# Express the following complex number in polar form. $2 + 2\sqrt{3}\mathit{i}$

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

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 $2+2\sqrt 3i=r(\cos \theta+i\sin \theta)$
Therefore $r\cos\theta=2$ and $r\sin \theta=2\sqrt 3$
$r^2=4+12=16$
$r=4$
Step 2:
$\alpha=\tan^{-1}\large\frac{2\sqrt 3}{2}$$=\tan^{-1}\sqrt 3 \Rightarrow \large\frac{\pi}{3} Now 2+2\sqrt 3 i represents a point in the quadrant 1. Therefore \theta=\alpha=\large\frac{\pi}{3} 2+2\sqrt 3i=4(\cos \large\frac{\pi}{3}$$+i\sin\large\frac{\pi}{3})$