Complex Numbers: Representation and Properties

Argand Plane (Geometric Representation of Complex Numbers)

Visualizing Complex Numbers

Real numbers can be easily visualized as points on a one-dimensional number line. However, a complex number $z = a + bi$ has two components: a real part ($a$) and an imaginary part ($b$). To represent these two independent components geometrically, we need a two-dimensional space. This space is called the Argand plane or the complex plane.

The Argand plane is essentially a standard Cartesian coordinate system but with specialized axes to represent the components of a complex number.

---

The Structure of the Argand Plane

The Argand plane is constructed as follows:

• The horizontal axis is used to represent the real part of a complex number. It is called the real axis. Points on this axis correspond to complex numbers where the imaginary part is zero (i.e., real numbers of the form $a + 0i$).
• The vertical axis is used to represent the imaginary part of a complex number. It is called the imaginary axis. Points on this axis correspond to purely imaginary numbers where the real part is zero (i.e., numbers of the form $0 + bi$, often written as $bi$). The points on this axis are typically labeled as $i, 2i, -i,$ etc.
• The intersection of the real axis and the imaginary axis is the origin, corresponding to the complex number $0 + 0i = 0$.

A complex number $z = a + bi$, where $a$ and $b$ are real numbers, is uniquely represented in the Argand plane by the point with coordinates $(a, b)$. The real part $a$ is the x-coordinate (abscissa), plotted along the real axis, and the imaginary part $b$ is the y-coordinate (ordinate), plotted along the imaginary axis.

The image shows the Argand plane setup. A complex number $a+bi$ is located by moving $a$ units along the real axis and $b$ units along the imaginary axis, identifying the point $(a, b)$.

---

Representing Complex Numbers as Points or Vectors

In the Argand plane, a complex number $z = a + bi$ can be visualized in two primary ways:

1. As a Point:

   The most common representation is simply the point $P(a, b)$ in the complex plane with coordinates $(a, b)$. For instance, the complex number $3 + 2i$ is represented by the point $(3, 2)$, and $-1 - 4i$ is represented by the point $(-1, -4)$. Real numbers are points on the real axis, and purely imaginary numbers are points on the imaginary axis.

2. As a Vector:

   A complex number $z = a + bi$ can also be represented by a position vector (a directed line segment) starting from the origin $(0, 0)$ and ending at the point $P(a, b)$. This vector corresponds to the displacement from the origin to the point $(a, b)$. This vector representation is particularly useful for visualizing the geometric effects of complex number operations like addition and subtraction.

The image illustrates a complex number $a+bi$ as a vector originating from $(0,0)$ and terminating at the point $(a,b)$.

---

Geometric Interpretation of Addition and Subtraction

The vector representation in the Argand plane provides a geometric interpretation for the addition and subtraction of complex numbers, which is analogous to vector addition and subtraction:

• Addition ($z_1 + z_2$):

  Let $z_1 = a + bi$ and $z_2 = c + di$. Their sum is $z_1 + z_2 = (a+c) + (b+d)i$. If $z_1$ is represented by the vector $\vec{v_1}$ from the origin to $(a, b)$, and $z_2$ by the vector $\vec{v_2}$ from the origin to $(c, d)$, then their sum $z_1 + z_2$ is represented by the vector from the origin to the point $(a+c, b+d)$. Geometrically, this corresponds to the vector sum $\vec{v_1} + \vec{v_2}$, which can be found using the parallelogram rule of vector addition (the sum vector is the diagonal of the parallelogram formed by $\vec{v_1}$ and $\vec{v_2}$).

• Subtraction ($z_1 - z_2$):

  Let $z_1 = a + bi$ and $z_2 = c + di$. Their difference is $z_1 - z_2 = (a-c) + (b-d)i$. Geometrically, this difference $z_1 - z_2$ is represented by the vector starting from the point representing $z_2$ and ending at the point representing $z_1$. If the vector representing $z_2$ from the origin is $\vec{v_2}$ and the vector representing $z_1$ from the origin is $\vec{v_1}$, then $z_1 - z_2$ is represented by the vector $\vec{v_1} - \vec{v_2}$.

The Argand plane is an invaluable tool for visualizing complex numbers, understanding their properties (like modulus and argument, discussed in the next section), and interpreting the geometric effects of complex arithmetic operations.

Modulus of a Complex Number and Its Properties

Defining the Modulus (Magnitude)

Just as the absolute value of a real number represents its distance from zero on the number line, the modulus of a complex number represents its distance from the origin $(0, 0)$ in the Argand plane. The modulus is also referred to as the magnitude or absolute value of the complex number. It is a non-negative real number.

Consider a complex number $z = a + bi$, where $a$ is the real part and $b$ is the imaginary part. This complex number is represented by the point $(a, b)$ in the Argand plane. The modulus of $z$, denoted by $|z|$, is the distance from the origin $(0, 0)$ to the point $(a, b)$.

Using the distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$, which is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$, the distance between $(0, 0)$ and $(a, b)$ is $\sqrt{(a-0)^2 + (b-0)^2} = \sqrt{a^2 + b^2}$.

Alternatively, consider the right triangle formed by the origin, the point $(a, 0)$ on the real axis, and the point $(a, b)$ in the complex plane. The lengths of the two perpendicular sides are $|a|$ and $|b|$. The length of the hypotenuse (the distance from the origin to $(a, b)$) is given by the Pythagorean theorem: $\sqrt{|a|^2 + |b|^2} = \sqrt{a^2 + b^2}$.

