathbf{2 - 7i}$.

To Express:

$\left( \frac{1}{5} +i\frac{2}{5}\right) - \left( 4+i\frac{5}{2}\right)$ in the form $a + bi$.

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Solution:

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

To subtract complex numbers, we subtract their real parts and their imaginary parts separately.

Let $z_1 = \frac{1}{5} + i\frac{2}{5}$ and $z_2 = 4 + i\frac{5}{2}$.

$z_1 - z_2 = \left(\frac{1}{5} - 4\right) + i\left(\frac{2}{5} - \frac{5}{2}\right)$

First, calculate the difference of the real parts:

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

Next, calculate the difference of the imaginary parts:

$\frac{2}{5} - \frac{5}{2}$

The least common multiple of 5 and 2 is 10. Convert the fractions to have a denominator of 10.

$\frac{2}{5} = \frac{2 \times 2}{5 \times 2} = \frac{4}{10}$

$\frac{5}{2} = \frac{5 \times 5}{2 \times 5} = \frac{25}{10}$

So, $\frac{2}{5} - \frac{5}{2} = \frac{4}{10} - \frac{25}{10} = \frac{4 - 25}{10} = -\frac{21}{10}$.

Now, combine the real and imaginary parts:

$\left(\frac{1}{5} - 4\right) + i\left(\frac{2}{5} - \frac{5}{2}\right) = -\frac{19}{5} + i\left(-\frac{21}{10}\right)$

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

The expression is in the form $a + bi$, where $a = -\frac{19}{5}$ and $b = -\frac{21}{10}$.

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

To Express:

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

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Solution:

We are given the expression $\left[ \left( \frac{1}{3} +i\frac{7}{3}\right)+\left( 4+i\frac{1}{3}\right) \right]-\left( -\frac{4}{3}+i \right)$.

First, we simplify the expression inside the square brackets by adding the two complex numbers:

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

Calculate the real part: $\frac{1}{3} + 4 = \frac{1}{3} + \frac{12}{3} = \frac{1+12}{3} = \frac{13}{3}$

Calculate the imaginary part: $\frac{7}{3} + \frac{1}{3} = \frac{7+1}{3} = \frac{8}{3}$

So, the expression inside the brackets simplifies to $\frac{13}{3} + \frac{8}{3}i$.

Now, subtract the complex number $\left(-\frac{4}{3}+i\right)$ from this result:

$\left(\frac{13}{3} + \frac{8}{3}i\right) - \left(-\frac{4}{3}+1i\right)$

Subtract the real parts: $\frac{13}{3} - \left(-\frac{4}{3}\right) = \frac{13}{3} + \frac{4}{3} = \frac{13+4}{3} = \frac{17}{3}$

Subtract the imaginary parts: $\frac{8}{3}i - 1i = \left(\frac{8}{3} - 1\right)i = \left(\frac{8}{3} - \frac{3}{3}\right)i = \frac{8-3}{3}i = \frac{5}{3}i$

Combining the real and imaginary parts, we get:

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

The expression is in the form $a + bi$, where $a = \frac{17}{3}$ and $b = \frac{5}{3}$.

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

To Express:

$(1 - i)^4$ in the form $a + bi$.

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Solution:

We need to express $(1 - i)^4$ in the form $a + bi$.

We can calculate this by first finding $(1 - i)^2$ and then squaring the result.

First, 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 have:

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

$= 1 - 2i - 1$

$= -2i$

Now, we need to calculate $(1 - i)^4 = ((1 - i)^2)^2$. Substitute the value of $(1 - i)^2$:

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

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

$= 4 \times i^2$

Again, substitute $i^2 = -1$:

$= 4 \times (-1)$

$= -4$

To express the result $-4$ in the form $a + bi$, we write it as:

$-4 = -4 + 0i$

The expression is in the form $a + bi$, where $a = -4$ and $b = 0$.

Thus, $(1 - i)^4$ in the form $a + bi$ is $\mathbf{-4 + 0i}$.

To Express:

$\left( \frac{1}{3}+3i \right)^{3}$ in the form $a + bi$.

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Solution:

We need to express $\left( \frac{1}{3}+3i \right)^{3}$ in the form $a + bi$.

We use the binomial expansion formula $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$.

Here, $x = \frac{1}{3}$ and $y = 3i$.

Substituting these values into the formula:

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

Calculate each term:

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

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

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

Since $i^2 = -1$, this term is $9(-1) = -9$.

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

Since $i^3 = i^2 \times i = (-1) \times i = -i$, this term is $27(-i) = -27i$.

Now, combine these terms:

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

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

Group the real and imaginary parts:

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

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

$= \left(\frac{1}{27} - \frac{243}{27}\right) + (-26)i$

$= \frac{1 - 243}{27} - 26i$

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

The expression is in the form $a + bi$, where $a = -\frac{242}{27}$ and $b = -26$.

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

To Express:

$\left( -2-\frac{1}{3}i\right)^{3}$ in the form $a + bi$.

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Solution:

We need to express $\left( -2-\frac{1}{3}i\right)^{3}$ in the form $a + bi$.

We use the binomial expansion formula $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$.

Here, $x = -2$ and $y = -\frac{1}{3}i$.

Substituting these values into the formula:

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

Calculate each term:

$(-2)^3 = -8$

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

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

Since $i^2 = -1$, we have: $-6\left(\frac{1}{9}(-1)\right) = -6\left(-\frac{1}{9}\right) = \frac{6}{9} = \frac{2}{3}$

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

Since $i^3 = i^2 \times i = (-1) \times i = -i$, we have: $-\frac{1}{27} (-i) = \frac{1}{27}i$

Now, combine these terms:

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

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

Group the real and imaginary parts:

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

Combine the real part: $-8 + \frac{2}{3} = -\frac{8 \times 3}{3} + \frac{2}{3} = \frac{-24 + 2}{3} = -\frac{22}{3}$

Combine the imaginary part: $-4 + \frac{1}{27} = -\frac{4 \times 27}{27} + \frac{1}{27} = \frac{-108 + 1}{27} = -\frac{107}{27}$

So, the expression is:

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

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

The expression is in the form $a + bi$, where $a = -\frac{22}{3}$ and $b = -\frac{107}{27}$.

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

Given:

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

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To Find:

The multiplicative inverse of $4 - 3i$.

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Solution:

Let the complex number be $z = 4 - 3i$.

The multiplicative inverse of a non-zero complex number $z = a + bi$ is given by the formula:

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

where $\overline{z}$ is the conjugate of $z$ and $|z|^2$ is the square of the modulus of $z$.

