# Express the following complex number in polar form. $-1 + \mathit{i}\sqrt{3}$

This is the second part of the multi-part Q6.

## 1 Answer

Toolbox:
• If $z=x+iy$ is written in exponential form as $z=r(\cos \theta+i\sin \theta),r=\sqrt{x^2+y^2}$ and the argument $\theta$ is given by the following rule
• $\theta=\pi-\alpha\Rightarrow \theta=\alpha$
• $\theta=-\pi+\alpha\Rightarrow \theta=-\alpha$
• Where $\alpha=\tan^{-1}\mid\large\frac{y}{x}\mid$ and $(x,y)$ lies in one of the four quadrants (or the axes).
Step 1:
Let $-1+i\sqrt 3=r(\cos \theta+i\sin \theta)$
Therefore $r\cos\theta=-1$ and $r\sin \theta=\sqrt 3$
Squaring and adding we get
$r^2=1+3=4$
$\Rightarrow r=2$
Step 2:
$\alpha=\tan^{-1}\mid\large\frac {\sqrt 3}{-1}\mid$$=\tan^{-1}\sqrt 3 \Rightarrow \large\frac{\pi}{3} Now the point representing -1+i\sqrt 3 in the quadrant 2. Therefore \theta=\pi-\alpha=\pi-\large\frac{\pi}{3}=\large\frac{2\pi}{3} Therefore -1+i\sqrt 3=4(\cos \large\frac{2\pi}{3}$$+i\sin\large\frac{2\pi}{3})$
answered Jun 10, 2013

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