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# Polar Form of Complex Numbers

A complex number is a number that can be put in the form x + iy, where x and y are real numbers and i is called the imaginary unit, where i2 = −1. The polar form of a complex number is a shorthand for the exponential form. The polar shorthand for Aei$\theta$ is A $\angle$$\theta$. In this page we are going to discuss about polar form of a complex number and operation on the complex numbers.

 Related Calculators Complex Numbers in Standard Form Calculator Polar Form to Rectangular Form Calculator Adding Complex Numbers Calculator Dividing Complex Numbers Calculator

## Complex Number to Polar Form

The Cartesian form of complex number is x + iy be the given complex number, where x is called as the real part and y is called as the imaginary part.

This can be written in the polar form as follows:

x + iy = r (cos θ + i sin θ) ………….. (A)

where, x = r cos θ, y = r sin θ [This we got by comparing the real and imaginary part on both sides]

Here, r cos θ is the real part and r sin θ is the imaginary part.

Therefore, cos θ = $\frac{x}{r}$

sin θ = $\frac{y}{r}$

Therefore, cos 2 θ + sin 2 θ = 1

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

$\frac{x ^2 + y ^2}{r ^2}$ = $1$

$r^2 = x^2 + y^2$

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

Now, $\frac{\sin \theta}{\cos \theta}$ = $\tan \theta$

$\tan \theta$ = $\frac{\frac{y}{r}}{\frac{x}{r}}$ = $\frac{y}{x}$

Therefore, θ = $\tan ^{-1} (\frac{y}{x})$ …………… (2)

Therefore, by using (1) and (2) in (A), we can represent x + iy in the polar form.

## Multiplying Complex Numbers in Polar Form

It is easy to multiply complex number in polar form as compares to algebraic form. For multiplying two complex numbers, let us assume the values of Z1 and Z2.

Suppose Z1 = r1(cos$\theta_1$ + isin$\theta_1$)

and Z2 = r2(cos$\theta_2$ + isin$\theta_2$).

The product Z1Z2 can be formed as follows:

Z1Z2 = r1r2 (cos$\theta_1$ + isin$\theta_1$) (cos$\theta_2$ + isin$\theta_2$)

= r1r2 (cos$\theta_1$cos$\theta_2$ - sin$\theta_1$sin$\theta_2$) + i(cos$\theta_1$sin$\theta_2$ + sin$\theta_1$cos$\theta_2$)

= r1r2 [cos($\theta_1$ + $\theta_2$) + i($\theta_1$ + $\theta_2$)]

Thus to form the product of two complex numbers written in polar form, we multiply the moduli and add the arguments.

### Solved Example

Question: If Z1 = 5(cos 60o + i sin 60o) and Z2 = 6(cos 70o + i sin 70o). Find Z1 Z2 in polar form.
Solution:
Given: Z1 = 5(cos 60o + i sin 60o) and Z2 = 6(cos 70o + i sin 70o)

Now
Z1 Z2 = 5(cos 60o + i sin 60o) * 6(cos 70o + i sin 70o)

= 5 * 6 [cos(60 + 70) + i sin(60 + 70)]

= 30[cos(130o) + i sin(130o)]

=> Z1 Z2 = 30[cos(130o) + i sin(130o)].

## Adding Complex Numbers in Polar Form

There is not a particular way to add complex numbers in polar form.  In order to add complex numbers in polar form, convert back to Cartesian form. Then add the rectangular form of complex numbers and convert it back into polar form. Let us see with the hep of example:

### Solved Example

Question: Add two complex numbers, where Z1 = 2(cos 60o + i sin 60o) and Z2 = 2(cos 30o + i sin 30o).
Solution:
Given complex numbers: Z1 = 2(cos 60o + i sin 60o) and Z2 = 2(cos 30o + i sin 30o).

Step 1:
Convert polar form to Cartesian form

Z1 = 2(cos 60o + i sin 60o)

On comparing with Z1 = x1 + iy1

=> x1 = 2 cos 60o and y1 = 2 sin 60o

or x1 = 2 * $\frac{1}{2}$ = 1

and y1 = 2 * $\frac{\sqrt{3}}{2}$ = $\sqrt{3}$

So Z1 = 1 + i $\sqrt 3$ ..........................(i)

Again

Z2 = 2(cos 30o + i sin 30o)

On comparing with Z2 = x2 + iy2

=> x2 = 2 cos 30o and y2 = 2 sin 60o

or x2 = 2 * $\frac{\sqrt{3}}{2}$ = $\sqrt{3}$

and y2 = 2 * $\frac{1}{2}$ = 1

So Z2 = $\sqrt 3$ + i ..........................(ii)

Step 2:

=> Z1 + Z2 = 1 + i $\sqrt 3$ + $\sqrt 3$ + i

= 1 + $\sqrt 3$ + i ( 1 + $\sqrt 3$)

Step 3:

Convert above result in complex polar form:

Let Z3 = 1 + $\sqrt 3$ + i ( 1 + $\sqrt 3$)

By comparing with the general form of the complex number, Z = x + iy

We have, x = 1 + $\sqrt 3$ and y = 1 + $\sqrt 3$.

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

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

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

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

θ = $\tan ^{-1} (\frac{y}{x})$ = $\frac{1 + \sqrt 3}{1 + \sqrt 3}$

= $\tan ^{-1} (1)$

= $\frac{\pi}{4}$

=> $\theta$ = $\frac{\pi}{4}$

Therefore required polar form is:

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

## Writing Complex Numbers in Polar Form

Given below are some of the examples in writing complex numbers in polar form.

### Solved Examples

Question 1: Find the polar form of $\sqrt 3$ + i
Solution:
The complex number is $\sqrt 3$ + i.

By comparing with the general form of the complex number, Z = x + iy

We have, x = $\sqrt 3$ and y = 1.

We know that polar form of the complex number is Z = r(cos$\theta$ + i sin $\theta$) ..................(i)

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

=> r = 2

θ = $\tan ^{-1} (\frac{y}{x}) = \tan ^{-1} (\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$

=> $\theta$ = $\frac{\pi}{6}$

Put the value of r and $\theta$ in (i)

=> $Z = 2 (\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$

Therefore, the polar form of $\sqrt{3} + 1$ is $2 (\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$

Question 2: Find the polar form of $1 + i \sqrt{3}$
Solution:
The complex number is $1 + i \sqrt{3}$

By comparing with the general form of the complex number, Z = x + iy

Here x = 1, y = $\sqrt{3}$

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

θ = $\tan ^{-1} (\frac{y}{x}) = \tan ^{-1} (\frac{\sqrt 3}{1}) = \frac{\pi}{3}$

Therefore, the polar form of $1 + i \sqrt{3}$ is $2 (\cos \frac{\pi}{3} + i \sin \frac{\pi}{3})$

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