For $z = 4 - 3i$, the real part is $a = 4$ and the imaginary part is $b = -3$.

The conjugate of $z$ is $\overline{z} = a - bi = 4 - (-3)i = 4 + 3i$.

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

Now, substitute these values into the formula for the multiplicative inverse:

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

To express this in the form $a + bi$, we separate the real and imaginary parts:

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

Thus, the multiplicative inverse of $4 - 3i$ is $\mathbf{\frac{4}{25} + \frac{3}{25}i}$.

Given:

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

---

To Find:

The multiplicative inverse of $\sqrt{5}+3i$.

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Solution:

Let the complex number be $z = \sqrt{5}+3i$.

The multiplicative inverse of a non-zero complex number $z = a + bi$ is given by the formula:

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

where $\overline{z}$ is the conjugate of $z$ and $|z|^2$ is the square of the modulus of $z$.

For $z = \sqrt{5}+3i$, the real part is $a = \sqrt{5}$ and the imaginary part is $b = 3$.

The conjugate of $z$ is $\overline{z} = a - bi = \sqrt{5} - 3i$.

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

Now, substitute these values into the formula for the multiplicative inverse:

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

To express this in the form $a + bi$, we separate the real and imaginary parts:

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

Thus, the multiplicative inverse of $\sqrt{5}+3i$ is $\mathbf{\frac{\sqrt{5}}{14} - \frac{3}{14}i}$.

Given:

The complex number $z = -i$.

---

To Find:

The multiplicative inverse of $-i$.

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Solution:

Let the complex number be $z = -i$.

We can write $z$ in the form $a + bi$: $z = 0 - 1i$.

The multiplicative inverse of a non-zero complex number $z = a + bi$ is given by the formula:

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

where $\overline{z}$ is the conjugate of $z$ and $|z|^2$ is the square of the modulus of $z$.

For $z = 0 - 1i$, the real part is $a = 0$ and the imaginary part is $b = -1$.

The conjugate of $z$ is $\overline{z} = a - bi = 0 - (-1)i = 0 + 1i = i$.

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

Now, substitute these values into the formula for the multiplicative inverse:

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

$z^{-1} = i$

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

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

Thus, the multiplicative inverse of $-i$ is $\mathbf{0 + i}$ (or simply $\mathbf{i}$).

Given:

The expression is $\frac{(3\;+\;i\sqrt{5})(3\;-\;i\sqrt{5})}{(\sqrt{3}\;+\;\sqrt{2}i)\;-\;(\sqrt{3}\;-\;i\sqrt{2})}$.

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To Express:

The given expression in the form $a + bi$, where $a$ and $b$ are real numbers and $i = \sqrt{-1}$.

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Solution:

Let the given expression be E.

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

---

First, we simplify the numerator. It is in the form $(a+b)(a-b)$, where $a=3$ and $b=i\sqrt{5}$.

Using the algebraic identity $(a+b)(a-b) = a^2 - b^2$:

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

Numerator $= 9 - (i^2 \cdot (\sqrt{5})^2)$

We know that $i^2 = -1$ and $(\sqrt{5})^2 = 5$.

Numerator $= 9 - (-1 \cdot 5)$

Numerator $= 9 - (-5)$

Numerator $= 9 + 5 = 14$

---

Next, we simplify the denominator. We remove the parentheses and combine like terms.

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

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

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

Combine the real parts ($\sqrt{3} - \sqrt{3}$) and the imaginary parts ($\sqrt{2}i + \sqrt{2}i$):

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

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

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Now, we substitute the simplified numerator and denominator back into the expression E:

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

We can cancel the common factor of 2 in the numerator and denominator:

$E = \frac{\cancel{14}^{7}}{\cancel{2}_{1}\sqrt{2}i} = \frac{7}{\sqrt{2}i}$

To express this complex number in the standard form $a + bi$, we need to remove the imaginary unit $i$ from the denominator. We do this by multiplying the numerator and denominator by $i$ (or by the conjugate of the denominator, which is $-2\sqrt{2}i$; multiplying by $i$ is simpler here).

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

$E = \frac{7i}{\sqrt{2} \cdot i^2}$

Substitute $i^2 = -1$ in the denominator:

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

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

To rationalize the denominator (remove the square root from the denominator), we multiply the numerator and denominator by $\sqrt{2}$:

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

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

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

---

The expression is now in the form of a purely imaginary number. The standard form of a complex number is $a + bi$, where $a$ is the real part and $b$ is the imaginary part.

In our result, $-\frac{7\sqrt{2}}{2}i$, the real part is 0.

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

Thus, the expression in the form $a + bi$ is:

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

Given:

The complex expression $z = \frac{(3 \;-\; 2i) (2 \;+\; 3i)} {(1 \;+\; 2i) (2 \;-\; i)}$.

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To Find:

The conjugate of the given expression in the form $a + bi$.

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Solution:

Let the given expression be $z$. We want to find the conjugate of $z$, denoted by $\overline{z}$.

Using the properties of complex conjugates, $\overline{\left(\frac{w_1 w_2}{w_3 w_4}\right)} = \frac{\overline{w_1} \overline{w_2}}{\overline{w_3} \overline{w_4}}$.

So, the conjugate of the given expression is:

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

The conjugates of the individual complex numbers are:

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

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

$\overline{1 + 2i} = 1 - 2i$

$\overline{2 - i} = 2 + i$

Substituting these conjugates back into the expression for $\overline{z}$:

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

Now, we simplify the numerator and the denominator.

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

Multiply the terms:

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

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

Since $i^2 = -1$, this becomes:

$= 6 - 5i - 6(-1)$

$= 6 - 5i + 6$

$= 12 - 5i$

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

Multiply the terms:

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

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

Since $i^2 = -1$, this becomes:

$= 2 - 3i - 2(-1)$

$= 2 - 3i + 2$

$= 4 - 3i$

So, the conjugate expression is $\overline{z} = \frac{12 - 5i}{4 - 3i}$.

To express this in the form $a + bi$, we multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of $4 - 3i$ is $4 + 3i$.

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

Multiply the numerators:

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

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

$= 48 + 16i - 15(-1)$

$= 48 + 16i + 15$

$= 63 + 16i$

Multiply the denominators (using $(a-bi)(a+bi) = a^2+b^2$):

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

$= 16 + 9$

$= 25$

So, $\overline{z} = \frac{63 + 16i}{25}$.