Here, $a = \text{Re}(z)$ and $b = \text{Im}(z)$. So, $|z| = \sqrt{(\text{Re}(z))^2 + (\text{Im}(z))^2}$.

Examples of calculating the modulus:

• For $z = 3 + 4i$: $|z| = |3 + 4i| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
• For $z = 1 - 2i$: $|z| = |1 - 2i| = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$.
• For $z = -5i$: This is $0 - 5i$. $|z| = |-5i| = \sqrt{0^2 + (-5)^2} = \sqrt{0 + 25} = \sqrt{25} = 5$. For a purely imaginary number $bi$, $|bi| = \sqrt{0^2 + b^2} = \sqrt{b^2} = |b|$.
• For $z = 7$: This is $7 + 0i$. $|z| = |7| = \sqrt{7^2 + 0^2} = \sqrt{49 + 0} = \sqrt{49} = 7$. For a real number $a$, $|a| = \sqrt{a^2 + 0^2} = \sqrt{a^2} = |a|$. The modulus of a real number is its absolute value.
• For $z = 0$: This is $0 + 0i$. $|z| = |0| = \sqrt{0^2 + 0^2} = \sqrt{0} = 0$.

The modulus provides a measure of the "size" or "length" of the complex number when viewed as a vector from the origin in the Argand plane.

---

Properties of the Modulus

The modulus of a complex number has several important properties. For any complex numbers $z_1, z_2 \in \mathbb{C}$:

1. Non-negativity:

   The modulus is always greater than or equal to zero. It is zero if and only if the complex number itself is zero.

   $|z| \ge 0$ for all $z \in \mathbb{C} $

   [Property 1]

   $|z| = 0 \iff z = 0 $

   [Property 2]

2. Modulus of Conjugate and Negative:

   A complex number, its conjugate, and its additive inverse all have the same modulus.

   $|z| = |\overline{z}| = |-z| $

   [Property 3]

   Proof: Let $z = a+bi$. $\overline{z} = a-bi$, $-z = -a-bi$. $|z| = \sqrt{a^2 + b^2}$. $|\overline{z}| = \sqrt{a^2 + (-b)^2} = \sqrt{a^2 + b^2}$. $|-z| = \sqrt{(-a)^2 + (-b)^2} = \sqrt{a^2 + b^2}$. All are equal.

3. Product with Conjugate:

   The product of a complex number and its conjugate is equal to the square of its modulus.

   $z \overline{z} = |z|^2 = a^2 + b^2 $

   [Property 4]

   Proof: This was shown in the previous section and the definition of modulus. $(a+bi)(a-bi) = a^2+b^2$, and $|z|^2 = (\sqrt{a^2+b^2})^2 = a^2+b^2$.

4. Modulus vs. Real/Imaginary Parts:

   The absolute value of the real part or the imaginary part of a complex number is less than or equal to its modulus.

   $|\text{Re}(z)| \le |z| \quad \text{and} \quad |\text{Im}(z)| \le |z| $

   [Property 5]

   Proof: Let $z=a+bi$. $|z|=\sqrt{a^2+b^2}$. Since $b^2 \ge 0$, $a^2 \le a^2+b^2$. Taking the square root of both sides (since both are non-negative), $\sqrt{a^2} \le \sqrt{a^2+b^2}$, which is $|a| \le |z|$. Similarly, since $a^2 \ge 0$, $b^2 \le a^2+b^2$, so $\sqrt{b^2} \le \sqrt{a^2+b^2}$, which is $|b| \le |z|$.

5. Modulus of a Product:

   The modulus of the product of two complex numbers is equal to the product of their moduli.

   $|z_1 z_2| = |z_1| |z_2| $

   [Property 6]

   Proof: $|z_1 z_2|^2 = (z_1 z_2) \overline{(z_1 z_2)} = (z_1 z_2)(\overline{z_1} \overline{z_2})$ (using conjugate of product identity) $= (z_1 \overline{z_1})(z_2 \overline{z_2})$ (using commutative and associative properties) $= |z_1|^2 |z_2|^2 = (|z_1| |z_2|)^2$. Since both $|z_1 z_2|$ and $|z_1| |z_2|$ are non-negative, taking the square root of both sides gives $|z_1 z_2| = |z_1| |z_2|$.

6. Modulus of a Quotient:

   The modulus of the quotient of two complex numbers is equal to the quotient of their moduli, provided the divisor is non-zero.

   $|\frac{z_1}{z_2}| = \frac{|z_1|}{|z_2|}$, provided $z_2 \neq 0 $

   [Property 7]

   Proof: Similar proof using properties of conjugates and modulus. $|\frac{z_1}{z_2}|^2 = \frac{z_1}{z_2} \overline{\left(\frac{z_1}{z_2}\right)} = \frac{z_1}{z_2} \frac{\overline{z_1}}{\overline{z_2}} = \frac{z_1 \overline{z_1}}{z_2 \overline{z_2}} = \frac{|z_1|^2}{|z_2|^2} = \left(\frac{|z_1|}{|z_2|}\right)^2$. Taking square roots gives the result.

