# Polar Form

You should be familiar with Complex Numbers.

The complex numbers can be represented as a point (or vector) in a 2-dimensional Cartesian Coordianate system, called the Argand Diagram. The complex number $z = x + iy$ is represented by the point $P_z = (x, y )$.

{{image PolarForm }}

The distance of the point to the origin $(O)$ is known as the absolute value (or modulus), and is equal to $r_z = \sqrt{x^2 + y^2 }$. The angle between line $P_z O$ and the positive real axis is known as the argument (or phase) of $z$, which we denote by $\theta_z$. From the definition, we easily see that $\theta_z$ is related to $\arctan \left( \frac{y}{x} \right)$, where slight care has to be taken when the angle is obtuse, $x=0$ or $y = 0$. Typically, $\theta_z$ is expressed in radians, and represented by the principal value in the interval $\left( - \pi, \pi \right]$.

With these values of $r_z$ and $\theta_z$, we see that

$z = x + iy = r_z ( \cos \theta_z + i \sin \theta_z ).$

This is known as the polar form of $z$, which is sometimes abbreviated to $z = r_z \text{ cis } \theta_z$.

Euler's formula: For any real number $x$, $e^{ix} = \cos x + i \sin x.$

As such, we may also express the complex number $z$ as $r_z e^{ i \theta_z }$. With this interpretation, multiplication and division of complex numbers follow the rules of exponentiation that we are used to. Given two complex numbers $z_1 = r_1 \text{ cis } \theta_1 = r_1 e^{i \theta_1}$ and $z_ 2 = r_2 \text{ cis } \theta_2 = r_2 e^{i \theta_2}$, we have

$z_1 z_2 = (r_1 r_2) e^{ i ( \theta_1 + \theta_2 ) } = ( r_1 r_2 ) \text{ cis } ( \theta_1 + \theta_2),$ $\frac{z_1}{z_2} = \frac{r_1}{r_2} e^{i (\theta_1 - \theta_2) } = \frac{r_1}{r_2} \text{ cis } (\theta_1 - \theta_2).$

While multiplication and division are greatly simplified, addition and subtraction of complex numbers in polar form do not follow a standard algorithm. Thus, in calculations involving complex numbers, it is important to be able to properly convert between the rectangular and polar forms.

## Worked Examples

### 1. Show that $z_1 z_2 = ( r_1 r_2 ) \text{ cis } ( \theta_1 + \theta_2).$

Solution: $\begin{array} {l l} z_1 z_2 & = r_1 \text{ cis } \theta_1 \times r_2 \text{ cis } \theta_2 \\ & = r_1 r_2 ( \cos \theta_1 + i \sin \theta_1 ) ( \cos \theta_2 + i \sin \theta_2 ) \\ & = r_1 r_2 \left( \cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2 + i ( \sin \theta_1 \cos \theta_2 + \cos \theta_1 \sin \theta_2 ) \right)\\ & = r_1 r_2 \left( \cos (\theta_1 + \theta_2) + i \sin ( \theta_1 + \theta_2) \right) \\ & = (r_1 r_2 ) \text{ cis } ( \theta_1 + \theta_2) \\ \end{array}$

The division of complex numbers in polar form can be dealt with in a similar manner, or by noticing that $\frac{1}{z_2} = \frac{1}{r_2} \text{ cis } (- \theta_2 )$.

### 2. Evaluate $( 2 + i) ( 3 + i )$. Hence, determine $\arctan \frac{1}{2} + \arctan \frac{1}{3}$.

Solution: Let $z_1 = 2 +i, z_2 = 3+i, z_3 = z_1 \times z_2$. Expanding $z_3$, we get $z_3 = (2+i)(3+i) = 6 + 2i + 3i + i^2 = 5 + 5i .$

By comparing the arguments on both sides, we get that $\theta_1 + \theta_2 = \theta_3$, which gives us

$\arctan \frac{1}{2} + \arctan \frac{1}{3} = \arctan \frac{5}{5} = \frac{\pi}{4}.$ Note by Arron Kau
5 years, 10 months ago

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i want explanation for second example can anyone help me

- 5 years, 7 months ago