To write this in the form $a + bi$, we separate the real and imaginary parts:

$\overline{z} = \frac{63}{25} + \frac{16}{25}i$

Thus, the conjugate of the given expression in the form $a + bi$ is $\mathbf{\frac{63}{25} + \frac{16}{25}i}$.

Given:

The equation $x + iy = \frac{a + ib}{a - ib}$, where $x, y, a, b$ are real numbers and $a - ib \neq 0$.

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To Prove:

$x^2 + y^2 = 1$

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Proof:

We are given the complex equation:

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

Take the modulus of both sides of the equation. The modulus of a complex number $z = c + id$ is $|z| = \sqrt{c^2 + d^2}$. The modulus of a quotient of complex numbers is the quotient of their moduli, i.e., $\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}$.

Calculate the modulus of the left side:

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

Calculate the modulus of the numerator on the right side:

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

Calculate the modulus of the denominator on the right side:

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

Substitute these moduli back into equation (i):

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

Assuming $a$ and $b$ are such that $a - ib \neq 0$, the denominator modulus $\sqrt{a^2 + b^2}$ is non-zero. Since the numerator modulus is also $\sqrt{a^2 + b^2}$, and it is non-zero, the fraction simplifies to 1.

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

Square both sides of the equation:

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

$x^2 + y^2 = 1$

Thus, we have proved that $x^2 + y^2 = 1$.

To Evaluate:

$\left[ i^{18} + \left( \frac{1}{i} \right)^{25} \right]^3$ in the form $a + bi$.

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Solution:

We need to evaluate the expression $\left[ i^{18} + \left( \frac{1}{i} \right)^{25} \right]^3$.

First, let's simplify the terms inside the bracket.

For $i^{18}$: Divide the exponent 18 by 4. $18 = 4 \times 4 + 2$. The remainder is 2. So, $i^{18} = i^2 = -1$.

For $\left( \frac{1}{i} \right)^{25}$: First, simplify $\frac{1}{i}$.

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

Now, evaluate $(-i)^{25}$.

$(-i)^{25} = (-1)^{25} \times i^{25}$

Since 25 is an odd number, $(-1)^{25} = -1$.

For $i^{25}$: Divide the exponent 25 by 4. $25 = 4 \times 6 + 1$. The remainder is 1. So, $i^{25} = i^1 = i$.

Therefore, $(-i)^{25} = (-1) \times i = -i$.

Now, substitute these simplified values back into the expression inside the bracket:

$i^{18} + \left( \frac{1}{i} \right)^{25} = (-1) + (-i) = -1 - i$

Finally, we need to evaluate the cube of this result:

$(-1 - i)^3$

We can factor out -1:

$(-1 - i)^3 = (-1(1 + i))^3 = (-1)^3 (1 + i)^3 = -1 (1 + i)^3$

Now, we expand $(1 + i)^3$ using the binomial formula $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$:

$(1 + i)^3 = (1)^3 + 3(1)^2(i) + 3(1)(i)^2 + (i)^3$

$= 1 + 3i + 3i^2 + i^3$

Substitute $i^2 = -1$ and $i^3 = -i$:

$= 1 + 3i + 3(-1) + (-i)$

$= 1 + 3i - 3 - i$

Combine the real and imaginary parts:

$= (1 - 3) + (3i - i)$

$= -2 + 2i$

Now, multiply this result by -1:

$(-1)(1 + i)^3 = -1(-2 + 2i)$

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

$= 2 - 2i$

The expression is in the form $a + bi$, where $a = 2$ and $b = -2$.

Thus, the evaluation of $\left[ i^{18} + \left( \frac{1}{i} \right)^{25} \right]^3$ in the form $a + bi$ is $\mathbf{2 - 2i}$.

Given:

Two complex numbers $z_1$ and $z_2$.

---

To Prove:

$Re (z_1 z_2 ) = Re z_1 Re z_2 – Imz_1 Imz_2$.

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Proof:

Let the two complex numbers be $z_1 = a + bi$ and $z_2 = c + di$, where $a, b, c, d$ are real numbers.

By definition, we have:

$Re(z_1) = a$

$Im(z_1) = b$

$Re(z_2) = c$

$Im(z_2) = d$

Now, let's find the product $z_1 z_2$:

$z_1 z_2 = (a + bi)(c + di)$

$z_1 z_2 = ac + adi + bci + bdi^2$

Since $i^2 = -1$, we substitute this into the expression:

$z_1 z_2 = ac + adi + bci + bd(-1)$

$z_1 z_2 = ac + adi + bci - bd$

Group the real and imaginary terms together:

$z_1 z_2 = (ac - bd) + (ad + bc)i$

The real part of the product $z_1 z_2$ is the term without $i$:

$Re(z_1 z_2) = ac - bd$

This is the Left-Hand Side (LHS) of the identity we need to prove.

Now, let's consider the Right-Hand Side (RHS) of the identity: $Re z_1 Re z_2 – Imz_1 Imz_2$.

Substitute the real and imaginary parts of $z_1$ and $z_2$ into the RHS expression:

$Re z_1 Re z_2 – Imz_1 Imz_2 = (a)(c) - (b)(d)$

$= ac - bd$

So, the RHS is $ac - bd$.

Comparing the LHS and the RHS, we see that:

$LHS = ac - bd$

$RHS = ac - bd$

Therefore, $LHS = RHS$.

$Re (z_1 z_2 ) = Re z_1 Re z_2 – Imz_1 Imz_2$.

Hence, the identity is proved.

Given:

The expression $\left( \frac{1}{1 - 4i} -\frac{2}{1 + i}\right) \left( \frac{3 - 4i}{5 + i} \right)$.

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To Reduce:

The given expression to the standard form $a + bi$.

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Solution:

We need to reduce the expression $\left( \frac{1}{1 - 4i} -\frac{2}{1 + i}\right) \left( \frac{3 - 4i}{5 + i} \right)$ to the standard form $a + bi$.

Let's first simplify the expression within the first parenthesis: $\left( \frac{1}{1 - 4i} - \frac{2}{1 + i}\right)$.

To subtract the complex fractions, we find a common denominator, which is $(1 - 4i)(1 + i)$.

Denominator $=(1 - 4i)(1 + i) = 1(1) + 1(i) - 4i(1) - 4i(i) = 1 + i - 4i - 4i^2 = 1 - 3i - 4(-1) = 1 - 3i + 4 = 5 - 3i$.