7. Triangle Inequality:

   The magnitude of the sum of two complex numbers is less than or equal to the sum of their magnitudes. Geometrically, this states that the length of one side of a triangle (the sum vector) is less than or equal to the sum of the lengths of the other two sides (the original vectors).

   $|z_1 + z_2| \le |z_1| + |z_2| $

   [Property 8 - Triangle Inequality]

   Proof: This property is typically proven by squaring both sides and using the fact that $z+\overline{z}=2\text{Re}(z) \le 2|z|$.

8. Reverse Triangle Inequality:

   The magnitude of the difference of two complex numbers is greater than or equal to the absolute value of the difference of their magnitudes.

   $|z_1 - z_2| \ge ||z_1| - |z_2|| $

   [Property 9 - Reverse Triangle Inequality]

---

Applications of the Modulus

The modulus of a complex number has various applications:

• Distance in the Argand Plane: The distance between the points representing two complex numbers $z_1$ and $z_2$ in the Argand plane is given by the modulus of their difference, $|z_1 - z_2|$. If $z_1 = a+bi$ and $z_2 = c+di$, then $z_1 - z_2 = (a-c) + (b-d)i$. The distance is $|z_1 - z_2| = \sqrt{(a-c)^2 + (b-d)^2}$, which is the familiar distance formula between points $(a,b)$ and $(c,d)$.
• Solving Equations and Inequalities: The modulus is used in equations and inequalities involving complex numbers (e.g., $|z - (1+i)| = 2$ represents a circle centered at $(1, 1)$ with radius 2 in the Argand plane).
• Polar Representation: The modulus is one of the two components (the other being the argument or angle) used to represent complex numbers in polar form, which is very useful for multiplication and division.
• Various Fields: The modulus has physical significance in fields like electrical engineering (impedance magnitude), signal processing, and quantum mechanics.

The modulus is a fundamental property that quantifies the size or distance of a complex number from the origin in the complex plane and plays a vital role in the theory and applications of complex numbers.

Conjugate of a Complex Number and Its Properties

Defining the Complex Conjugate

For any complex number $z = a + bi$, where $a$ and $b$ are real numbers, its complex conjugate is a related complex number that is formed by keeping the real part the same and changing the sign of the imaginary part. The complex conjugate of $z$ is denoted by $\overline{z}$ (read as "z-bar") or sometimes by $z^*$.

Geometrically, in the Argand plane, if a complex number $z=a+bi$ is represented by the point $(a, b)$, then its complex conjugate $\overline{z}=a-bi$ is represented by the point $(a, -b)$. This means that the conjugate $\overline{z}$ is the reflection of the point representing $z$ across the real axis (the horizontal axis).

Examples of finding the complex conjugate:

• The conjugate of $5 + 2i$ is $5 - 2i$.
• The conjugate of $1 - 3i$ is $1 + 3i$.
• The conjugate of $-4i$: This is a purely imaginary number, which can be written as $0 - 4i$. The conjugate is $0 - (-4)i = 0 + 4i = 4i$. In general, the conjugate of $bi$ is $-bi$.
• The conjugate of $6$: This is a real number, which can be written as $6 + 0i$. The conjugate is $6 - 0i = 6$. The conjugate of any real number $a$ is $a$ itself.
• The conjugate of $0$: This is $0 + 0i$. The conjugate is $0 - 0i = 0$.

---

Properties of the Complex Conjugate

The complex conjugate operation has several useful properties that relate the conjugate of sums, differences, products, and quotients to the conjugates of the individual complex numbers. For any complex numbers $z_1, z_2 \in \mathbb{C}$:

1. Conjugate of a Conjugate:

   Taking the conjugate of a complex number twice results in the original complex number.

   $\overline{\overline{z}} = z $

   [Property 1]

   Proof: Let $z = a+bi$. Then $\overline{z} = a-bi$. The conjugate of $\overline{z}$ is $\overline{\overline{z}} = \overline{a-bi} = a - (-b)i = a + bi$, which is the original complex number $z$.

2. Real and Purely Imaginary Conditions:

   A complex number is real if and only if it is equal to its conjugate. A complex number is purely imaginary if and only if it is equal to the negative of its conjugate (for non-zero numbers).

   $z \in \mathbb{R} \iff z = \overline{z} $

   [Property 2]

   Proof: If $z = a+bi$, then $z = \overline{z} \implies a+bi = a-bi$. Subtracting $a$ from both sides gives $bi = -bi$, which means $2bi = 0$. This is true if and only if $b=0$. If $b=0$, then $z=a+0i=a$, which is a real number.

   $z \in i\mathbb{R}, z \neq 0 \iff z = -\overline{z} $

   [Property 3]

   Proof: If $z = a+bi$, then $z = -\overline{z} \implies a+bi = -(a-bi) = -a+bi$. Subtracting $bi$ from both sides gives $a = -a$, which means $2a = 0$. This is true if and only if $a=0$. If $a=0$ and $z \neq 0$, then $z=0+bi=bi$ with $b \neq 0$, which is a purely imaginary number.

3. Sum and Difference in terms of Real/Imaginary Parts:

   The sum and difference of a complex number and its conjugate can be used to isolate the real and imaginary parts.

   $z + \overline{z} = 2 \text{Re}(z) = 2a $

   [Property 4]

   Proof: $(a + bi) + (a - bi) = (a+a) + (b-b)i = 2a + 0i = 2a$.

