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Home  >>  TN XII Math  >>  Complex Numbers
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Express the following complex number in polar form. $2 + 2\sqrt{3}\mathit{i}$

This is the first part of the multi-part Q6.
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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 $r\sin \theta=2\sqrt 3$
Squaring and adding we get
$r^2=4+12=16$
$r=4$
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})$
answered Jun 10, 2013 by sreemathi.v
 

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