Now, rewrite the fractions with the common denominator:

$\frac{1}{1 - 4i} = \frac{1 \times (1 + i)}{(1 - 4i)(1 + i)} = \frac{1 + i}{5 - 3i}$

$\frac{2}{1 + i} = \frac{2 \times (1 - 4i)}{(1 + i)(1 - 4i)} = \frac{2 - 8i}{5 - 3i}$

Subtract the fractions:

$\frac{1 + i}{5 - 3i} - \frac{2 - 8i}{5 - 3i} = \frac{(1 + i) - (2 - 8i)}{5 - 3i} = \frac{1 + i - 2 + 8i}{5 - 3i} = \frac{(1 - 2) + (1 + 8)i}{5 - 3i} = \frac{-1 + 9i}{5 - 3i}$

To express $\frac{-1 + 9i}{5 - 3i}$ in the form $a + bi$, we multiply the numerator and the denominator by the conjugate of the denominator, which is $5 + 3i$.

$\frac{-1 + 9i}{5 - 3i} \times \frac{5 + 3i}{5 + 3i} = \frac{(-1 + 9i)(5 + 3i)}{(5 - 3i)(5 + 3i)}$

Numerator: $(-1 + 9i)(5 + 3i) = -1(5) + (-1)(3i) + 9i(5) + 9i(3i) = -5 - 3i + 45i + 27i^2 = -5 + 42i + 27(-1) = -5 + 42i - 27 = -32 + 42i$.

Denominator: $(5 - 3i)(5 + 3i) = 5^2 - (3i)^2 = 25 - 9i^2 = 25 - 9(-1) = 25 + 9 = 34$.

So, the first parenthesis simplifies to $\frac{-32 + 42i}{34} = \frac{-32}{34} + \frac{42}{34}i = -\frac{16}{17} + \frac{21}{17}i$.

Next, let's simplify the second parenthesis: $\left( \frac{3 - 4i}{5 + i} \right)$.

Multiply the numerator and the denominator by the conjugate of the denominator, which is $5 - i$.

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

Numerator: $(3 - 4i)(5 - i) = 3(5) + 3(-i) - 4i(5) + (-4i)(-i) = 15 - 3i - 20i + 4i^2 = 15 - 23i + 4(-1) = 15 - 23i - 4 = 11 - 23i$.

Denominator: $(5 + i)(5 - i) = 5^2 - i^2 = 25 - (-1) = 25 + 1 = 26$.

So, the second parenthesis simplifies to $\frac{11 - 23i}{26}$.

Finally, we multiply the simplified results of the two parentheses:

$\left(-\frac{16}{17} + \frac{21}{17}i\right) \left(\frac{11}{26} - \frac{23}{26}i\right)$

Multiply the complex numbers $(a+bi)(c+di) = (ac-bd) + (ad+bc)i$. Here, $a = -\frac{16}{17}$, $b = \frac{21}{17}$, $c = \frac{11}{26}$, $d = -\frac{23}{26}$.

Real part $= ac - bd = \left(-\frac{16}{17}\right) \left(\frac{11}{26}\right) - \left(\frac{21}{17}\right) \left(-\frac{23}{26}\right) = -\frac{176}{442} - \left(-\frac{483}{442}\right) = -\frac{176}{442} + \frac{483}{442} = \frac{-176 + 483}{442} = \frac{307}{442}$.

Imaginary part $= ad + bc = \left(-\frac{16}{17}\right) \left(-\frac{23}{26}\right) + \left(\frac{21}{17}\right) \left(\frac{11}{26}\right) = \frac{368}{442} + \frac{231}{442} = \frac{368 + 231}{442} = \frac{599}{442}$.

The product is $\frac{307}{442} + \frac{599}{442}i$.

The expression is in the standard form $a + bi$, where $a = \frac{307}{442}$ and $b = \frac{599}{442}$.

Thus, the given expression reduced to the standard form is $\mathbf{\frac{307}{442} + \frac{599}{442}i}$.

Given:

The equation $x - iy = \sqrt{\frac{a - ib}{c - id}}$, where $x, y, a, b, c, d$ are real numbers.

---

To Prove:

$(x^2 + y^2)^2 = \frac{a^2 + b^2}{c^2 + d^2}$.

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Proof:

We are given the equation:

$x - iy = \sqrt{\frac{a - ib}{c - id}}$

Take the modulus of both sides of the equation. The modulus of a complex number $z = p + q i$ is $|z| = \sqrt{p^2 + q^2}$. Also, the modulus of a square root is $|\sqrt{z}| = \sqrt{|z|}$, and the modulus of a quotient is $\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}$.

Applying the modulus on both sides:

Using the property $|\sqrt{z}| = \sqrt{|z|}$ on the right side:

$\left|\sqrt{\frac{a - ib}{c - id}}\right| = \sqrt{\left|\frac{a - ib}{c - id}\right|}$

Using the property $\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}$ on the right side:

$\sqrt{\left|\frac{a - ib}{c - id}\right|} = \sqrt{\frac{|a - ib|}{|c - id|}}$

Now, calculate the modulus of each complex number:

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

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

$|c - id| = \sqrt{c^2 + (-d)^2} = \sqrt{c^2 + d^2}$

Substitute these modulus values back into the equation derived from (i):

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

$\sqrt{x^2 + y^2} = \sqrt{\left(\frac{a^2 + b^2}{c^2 + d^2}\right)^{1/2}}$

Using the property $\sqrt[n]{\sqrt[m]{k}} = \sqrt[nm]{k}$: $\sqrt{(...)^{1/2}} = ((...)^{1/2})^{1/2} = (...)^{1/4}$.

$\sqrt{x^2 + y^2} = \left(\frac{a^2 + b^2}{c^2 + d^2}\right)^{1/4}$

Square both sides of the equation:

$(\sqrt{x^2 + y^2})^2 = \left(\left(\frac{a^2 + b^2}{c^2 + d^2}\right)^{1/4}\right)^2$

$x^2 + y^2 = \left(\frac{a^2 + b^2}{c^2 + d^2}\right)^{2/4}$

$x^2 + y^2 = \left(\frac{a^2 + b^2}{c^2 + d^2}\right)^{1/2}$

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

Square both sides again to get the expression $(x^2+y^2)^2$:

$(x^2 + y^2)^2 = \left(\sqrt{\frac{a^2 + b^2}{c^2 + d^2}}\right)^2$

$(x^2 + y^2)^2 = \frac{a^2 + b^2}{c^2 + d^2}$

This is the required identity.

Hence, proved.