   $z - \overline{z} = 2i \text{Im}(z) = 2bi $

   [Property 5]

   Proof: $(a + bi) - (a - bi) = (a-a) + (b-(-b))i = 0 + (b+b)i = 2bi$.

4. Conjugate of Sum/Difference:

   The conjugate of a sum (or difference) is the sum (or difference) of the conjugates.

   $\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2} $

   [Property 6]

   Proof: Let $z_1=a+bi, z_2=c+di$. $\overline{z_1+z_2} = \overline{(a+c)+(b+d)i} = (a+c)-(b+d)i$. $\overline{z_1}+\overline{z_2} = (a-bi)+(c-di) = (a+c)+(-b-d)i = (a+c)-(b+d)i$.

   $\overline{z_1 - z_2} = \overline{z_1} - \overline{z_2} $

   [Property 7]

   Proof: Similar to Property 6.

5. Conjugate of Product/Quotient:

   The conjugate of a product (or quotient) is the product (or quotient) of the conjugates.

   $\overline{z_1 z_2} = \overline{z_1} \overline{z_2} $

   [Property 8]

   Proof: Let $z_1=a+bi, z_2=c+di$. $z_1 z_2 = (ac-bd) + (ad+bc)i$. $\overline{z_1 z_2} = (ac-bd) - (ad+bc)i$. $\overline{z_1} \overline{z_2} = (a-bi)(c-di) = (ac-(-b)(-d)) + (a(-d)+(-b)c)i = (ac-bd) + (-ad-bc)i = (ac-bd) - (ad+bc)i$.

   $\overline{\left(\frac{z_1}{z_2}\right)} = \frac{\overline{z_1}}{\overline{z_2}}$, provided $z_2 \neq 0 $

   [Property 9]

   Proof: Let $w = z_1/z_2$. Then $z_1 = w z_2$. Taking conjugates of both sides: $\overline{z_1} = \overline{w z_2} = \overline{w} \overline{z_2}$ (using Property 8). Divide by $\overline{z_2}$ (since $z_2 \neq 0$, $\overline{z_2} \neq 0$): $\overline{w} = \frac{\overline{z_1}}{\overline{z_2}}$. Substitute back $w = z_1/z_2$.

6. Conjugate of a Power:

   The conjugate of a power of a complex number is the power of the conjugate.

   $\overline{z^n} = (\overline{z})^n $ for any integer $n$.

   [Property 10]

   Proof: This can be proven by induction for positive integers $n$, and then extended to $n=0$ and negative integers. For $n=2$, $\overline{z^2} = \overline{z \times z} = \overline{z} \times \overline{z} = (\overline{z})^2$ using Property 8.

7. Relationship with Modulus:

   The product of a complex number and its conjugate is the square of its modulus. Also, the modulus of a number is equal to the modulus of its conjugate.

   $z \overline{z} = |z|^2 = |\overline{z}|^2 = a^2 + b^2 $

   [Property 11]

   Proof: Already shown in the previous section and Property 4.

---

Applications of the Conjugate

The complex conjugate is a powerful tool with numerous applications:

• Complex Division: The most common application is in performing division of complex numbers. By multiplying the numerator and denominator by the conjugate of the denominator, we make the denominator a real number, allowing the result to be written in standard $a+bi$ form.
• Finding Modulus: The modulus $|z|$ can be calculated as $\sqrt{z \overline{z}}$.
• Finding Real and Imaginary Parts: The real and imaginary parts of a complex number can be expressed using conjugates: $\text{Re}(z) = \frac{z + \overline{z}}{2}$ and $\text{Im}(z) = \frac{z - \overline{z}}{2i}$.
• Roots of Polynomials: For a polynomial equation with real coefficients, if a complex number $z$ is a root, then its complex conjugate $\overline{z}$ is also a root. Complex roots of polynomials with real coefficients always appear in conjugate pairs. This is a consequence of Identity 8.
• Simplification and Proofs: The properties of conjugates are used extensively in simplifying complex expressions and proving theorems in complex analysis.
• Engineering and Physics: In fields like electrical engineering (AC circuit analysis), quantum mechanics, and signal processing, complex numbers are used to represent physical quantities, and the conjugate often has a specific physical meaning.

The complex conjugate is a fundamental concept for understanding the structure and properties of the complex number system.

Polar Representation of Complex Numbers

Representing Complex Numbers Geometrically by Magnitude and Direction

We have learned to represent complex numbers in the form $z = a + bi$, where $a$ and $b$ are real numbers and $i$ is the imaginary unit ($\sqrt{-1}$). This is known as the rectangular form or Cartesian form. When we plot a complex number on the Argand plane, $a$ represents the position along the real axis and $b$ represents the position along the imaginary axis, similar to plotting a point $(a, b)$ in the Cartesian coordinate system.

The polar representation provides an alternative way to describe the location of a complex number on the Argand plane. Instead of using its horizontal and vertical components $(a, b)$, it uses the distance of the point from the origin and the angle that the line connecting the origin to the point makes with the positive real axis.

This geometric perspective offered by the polar form is particularly powerful and simplifies many operations involving complex numbers, especially multiplication, division, and raising complex numbers to a power.

