Operations on Complex Numbers in Polar Form

 

 

1. Addition and Subtraction:

 

When performing addition or subtraction with complex numbers in polar form:

a) If the absolute values (moduli) of the complex numbers are the same, the addition or subtraction can be performed by using trigonometric sum-to-product identities.

b) If the absolute values (moduli) of the complex numbers are different, the operation is carried out by first converting the complex numbers into rectangular/algebraic form \( (z = x + yi) \).

 

Example:

 

Given \( z_1 = 3(\cos 72^\circ + i \sin 72^\circ) \) and \( z_2 = 3(\cos 18^\circ + i \sin 18^\circ) \), let us find the sum \( z_1 + z_2 \).

Group the terms:
\[ z_1 + z_2 = 3[\cos 72^\circ + \cos 18^\circ + i(\sin 72^\circ + \sin 18^\circ)] \]

Applying the sum-to-product identities:
\[ = 3(2 \cos 45^\circ \cos 27^\circ + i \, 2 \sin 45^\circ \cos 27^\circ) \]
\[ = 3 \cdot 2 \cos 27^\circ (\cos 45^\circ + i \sin 45^\circ) \]
\[ = 6 \cos 27^\circ \left( \displaystyle \frac{\sqrt{2}}{2} + i \displaystyle \frac{\sqrt{2}}{2} \right) \]
\[ = 3\sqrt{2} \cos 27^\circ (1 + i) \]

 

Example:

 

Given \( z_1 = \cos 20^\circ + i \sin 20^\circ \) and \( z_2 = \cos 10^\circ + i \sin 10^\circ \), let us find the difference \( z_1 – z_2 \).

\[ z_1 \ – \ z_2 = (\cos 20^\circ + i \sin 20^\circ) \ – \ (\cos 10^\circ + i \sin 10^\circ) \]
\[ = \cos 20^\circ – \cos 10^\circ + i(\sin 20^\circ – \sin 10^\circ) \]

Applying the sum-to-product identities:
\[ = -2 \sin 15^\circ \sin 5^\circ + i \, 2 \cos 15^\circ \sin 5^\circ \]
\[ = 2 \sin 5^\circ (-\sin 15^\circ + i \cos 15^\circ) \]
\[ = 2 \sin 5^\circ [\sin(-15^\circ) + i \cos(-15^\circ)] \]
\[ = 2 \sin 5^\circ [\cos(90^\circ \ – \ (-15^\circ)) + i \sin(90^\circ \ – \ (-15^\circ))] \]
\[ = 2 \sin 5^\circ (\cos 105^\circ + i \sin 105^\circ) \]

 

Example:

 

Given \( z_1 = 2(\cos 30^\circ + i \sin 30^\circ) \) and \( z_2 = 3(\cos 60^\circ + i \sin 60^\circ) \), let us find the sum \( z_1 + z_2 \).

Convert both to algebraic form:
\[ z_1 = 2\left( \displaystyle \frac{\sqrt{3}}{2} + \displaystyle \frac{1}{2}i \right) \quad \text{and} \quad z_2 = 3\left( \displaystyle \frac{1}{2} + \displaystyle \frac{\sqrt{3}}{2}i \right) \]

Add the real and imaginary components:
\[ z_1 + z_2 = \sqrt{3} + \displaystyle \frac{3}{2} + \left( 1 + \displaystyle \frac{3\sqrt{3}}{2} \right)i \]

 

Note:

 

A complex number \( z = a + bi \) can be represented in exponential form using Euler’s formula:

\[ z = |z|(\cos \theta + i \sin \theta) = |z|e^{i\theta} \]
\[ \overline{z} = |z|(\cos \theta \ – \ i \sin \theta) = |z|e^{-i\theta} \quad (\text{where } e \approx 2.7182) \]

 

For example:
\[ z = 2\left( \displaystyle \frac{\sqrt{2}}{2} + \displaystyle \frac{\sqrt{2}}{2}i \right) = 2 \, \text{cis} \displaystyle \frac{\pi}{4} = 2e^{\displaystyle \frac{\pi}{4}i} \]

By utilizing the exponential representation, we can easily derive formulas for multiplying, dividing, and finding powers of complex numbers in polar form.

 

2. Multiplication:

 

Let two complex numbers be:
\[ z_1 = |z_1|(\cos \theta_1 + i \sin \theta_1) \quad \text{and} \quad z_2 = |z_2|(\cos \theta_2 + i \sin \theta_2) \]

In exponential form, they are expressed as:
\[ z_1 = |z_1|e^{i\theta_1} \quad \text{and} \quad z_2 = |z_2|e^{i\theta_2} \]

Their product becomes:
\[ z_1 \cdot z_2 = |z_1| \cdot |z_2| \, e^{i(\theta_1 + \theta_2)} \]

Therefore, when multiplying two complex numbers in polar form, we multiply their moduli and add their arguments.

\[ \mathbf{z_1 \cdot z_2 = |z_1| \cdot |z_2| [\cos(\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2)]} \]
\[ \mathbf{\text{arg}(z_1 \cdot z_2) = \text{arg}(z_1) + \text{arg}(z_2)} \]

 

Example:

 

Given \( z_1 = 3 \ – \ 3\sqrt{3}i \) and \( z_2 = 1 + i \), let us find the polar form of their product \( z_1 \cdot z_2 \).

First, convert each to polar form:
\[ z_1 = 6(\cos 300^\circ + i \sin 300^\circ) \]
\[ z_2 = \sqrt{2}(\cos 45^\circ + i \sin 45^\circ) \]

Multiply the moduli and add the angles:
\[ z_1 \cdot z_2 = 6 \cdot \sqrt{2} \, [\cos(300^\circ + 45^\circ) + i \sin(300^\circ + 45^\circ)] \]
\[ = 6\sqrt{2}(\cos 345^\circ + i \sin 345^\circ) \]