Given:

The complex numbers $z_1 = 2 - i$ and $z_2 = 1 + i$.

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To Find:

The value of $\left| \frac{z_{1}\;+\;z_{2}\;+\;1}{z_{1}\;-\;z_{2}\;+\;1} \right|$.

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Solution:

We need to find the modulus of the complex expression $\frac{z_{1}\;+\;z_{2}\;+\;1}{z_{1}\;-\;z_{2}\;+\;1}$.

First, calculate the numerator $z_1 + z_2 + 1$:

$z_1 + z_2 + 1 = (2 - i) + (1 + i) + 1$

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

$= 4 + 0i = 4$

Next, calculate the denominator $z_1 - z_2 + 1$:

$z_1 - z_2 + 1 = (2 - i) - (1 + i) + 1$

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

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

Now, substitute these values into the expression:

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

To simplify this complex fraction, multiply the numerator and denominator by the conjugate of the denominator. The conjugate of $2 - 2i$ is $2 + 2i$.

$\frac{4}{2 - 2i} = \frac{4}{2 - 2i} \times \frac{2 + 2i}{2 + 2i}$

Numerator: $4(2 + 2i) = 8 + 8i$

Denominator: $(2 - 2i)(2 + 2i) = (2)^2 + (-2)^2 = 4 + 4 = 8$

So, the complex number is $\frac{8 + 8i}{8} = \frac{8}{8} + \frac{8i}{8} = 1 + i$.

Now, we need to find the modulus of $1 + i$.

The modulus of a complex number $a + bi$ is $\sqrt{a^2 + b^2}$.

For $1 + i$, the real part is $a = 1$ and the imaginary part is $b = 1$.

$\left| \frac{z_{1}\;+\;z_{2}\;+\;1}{z_{1}\;-\;z_{2}\;+\;1} \right| = |1 + i| = \sqrt{(1)^2 + (1)^2}$

$= \sqrt{1 + 1}$

$= \sqrt{2}$

Thus, the value of $\left| \frac{z_{1}\;+\;z_{2}\;+\;1}{z_{1}\;-\;z_{2}\;+\;1} \right|$ is $\mathbf{\sqrt{2}}$.

Given:

The complex number equation $a + ib = \frac{(x\;+\;i)^{2}}{2x^{2}\;+\;1}$, where $a, b, x$ are real numbers and $i = \sqrt{-1}$.

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To Prove:

$a^2 + b^2 = \frac{(x^2\;+\;1)^{2}}{(2x^{2}\;+\;1)^2}$

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Proof:

We are given the equation:

First, let's expand the numerator $(x+i)^2$ using the formula $(p+q)^2 = p^2 + 2pq + q^2$ where $p=x$ and $q=i$:

Since $i^2 = -1$, we have:

Rearranging the terms to separate the real and imaginary parts:

Now substitute this back into the given equation:

Since the denominator $2x^2+1$ is a real number (assuming x is real, $2x^2 \ge 0$, so $2x^2+1 \ge 1 \neq 0$), we can separate the real and imaginary parts of the right-hand side:

By comparing the real and imaginary parts of both sides of the equation, we get the expressions for $a$ and $b$:

We need to find $a^2 + b^2$. Square the expressions for $a$ and $b$:

Now add $a^2$ and $b^2$:

Since the denominators are the same, we can add the numerators:

Expand the term $(x^2 - 1)^2$ in the numerator using $(p-q)^2 = p^2 - 2pq + q^2$, where $p=x^2$ and $q=1$:

Substitute this back into the numerator of $a^2 + b^2$:

Combine the like terms in the numerator:

Notice that the numerator $x^4 + 2x^2 + 1$ is a perfect square trinomial, $(x^2)^2 + 2(x^2)(1) + 1^2$, which is equal to $(x^2 + 1)^2$.

Substitute this back into the expression for $a^2 + b^2$:

This is the expression we were required to prove.

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Alternate Approach (Using Modulus):

We are given $a + ib = \frac{(x\;+\;i)^{2}}{2x^{2}\;+\;1}$.

We know that for a complex number $z = a + ib$, $|z| = \sqrt{a^2 + b^2}$, and $|z|^2 = a^2 + b^2$.

Also, for complex numbers $z_1$ and $z_2$, $|z_1 z_2| = |z_1| |z_2|$ and $\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}$ (where $z_2 \neq 0$).

Let $z = a + ib$. Then $|z|^2 = a^2 + b^2$.

$|z|^2 = \left|\frac{(x\;+\;i)^{2}}{2x^{2}\;+\;1}\right|^2$

Using the properties of modulus:

Alternatively, using $|z^n| = |z|^n$:

For a complex number $w = c + id$, $|w| = \sqrt{c^2 + d^2}$, so $|w|^2 = c^2 + d^2$.

For $x+i$, the real part is $x$ and the imaginary part is 1. So, $|x+i|^2 = x^2 + 1^2 = x^2 + 1$.

For $2x^2+1$, this is a real number. The modulus of a real number is its absolute value. Since $2x^2+1 \ge 1$, $|2x^2+1| = 2x^2+1$. Thus, $|2x^2+1|^2 = (2x^2+1)^2$.

Substitute these values back into the equation for $|z|^2$:

This matches the expression we needed to prove.

Hence, proved.

Given:

Complex numbers $z_1 = 2 - i$ and $z_2 = -2 + i$.

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To Find:

(i) The real part of $\frac{z_{1}z_{2}}{\overline{z_{1}}}$.

(ii) The imaginary part of $\frac{1}{z_{1}\overline{z_{1}}}$.

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Solution:

First, we find the conjugate of $z_1$, denoted by $\overline{z_{1}}$.

If $z_1 = 2 - i$, then $\overline{z_{1}} = 2 + i$.

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(i) Find $Re \left(\frac{z_{1}z_{2}}{\overline{z_{1}}} \right)$:

We first calculate the product $z_1 z_2$:

$z_1 z_2 = (2 - i)(-2 + i)$

Using the distributive property (FOIL method):

$z_1 z_2 = (2)(-2) + (2)(i) + (-i)(-2) + (-i)(i)$

$z_1 z_2 = -4 + 2i + 2i - i^2$

Since $i^2 = -1$:

$z_1 z_2 = -4 + 4i - (-1)$

$z_1 z_2 = -4 + 4i + 1$

$z_1 z_2 = -3 + 4i$

Now, we compute 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 this complex fraction, we multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of $2+i$ is $2-i$.