---

Components of Polar Form

Consider a complex number $z = a + bi$, corresponding to the point $(a, b)$ in the Argand plane. Let the distance of this point from the origin $(0, 0)$ be $r$, and let the angle measured counterclockwise from the positive real axis to the line segment connecting the origin to $(a, b)$ be $\theta$.

Using basic trigonometry in the right-angled triangle formed by the origin, the point $(a, 0)$, and the point $(a, b)$, we can relate the rectangular coordinates $(a, b)$ to the polar coordinates $(r, \theta)$:

$$ a = r \cos \theta $$

$$ b = r \sin \theta $$

Substituting these expressions for $a$ and $b$ into the rectangular form $z = a + bi$, we get:

$$ z = (r \cos \theta) + i (r \sin \theta) $$

Factoring out $r$, we obtain the polar form of the complex number $z$:

The components of the polar form are:

• Modulus ($r$): The modulus of $z$, denoted by $|z|$ or $r$, is the distance of the point $(a, b)$ from the origin. It is always a non-negative real number. Using the Pythagorean theorem, we have: $$ r = |z| = \sqrt{a^2 + b^2} $$ For any non-zero complex number $z$, the modulus $r$ is positive ($r > 0$). If $z = 0$, then $r=0$.
• Argument ($\theta$): The argument of $z$, denoted by $\arg(z)$ or $\theta$, is the angle that the line segment from the origin to the point $(a, b)$ makes with the positive real axis, measured in the counterclockwise direction. The angle $\theta$ is typically measured in radians, but degrees can also be used. The value of the argument is not unique; if $\theta$ is an argument, then $\theta + 2\pi k$ (or $\theta + 360^\circ k$) for any integer $k$ is also an argument.

To ensure a unique value for the argument, we usually use the principal argument, denoted by $\text{Arg}(z)$ or $\theta$. The principal argument is the unique value of the argument that lies within a specific interval. The most common interval is $(-\pi, \pi]$ (i.e., $-180^\circ < \theta \leq 180^\circ$), but $[0, 2\pi)$ is also sometimes used.

To find the argument $\theta$ from the rectangular coordinates $(a, b)$, we use the relations:

$$ \cos \theta = \frac{a}{r} \quad \text{and} \quad \sin \theta = \frac{b}{r} $$

Alternatively, if $a \neq 0$, we can use the ratio:

$$ \tan \theta = \frac{b}{a} $$

Using $\tan \theta = b/a$ requires careful consideration of the quadrant of the point $(a, b)$ to determine the correct value of $\theta$. Let $\alpha = \tan^{-1}\left(\left|\frac{b}{a}\right|\right)$ be the reference angle in the first quadrant, where $\tan^{-1}$ gives a value in $[0, \pi/2)$. We then find $\theta$ based on the quadrant:

• If $(a, b)$ is in the first quadrant ($a > 0, b > 0$): $\theta = \alpha$.
• If $(a, b)$ is in the second quadrant ($a < 0, b > 0$): $\theta = \pi - \alpha$ (or $180^\circ - \alpha$).
• If $(a, b)$ is in the third quadrant ($a < 0, b < 0$): $\theta = \pi + \alpha$ (or $180^\circ + \alpha$) or $\theta = \alpha - \pi$ (or $\alpha - 180^\circ$) for the principal argument in $(-\pi, \pi]$.
• If $(a, b)$ is in the fourth quadrant ($a > 0, b < 0$): $\theta = 2\pi - \alpha$ (or $360^\circ - \alpha$) or $\theta = -\alpha$ for the principal argument in $(-\pi, \pi]$.

For points on the axes:

• If $a > 0, b = 0$ (positive real axis): $\theta = 0$.
• If $a < 0, b = 0$ (negative real axis): $\theta = \pi$ (or $180^\circ$).
• If $a = 0, b > 0$ (positive imaginary axis): $\theta = \pi/2$ (or $90^\circ$).
• If $a = 0, b < 0$ (negative imaginary axis): $\theta = 3\pi/2$ (or $270^\circ$) or $\theta = -\pi/2$ (or $-90^\circ$) for the principal argument in $(-\pi, \pi]$.
• If $a = 0, b = 0$ (the origin $z=0$): The modulus $r=0$, and the argument $\theta$ is undefined. The polar form is simply $z = 0$.

It is often helpful to visualize the point $(a, b)$ in the Argand plane to determine the correct quadrant and angle.

---

Answer:

---

Answer:

---

Answer:

---

Answer:

---

Answer:

---

Answer:

---

Operations in Polar Form

One of the main advantages of the polar form is how it simplifies multiplication, division, and exponentiation of complex numbers. Let $z_1$ and $z_2$ be two complex numbers in polar form:

Let $z_1 = r_1(\cos \theta_1 + i \sin \theta_1)$ and $z_2 = r_2(\cos \theta_2 + i \sin \theta_2)$.

Multiplication of Complex Numbers in Polar Form

To multiply two complex numbers in polar form, we multiply their moduli and add their arguments.

$$ z_1 z_2 = (r_1 r_2) (\cos(\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2)) $$

This means $|z_1 z_2| = |z_1| |z_2| = r_1 r_2$ and $\arg(z_1 z_2) = \arg(z_1) + \arg(z_2) = \theta_1 + \theta_2$ (up to a multiple of $2\pi$).

