Question

express the number in trigonometric form. -2 + 2 √3 i

Answer 1

Given:

z = -2 + 2 √3 i​

The objective is to write the number in trigonometric form.

The trigonometric form can be represented as,

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

The general form of complex number is, z = x+iy.

By comparing with the given equation.

x = -2

y = 2 √3

The value of r can be calculated as,

$\begin{gathered} r=\sqrt[]{x^2+y^2} \\ r=\sqrt[]{(-2)^2+(2\sqrt[]{3})^2} \\ r=\sqrt[]{4+4\cdot3} \\ r=\sqrt[]{4+12} \\ r=\sqrt[]{16} \\ r=4 \end{gathered}$

The value of theta can be calculated as,

$\begin{gathered} \theta=\tan ^{-1}(\frac{y}{x}) \\ \theta=\tan ^{-1}(\frac{2\sqrt[]{3}}{-2}) \\ \theta=\tan ^{-1}(-\sqrt[]{3}) \\ \theta=-^{}60^0 \\ \theta=-\frac{\pi}{3} \end{gathered}$

Now, substitute the obtained values of r and theta in the general equation of trigonometric form.

$z=4\lbrack\cos (-\frac{\pi}{3})+i\sin (-\frac{\pi}{3})\rbrack$

Since,

$\begin{gathered} \cos (-\theta)=\cos \theta \\ \sin (-\theta)=-\sin \theta \end{gathered}$

Then,

$z=4\lbrack\cos \frac{\pi}{3}-i\sin \frac{\pi}{3}\rbrack$

Hence, the required trigonometric form is obtained.