$\frac{-3 + 4i}{2 + i} \times \frac{2 - i}{2 - i} = \frac{(-3 + 4i)(2 - i)}{(2 + i)(2 - i)}$

Calculate the numerator:

$(-3 + 4i)(2 - i) = (-3)(2) + (-3)(-i) + (4i)(2) + (4i)(-i)$

$= -6 + 3i + 8i - 4i^2$

$= -6 + 11i - 4(-1)$

$= -6 + 11i + 4$

$= -2 + 11i$

Calculate the denominator:

$(2 + i)(2 - i) = 2^2 - i^2 = 4 - (-1) = 4 + 1 = 5$

So, the fraction is:

$\frac{-2 + 11i}{5} = -\frac{2}{5} + \frac{11}{5}i$

This is in the form $a+bi$, where the real part is $a = -\frac{2}{5}$ and the imaginary part is $b = \frac{11}{5}$.

The real part of $\frac{z_{1}z_{2}}{\overline{z_{1}}}$ is the real part of $-\frac{2}{5} + \frac{11}{5}i$.

Thus, $Re \left(\frac{z_{1}z_{2}}{\overline{z_{1}}} \right) = -\frac{2}{5}$.

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(ii) Find $Im \left(\frac{1}{z_{1}\overline{z_{1}}} \right)$:

We first calculate the product $z_{1}\overline{z_{1}}$.

$z_1 \overline{z_{1}} = (2 - i)(2 + i)$

Using the identity $(a-b)(a+b) = a^2 - b^2$:

$z_1 \overline{z_{1}} = 2^2 - i^2$

$z_1 \overline{z_{1}} = 4 - (-1)$

$z_1 \overline{z_{1}} = 4 + 1 = 5$

Now, we compute the fraction $\frac{1}{z_{1}\overline{z_{1}}}$:

$\frac{1}{z_{1}\overline{z_{1}}} = \frac{1}{5}$

This is a real number. To express this in the form $a+bi$, we can write it as $\frac{1}{5} + 0i$.

Here, the real part is $a = \frac{1}{5}$ and the imaginary part is $b = 0$.

The imaginary part of $\frac{1}{z_{1}\overline{z_{1}}}$ is the imaginary part of $\frac{1}{5} + 0i$.

Thus, $Im \left(\frac{1}{z_{1}\overline{z_{1}}} \right) = 0$.

Given:

The complex number $(x - iy) (3 + 5i)$ is the conjugate of $-6 - 24i$, where $x$ and $y$ are real numbers.

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To Find:

The real numbers $x$ and $y$.

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Solution:

We are given that $(x - iy) (3 + 5i)$ is the conjugate of $-6 - 24i$.

First, let's find the conjugate of $-6 - 24i$. The conjugate of a complex number $a + bi$ is $a - bi$.

The conjugate of $-6 - 24i$ is $-6 - (-24i) = -6 + 24i$.

Now, we set the given expression equal to this conjugate:

$(x - iy) (3 + 5i) = -6 + 24i$

Next, we expand the left side of the equation by multiplying the two complex numbers:

$(x - iy)(3 + 5i) = x(3) + x(5i) - iy(3) - iy(5i)$

$= 3x + 5xi - 3yi - 5yi^2$

Since $i^2 = -1$, we substitute this into the expression:

$= 3x + 5xi - 3yi - 5y(-1)$

$= 3x + 5xi - 3yi + 5y$

Group the real and imaginary terms together:

$= (3x + 5y) + (5x - 3y)i$

Now, the equation is:

$(3x + 5y) + (5x - 3y)i = -6 + 24i$

For two complex numbers to be equal, their real parts must be equal and their imaginary parts must be equal.

Equating the real parts:

Equating the imaginary parts:

We now have a system of two linear equations. We can solve this system using the elimination method.

Multiply equation (1) by 3 and equation (2) by 5 to eliminate $y$:

$3 \times (3x + 5y) = 3 \times (-6) \Rightarrow 9x + 15y = -18$

$5 \times (5x - 3y) = 5 \times (24) \Rightarrow 25x - 15y = 120$

Add the two resulting equations:

$(9x + 15y) + (25x - 15y) = -18 + 120$

$9x + 25x + 15y - 15y = 102$

$34x = 102$

Solve for $x$:

$x = \frac{102}{34}$

$x = 3$

Now, substitute the value of $x = 3$ into equation (1):

$3x + 5y = -6$

$3(3) + 5y = -6$

$9 + 5y = -6$

$5y = -6 - 9$

$5y = -15$

Solve for $y$:

$y = \frac{-15}{5}$

$y = -3$

Thus, the real numbers are $\mathbf{x = 3}$ and $\mathbf{y = -3}$.

To Find:

The modulus of $\frac{1 \;+\; i}{1 \;-\; i} - \frac{1 \;-\; i}{1 \;+\; i}$.

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Solution:

Let the given expression be $Z$.

$Z = \frac{1 + i}{1 - i} - \frac{1 - i}{1 + i}$

First, simplify the term $\frac{1 + i}{1 - i}$. Multiply the numerator and denominator by the conjugate of the denominator, which is $1 + i$.

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

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

Using the formula $(a+b)^2 = a^2 + 2ab + b^2$ and $i^2 = -1$:

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

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

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

$= i$

Next, simplify the term $\frac{1 - i}{1 + i}$. Multiply the numerator and denominator by the conjugate of the denominator, which is $1 - i$.

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

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

Using the formula $(a-b)^2 = a^2 - 2ab + b^2$ and $i^2 = -1$:

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

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

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

$= -i$

Now, substitute these simplified terms back into the expression for $Z$:

$Z = (i) - (-i)$

$Z = i + i$

$Z = 2i$

The complex number is $Z = 2i$. To find its modulus, we write it in the standard form $a + bi$, which is $0 + 2i$.

The modulus of a complex number $a + bi$ is given by the formula $|a + bi| = \sqrt{a^2 + b^2}$.

For $Z = 0 + 2i$, we have $a = 0$ and $b = 2$.

The modulus is $|Z| = |0 + 2i| = \sqrt{(0)^2 + (2)^2}$

$= \sqrt{0 + 4}$

$= \sqrt{4}$

$= 2$

Thus, the modulus of $\frac{1 + i}{1 - i} - \frac{1 - i}{1 + i}$ is $\mathbf{2}$.

Given:

The equation $(x + iy)^3 = u + iv$, where $x, y, u, v$ are real numbers.

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To Show:

$\frac{u}{x} + \frac{v}{y} = 4 (x^2 - y^2)$, assuming $x \neq 0$ and $y \neq 0$ for the terms $\frac{u}{x}$ and $\frac{v}{y}$ to be defined.