Derivation:

Let's multiply $z_1$ and $z_2$ directly:

$$ z_1 z_2 = [r_1(\cos \theta_1 + i \sin \theta_1)] [r_2(\cos \theta_2 + i \sin \theta_2)] $$

$$ z_1 z_2 = r_1 r_2 (\cos \theta_1 + i \sin \theta_1) (\cos \theta_2 + i \sin \theta_2) $$

Expand the product of the terms in parentheses:

$$ (\cos \theta_1 + i \sin \theta_1) (\cos \theta_2 + i \sin \theta_2) = \cos \theta_1 \cos \theta_2 + i \cos \theta_1 \sin \theta_2 + i \sin \theta_1 \cos \theta_2 + i^2 \sin \theta_1 \sin \theta_2 $$

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

$$ (\cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2) + i (\cos \theta_1 \sin \theta_2 + \sin \theta_1 \cos \theta_2) $$

Using the trigonometric identities for the sum of angles:

$$ \cos(A+B) = \cos A \cos B - \sin A \sin B $$

$$ \sin(A+B) = \sin A \cos B + \cos A \sin B $$

With $A = \theta_1$ and $B = \theta_2$, the expression becomes:

$$ \cos(\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2) $$

Substituting this back into the expression for $z_1 z_2$:

$$ z_1 z_2 = r_1 r_2 [\cos(\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2)] $$

This completes the derivation.

Answer:

Division of Complex Numbers in Polar Form

To divide two complex numbers in polar form (where the denominator is non-zero), we divide their moduli and subtract their arguments.

$$ \frac{z_1}{z_2} = \frac{r_1}{r_2} (\cos(\theta_1 - \theta_2) + i \sin(\theta_1 - \theta_2)) \quad (r_2 \neq 0) $$

This means $\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|} = \frac{r_1}{r_2}$ and $\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2) = \theta_1 - \theta_2$ (up to a multiple of $2\pi$).

Derivation:

We want to find $\frac{z_1}{z_2} = \frac{r_1(\cos \theta_1 + i \sin \theta_1)}{r_2(\cos \theta_2 + i \sin \theta_2)}$.

Multiply the numerator and denominator by the conjugate of $\cos \theta_2 + i \sin \theta_2$, which is $\cos \theta_2 - i \sin \theta_2$.

$$ \frac{z_1}{z_2} = \frac{r_1}{r_2} \cdot \frac{(\cos \theta_1 + i \sin \theta_1)(\cos \theta_2 - i \sin \theta_2)}{(\cos \theta_2 + i \sin \theta_2)(\cos \theta_2 - i \sin \theta_2)} $$

The denominator is $(\cos \theta_2)^2 - (i \sin \theta_2)^2 = \cos^2 \theta_2 - i^2 \sin^2 \theta_2 = \cos^2 \theta_2 - (-1)\sin^2 \theta_2 = \cos^2 \theta_2 + \sin^2 \theta_2 = 1$.

The numerator is $(\cos \theta_1 + i \sin \theta_1)(\cos \theta_2 - i \sin \theta_2) = \cos \theta_1 \cos \theta_2 - i \cos \theta_1 \sin \theta_2 + i \sin \theta_1 \cos \theta_2 - i^2 \sin \theta_1 \sin \theta_2$.

Since $i^2 = -1$, the numerator is $(\cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2) + i (\sin \theta_1 \cos \theta_2 - \cos \theta_1 \sin \theta_2)$.

Using the trigonometric identities for the difference of angles:

$$ \cos(A-B) = \cos A \cos B + \sin A \sin B $$

$$ \sin(A-B) = \sin A \cos B - \cos A \sin B $$

With $A = \theta_1$ and $B = \theta_2$, the numerator becomes:

$$ \cos(\theta_1 - \theta_2) + i \sin(\theta_1 - \theta_2) $$

Substituting the numerator and denominator back into the expression for $\frac{z_1}{z_2}$:

$$ \frac{z_1}{z_2} = \frac{r_1}{r_2} \cdot \frac{\cos(\theta_1 - \theta_2) + i \sin(\theta_1 - \theta_2)}{1} = \frac{r_1}{r_2} (\cos(\theta_1 - \theta_2) + i \sin(\theta_1 - \theta_2)) $$

This completes the derivation.

Answer:

Exponentiation of Complex Numbers (De Moivre's Theorem)

For a complex number $z = r(\cos \theta + i \sin \theta)$ and any integer $n$, the power $z^n$ is found by raising the modulus to the power $n$ and multiplying the argument by $n$. This is known as De Moivre's Theorem.

$$ z^n = r^n (\cos(n\theta) + i \sin(n\theta)) $$

This means $|z^n| = |z|^n = r^n$ and $\arg(z^n) = n \cdot \arg(z) = n\theta$ (up to a multiple of $2\pi$).

Derivation (for positive integer $n$):

We can derive De Moivre's Theorem for positive integers using the principle of mathematical induction and the multiplication rule.

Let the statement $P(n)$ be $z^n = r^n (\cos(n\theta) + i \sin(n\theta))$.

Base Case (n=1):

For $n=1$, $z^1 = r^1 (\cos(1\theta) + i \sin(1\theta)) = r(\cos \theta + i \sin \theta)$. This is the original polar form of $z$, so $P(1)$ is true.