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Proof:

We are given the equation $(x + iy)^3 = u + iv$.

Let's expand the left side of the equation using the binomial formula $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$.

Here, $a = x$ and $b = iy$.

$(x + iy)^3 = x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3$

Simplify the terms involving $i$:

$= x^3 + 3ix^2y + 3x(i^2y^2) + i^3y^3$

Substitute the values $i^2 = -1$ and $i^3 = -i$:

$= x^3 + 3ix^2y + 3x(-1)y^2 + (-i)y^3$

$= x^3 + 3ix^2y - 3xy^2 - iy^3$

Group the real and imaginary parts:

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

We are given that $(x + iy)^3 = u + iv$.

By equating the real and imaginary parts of $(x + iy)^3$ and $u + iv$, we get:

Now, we need to form the terms $\frac{u}{x}$ and $\frac{v}{y}$. Since these terms appear in the statement to be proved, we assume $x \neq 0$ and $y \neq 0$.

Divide $u$ by $x$:

$\frac{u}{x} = \frac{x^3 - 3xy^2}{x}$

Factor out $x$ from the numerator:

$\frac{u}{x} = \frac{x(x^2 - 3y^2)}{x}$

Cancel $x$ (since $x \neq 0$):

Divide $v$ by $y$:

$\frac{v}{y} = \frac{3x^2y - y^3}{y}$

Factor out $y$ from the numerator:

$\frac{v}{y} = \frac{y(3x^2 - y^2)}{y}$

Cancel $y$ (since $y \neq 0$):

Now, add equations (1) and (2) to get the left side of the identity we need to show:

$\frac{u}{x} + \frac{v}{y} = (x^2 - 3y^2) + (3x^2 - y^2)$

Remove the parentheses and group like terms:

$= x^2 - 3y^2 + 3x^2 - y^2$

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

$= 4x^2 - 4y^2$

Factor out 4:

$= 4(x^2 - y^2)$

This is the right side of the identity we were asked to show.

Therefore, we have shown that $\frac{u}{x} + \frac{v}{y} = 4 (x^2 - y^2)$.

Hence, proved.

Given:

$\alpha$ and $\beta$ are different complex numbers ($\alpha \neq \beta$), and $|\beta| = 1$.

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To Find:

The value of $\left| \frac{\beta\;-\;\alpha}{1\;-\;\overline{\alpha}\beta} \right|$.

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Solution:

Let the given complex expression be $Z = \frac{\beta\;-\;\alpha}{1\;-\;\overline{\alpha}\beta}$. We need to find the modulus $|Z|$.

Using the property that the modulus of a quotient is the quotient of the moduli, i.e., $\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}$ (provided $|z_2| \neq 0$), we have:

To evaluate this, we can calculate the square of the modulus of the numerator and the denominator separately, using the property that for any complex number $z$, $|z|^2 = z\overline{z}$.

Consider the square of the modulus of the numerator:

$|\beta - \alpha|^2 = (\beta - \alpha) \overline{(\beta - \alpha)}$

Using the property that the conjugate of a difference is the difference of the conjugates ($\overline{z_1 - z_2} = \overline{z_1} - \overline{z_2}$), we get:

$|\beta - \alpha|^2 = (\beta - \alpha)(\overline{\beta} - \overline{\alpha})$

Expand this product:

$|\beta - \alpha|^2 = \beta\overline{\beta} - \beta\overline{\alpha} - \alpha\overline{\beta} + \alpha\overline{\alpha}$

Using the property that $z\overline{z} = |z|^2$, this becomes:

$|\beta - \alpha|^2 = |\beta|^2 - \beta\overline{\alpha} - \alpha\overline{\beta} + |\alpha|^2$

We are given that $|\beta| = 1$. Substituting $|\beta|^2 = 1^2 = 1$:

Now, consider the square of the modulus of the denominator:

$|1 - \overline{\alpha}\beta|^2 = (1 - \overline{\alpha}\beta) \overline{(1 - \overline{\alpha}\beta)}$

Using conjugate properties: $\overline{1} = 1$, $\overline{z_1 z_2} = \overline{z_1} \overline{z_2}$, and $\overline{\overline{z}} = z$:

The conjugate of the denominator is $\overline{(1 - \overline{\alpha}\beta)} = \overline{1} - \overline{(\overline{\alpha}\beta)} = 1 - \overline{\overline{\alpha}}\overline{\beta} = 1 - \alpha\overline{\beta}$.

So, $|1 - \overline{\alpha}\beta|^2 = (1 - \overline{\alpha}\beta)(1 - \alpha\overline{\beta})$

Expand this product:

$|1 - \overline{\alpha}\beta|^2 = 1 \cdot 1 + 1 \cdot (-\alpha\overline{\beta}) - (\overline{\alpha}\beta) \cdot 1 - (\overline{\alpha}\beta)(-\alpha\overline{\beta})$

$= 1 - \alpha\overline{\beta} - \overline{\alpha}\beta + \overline{\alpha}\beta\alpha\overline{\beta}$

Rearrange the terms in the last part:

$= 1 - \alpha\overline{\beta} - \overline{\alpha}\beta + (\overline{\alpha}\alpha)(\beta\overline{\beta})$

Using the property $z\overline{z} = |z|^2$, this becomes:

$= 1 - \alpha\overline{\beta} - \overline{\alpha}\beta + |\alpha|^2 |\beta|^2$

Using the given condition $|\beta| = 1$, so $|\beta|^2 = 1$:

$= 1 - \alpha\overline{\beta} - \overline{\alpha}\beta + |\alpha|^2 (1)$

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Now, we compare the expressions for the square of the modulus of the numerator and the denominator (from equation (1) and equation (2)).

We see that $|\beta - \alpha|^2 = |1 - \overline{\alpha}\beta|^2$.

Before calculating the ratio, we need to ensure the denominator is not zero.

Since $\alpha \neq \beta$ and $|\beta|=1$, the denominator $1 - \overline{\alpha}\beta$ must be non-zero.

If $1 - \overline{\alpha}\beta = 0$, then $\overline{\alpha}\beta = 1$.

Taking the modulus on both sides:

Using $|\overline{\alpha}||\beta| = 1$ and $|\overline{\alpha}|=|\alpha|$ and $|\beta|=1$:

This implies $|\alpha| = 1$.

If $|\alpha|=1$ and $\overline{\alpha}\beta=1$, then $\beta = \frac{1}{\overline{\alpha}}$.