Inductive Step:

Assume $P(k)$ is true for some positive integer $k$. That is, assume $z^k = r^k (\cos(k\theta) + i \sin(k\theta))$.

Now consider $P(k+1)$. We need to show that $z^{k+1} = r^{k+1} (\cos((k+1)\theta) + i \sin((k+1)\theta))$.

We can write $z^{k+1}$ as $z^k \cdot z$. Using the inductive hypothesis for $z^k$ and the original form for $z$, we have:

$$ z^{k+1} = [r^k(\cos(k\theta) + i \sin(k\theta))] \cdot [r(\cos \theta + i \sin \theta)] $$

Using the multiplication rule for complex numbers in polar form (multiply moduli, add arguments):

The new modulus is $r^k \cdot r = r^{k+1}$.

The new argument is $k\theta + \theta = (k+1)\theta$.

So, $z^{k+1} = r^{k+1} (\cos((k+1)\theta) + i \sin((k+1)\theta))$.

Thus, $P(k+1)$ is true whenever $P(k)$ is true.

By the Principle of Mathematical Induction, the formula $z^n = r^n (\cos(n\theta) + i \sin(n\theta))$ is true for all positive integers $n$.

De Moivre's Theorem can also be shown to hold for negative integers and rational exponents, which is crucial for finding the roots of complex numbers.

Answer:

Square Root of a Complex Number

Finding Roots in the Complex Plane

Just as positive real numbers have two square roots (e.g., $\sqrt{4} = \pm 2$), non-zero complex numbers also have square roots. However, the concept extends to the entire complex plane. Every non-zero complex number $z$ has exactly two distinct square roots. These two square roots are always opposites of each other, meaning if $w$ is a square root of $z$, then $-w$ is also a square root of $z$.

We can find the square roots of a complex number $z = a + bi$ using a couple of methods: by directly solving equations derived from the rectangular form, or by using the polar form and De Moivre's Theorem.

---

Method 1: Using the Rectangular Form

This method involves algebraic manipulation by assuming the square root is in the rectangular form $x+yi$.

Steps to find the square roots of $z = a + bi$:

1. Assume the form of the square root: Let the square root of $z = a + bi$ be $w = x + yi$, where $x$ and $y$ are real numbers that we need to determine.
2. Square the assumed root and equate to $z$: If $w$ is the square root of $z$, then squaring $w$ must give $z$. So, we have the equation $w^2 = z$. $$ (x + yi)^2 = a + bi $$
3. Expand and simplify the left side: $$ (x + yi)^2 = x^2 + 2(x)(yi) + (yi)^2 $$ $$ = x^2 + 2xyi + y^2 i^2 $$ $$ = x^2 + 2xyi - y^2 \quad \text{(since } i^2 = -1) $$ $$ = (x^2 - y^2) + (2xy)i $$
4. Equate real and imaginary parts: Now we equate the real and imaginary parts of the expanded form $(x^2 - y^2) + (2xy)i$ with the real and imaginary parts of $a + bi$. Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.

   $x^2 - y^2 = a $

   ... (1)

   $2xy = b $

   ... (2)

5. Use the modulus property: We know that for any complex number $w$, $|w^2| = |w|^2$. Also, since $w^2 = z$, we have $|w^2| = |z|$. Therefore, $|w|^2 = |z|$. The modulus of $w = x + yi$ is $|w| = \sqrt{x^2 + y^2}$, so $|w|^2 = (\sqrt{x^2 + y^2})^2 = x^2 + y^2$. The modulus of $z = a + bi$ is $|z| = \sqrt{a^2 + b^2}$. Equating $|w|^2$ and $|z|$:

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

   ... (3)

   (Note: $\sqrt{a^2+b^2}$ here means the non-negative real square root).
6. Solve the system of equations for $x^2$ and $y^2$: We now have a system of two linear equations involving $x^2$ and $y^2$: $$ x^2 - y^2 = a $$ $$ x^2 + y^2 = \sqrt{a^2 + b^2} $$ Adding equation (1) and equation (3): $$ (x^2 - y^2) + (x^2 + y^2) = a + \sqrt{a^2 + b^2} $$ $$ 2x^2 = a + \sqrt{a^2 + b^2} $$ $$ x^2 = \frac{a + \sqrt{a^2 + b^2}}{2} $$ Subtracting equation (1) from equation (3): $$ (x^2 + y^2) - (x^2 - y^2) = \sqrt{a^2 + b^2} - a $$ $$ 2y^2 = \sqrt{a^2 + b^2} - a $$ $$ y^2 = \frac{\sqrt{a^2 + b^2} - a}{2} $$
7. Find $|x|$ and $|y|$: Taking the square root of the expressions for $x^2$ and $y^2$: $$ |x| = \sqrt{\frac{a + \sqrt{a^2 + b^2}}{2}} $$ $$ |y| = \sqrt{\frac{\sqrt{a^2 + b^2} - a}{2}} $$ (Note: $\sqrt{\text{expression}}$ here means the non-negative real square root).
8. Determine the signs of $x$ and $y$: We use equation (2), $2xy = b$, or $xy = b/2$. This equation tells us about the relationship between the signs of $x$ and $y$.
   • If $b$ is positive ($b > 0$), then $xy$ must be positive. This means $x$ and $y$ must have the same sign (both positive or both negative).
   • If $b$ is negative ($b < 0$), then $xy$ must be negative. This means $x$ and $y$ must have opposite signs (one positive, one negative).
   • If $b = 0$, then $xy = 0$. This means either $x=0$ or $y=0$. If $b=0$ and $a>0$, $z=a$ (a positive real number), the roots are $\pm \sqrt{a}$ ($y=0$). If $b=0$ and $a<0$, $z=a$ (a negative real number), the roots are $\pm i\sqrt{|a|}$ ($x=0$).
   Based on the determined sign relationship and the values of $|x|$ and $|y|$, we find the possible pairs of $(x, y)$.
9. Write down the square roots: The two square roots are $w_1 = x+yi$ (using one valid pair of signs for $x$ and $y$) and $w_2 = -x-yi$ (using the other valid pair of signs, which will be the negative of the first root).