Since $|\alpha|=1$, we have $\alpha\overline{\alpha} = |\alpha|^2 = 1$, which means $\overline{\alpha} = \frac{1}{\alpha}$.

Substituting this into the expression for $\beta$:

This result $\beta = \alpha$ contradicts the given condition that $\alpha$ and $\beta$ are different complex numbers ($\alpha \neq \beta$).

Therefore, the denominator $1 - \overline{\alpha}\beta$ cannot be 0.

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Since the denominator is non-zero, we can proceed with the division of the squared moduli:

$|Z|^2 = \frac{|\beta - \alpha|^2}{|1 - \overline{\alpha}\beta|^2} = \frac{1 - \beta\overline{\alpha} - \alpha\overline{\beta} + |\alpha|^2}{1 - \alpha\overline{\beta} - \overline{\alpha}\beta + |\alpha|^2}$

The numerator and the denominator are identical, and since the denominator is non-zero, their ratio is 1.

Taking the square root of both sides (the modulus must be non-negative):

Thus, the value of the given expression is 1.

$\left| \frac{\beta\;-\;\alpha}{1\;-\;\overline{\alpha}\beta} \right| = \mathbf{1}$.

Given:

The equation $|1 - i|^x = 2^x$.

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To Find:

The number of non-zero integral solutions for $x$ in the equation $|1 - i|^x = 2^x$.

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Solution:

We are given the equation $|1 - i|^x = 2^x$.

First, we need to find the modulus of the complex number $1 - i$.

The modulus of a complex number $a + bi$ is given by $|a + bi| = \sqrt{a^2 + b^2}$.

For the complex number $1 - i$, the real part is $a = 1$ and the imaginary part is $b = -1$.

So, the modulus of $1 - i$ is:

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

Now, substitute this value back into the given equation:

$(\sqrt{2})^x = 2^x$

Rewrite $\sqrt{2}$ as a power of 2:

$\sqrt{2} = 2^{1/2}$

Substitute this into the equation:

$(2^{1/2})^x = 2^x$

Using the exponent rule $(a^m)^n = a^{mn}$, we simplify the left side:

$2^{(1/2) \times x} = 2^x$

$2^{x/2} = 2^x$

Since the bases are equal (and positive and not equal to 1), the exponents must be equal:

To solve for $x$, multiply both sides by 2:

$x = 2x$

Subtract $x$ from both sides:

$0 = 2x - x$

$0 = x$

The only solution to the equation $|1 - i|^x = 2^x$ is $x = 0$.

The question asks for the number of non-zero integral solutions.

Since the only solution we found is $x = 0$, and 0 is not a non-zero integer, there are no non-zero integral solutions.

The number of non-zero integral solutions is $\mathbf{0}$.

Given:

The equation $(a + ib) (c + id) (e + if) (g + ih) = A + iB$, where $a, b, c, d, e, f, g, h, A, B$ are real numbers.

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To Show:

$(a^2 + b^2 ) (c^2 + d^2 ) (e^2 + f^2 ) (g^2 + h^2 ) = A^2 + B^2$.

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Proof:

We are given the equation:

$(a + ib) (c + id) (e + if) (g + ih) = A + iB$

Take the modulus of both sides of the equation. The modulus of a complex number $z = x + yi$ is $|z| = \sqrt{x^2 + y^2}$. Also, the modulus of a product of complex numbers is equal to the product of their moduli, i.e., $|z_1 z_2 ... z_n| = |z_1| |z_2| ... |z_n|$.

Applying the modulus on both sides:

$|(a + ib) (c + id) (e + if) (g + ih)| = |A + iB|$

Using the property of the modulus of a product:

$|a + ib| \cdot |c + id| \cdot |e + if| \cdot |g + ih| = |A + iB|$

Now, calculate the modulus of each complex number:

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

$|c + id| = \sqrt{c^2 + d^2}$

$|e + if| = \sqrt{e^2 + f^2}$

$|g + ih| = \sqrt{g^2 + h^2}$

$|A + iB| = \sqrt{A^2 + B^2}$

Substitute these modulus values back into the equation:

$\sqrt{a^2 + b^2} \cdot \sqrt{c^2 + d^2} \cdot \sqrt{e^2 + f^2} \cdot \sqrt{g^2 + h^2} = \sqrt{A^2 + B^2}$

Square both sides of the equation:

$\left(\sqrt{a^2 + b^2} \cdot \sqrt{c^2 + d^2} \cdot \sqrt{e^2 + f^2} \cdot \sqrt{g^2 + h^2}\right)^2 = \left(\sqrt{A^2 + B^2}\right)^2$

Using the property $(\sqrt{x})^2 = x$ and $(xyzw)^2 = x^2y^2z^2w^2$, we have:

$(\sqrt{a^2 + b^2})^2 (\sqrt{c^2 + d^2})^2 (\sqrt{e^2 + f^2})^2 (\sqrt{g^2 + h^2})^2 = A^2 + B^2$

$(a^2 + b^2) (c^2 + d^2) (e^2 + f^2) (g^2 + h^2) = A^2 + B^2$

This is the required identity.

Hence, proved.

Given:

The equation $\left( \frac{1 \;+\; i}{1 \;-\; i} \right)^{m} = 1$, where $m$ is a positive integer.

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To Find:

The least positive integral value of $m$.

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Solution:

We are given the equation $\left( \frac{1 \;+\; i}{1 \;-\; i} \right)^{m} = 1$.

First, simplify the complex number inside the parenthesis: $\frac{1 + i}{1 - i}$.

Multiply the numerator and the denominator by the conjugate of the denominator, which is $1 + i$:

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

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

Expand the numerator and simplify the denominator:

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

Since $i^2 = -1$, this becomes:

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

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

$= i$

Now, substitute the simplified value back into the given equation:

$(i)^{m} = 1$

We need to find the least positive integer $m$ such that $i^m = 1$.

Let's list the first few positive integral powers of $i$:

$i^1 = i$

$i^2 = -1$

$i^3 = -i$

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

$i^5 = i^4 \times i = 1 \times i = i$

The powers of $i$ repeat in a cycle of 4: $i, -1, -i, 1$.

The equation $i^m = 1$ is satisfied when the exponent $m$ is a positive multiple of 4.

The positive integral values of $m$ for which $i^m = 1$ are $4, 8, 12, 16, \dots$.

The least of these positive integral values is 4.

Thus, the least positive integral value of $m$ is $\mathbf{4}$.