Answer:

---

Method 2: Using the Polar Form

This method leverages De Moivre's Theorem, which is particularly useful for finding any $n$-th root of a complex number.

Steps to find the square roots of $z = a + bi$:

1. Convert the complex number to polar form: First, express the given complex number $z = a + bi$ in its polar form $z = r(\cos \theta + i \sin \theta)$. $$ r = |z| = \sqrt{a^2 + b^2} $$ The argument $\theta$ is found by considering the quadrant of $(a, b)$ and using $\tan \theta = b/a$ (or by considering points on the axes). Remember that the argument is periodic, so any angle $\theta + 2k\pi$ for an integer $k$ is also a valid argument.
2. Assume the form of the square root in polar coordinates: Let the square root of $z$ be $w = \rho (\cos \phi + i \sin \phi)$, where $\rho$ is the modulus of $w$ and $\phi$ is the argument of $w$.
3. Square $w$ using De Moivre's Theorem: $$ w^2 = [\rho (\cos \phi + i \sin \phi)]^2 = \rho^2 (\cos(2\phi) + i \sin(2\phi)) $$
4. Equate $w^2$ and $z$ in polar form: We set the squared root equal to the original complex number, using the general form of the argument for $z$: $$ \rho^2 (\cos(2\phi) + i \sin(2\phi)) = r (\cos (\theta + 2k\pi) + i \sin (\theta + 2k\pi)) $$ where $k$ is an integer.
5. Equate moduli and arguments: For these two polar forms to be equal, their moduli must be equal, and their arguments must be equal up to multiples of $2\pi$.
   • Equating moduli: $\rho^2 = r$. Since $\rho$ must be non-negative (it's a distance), we take the principal square root: $$ \rho = \sqrt{r} = \sqrt{\sqrt{a^2 + b^2}} $$
   • Equating arguments: $2\phi = \theta + 2k\pi$. Solving for $\phi$: $$ \phi = \frac{\theta + 2k\pi}{2} = \frac{\theta}{2} + k\pi $$
6. Find the distinct square roots by choosing values for $k$: We substitute different integer values for $k$ into the expression for $\phi$. For square roots (finding 2 roots), we only need to use $k=0$ and $k=1$. Any other integer value of $k$ will produce an argument $\phi$ that differs from $\phi_0$ or $\phi_1$ by a multiple of $2\pi$, resulting in the same complex number.
   • For $k=0$: $$ \phi_0 = \frac{\theta}{2} + (0)\pi = \frac{\theta}{2} $$ The first square root is $w_0 = \rho (\cos \phi_0 + i \sin \phi_0) = \sqrt{r}\left(\cos \frac{\theta}{2} + i \sin \frac{\theta}{2}\right)$.
   • For $k=1$: $$ \phi_1 = \frac{\theta}{2} + (1)\pi = \frac{\theta}{2} + \pi $$ The second square root is $w_1 = \rho (\cos \phi_1 + i \sin \phi_1) = \sqrt{r}\left(\cos \left(\frac{\theta}{2} + \pi\right) + i \sin \left(\frac{\theta}{2} + \pi\right)\right)$. Using the trigonometric identities $\cos(\alpha + \pi) = -\cos \alpha$ and $\sin(\alpha + \pi) = -\sin \alpha$, we can see that: $$ w_1 = \sqrt{r}\left(-\cos \frac{\theta}{2} - i \sin \frac{\theta}{2}\right) = -\sqrt{r}\left(\cos \frac{\theta}{2} + i \sin \frac{\theta}{2}\right) = -w_0 $$ So, the two square roots are indeed $w_0$ and $-w_0$.
7. Convert the roots back to rectangular form (optional): If required, use $w = \rho(\cos \phi + i \sin \phi) = (\rho \cos \phi) + i (\rho \sin \phi)$ to convert the roots back to the $x+yi$ form.

Answer:

Both the rectangular and polar methods are valid for finding the square roots of a complex number. The rectangular method can sometimes be faster for numbers with 'nice' integer or simple fractional components, while the polar method is often more systematic and easily generalizable to finding any $n$-th root of a complex number using the formula derived from De Moivre's Theorem: The $n$-th roots of $z = r(\cos \theta + i \sin \theta)$ are given by $w_k = \sqrt[n]{r}\left(\cos \left(\frac{\theta + 2k\pi}{n}\right) + i \sin \left(\frac{\theta + 2k\pi}{n}\right)\right)$ for $k = 0, 1, 2, \ldots, n-1$.