Question

Find the modulus and argument of mentioned below complex number and hence express in polar form

$\frac{-16}{1+i\sqrt{3}}$

Answer 1

Let $z=\frac{-16}{1+i\sqrt{3}}$

$\Rightarrow z=\frac{-16}{1+i\sqrt{3}}\times \frac{1-i\sqrt{3}}{1-i\sqrt{3}}$

$\Rightarrow z=\frac{-16+16\sqrt{3}i}{1^2-3i^2}$

$\Rightarrow z=-4+4\sqrt{3}i\dots 1$

$\because z=-4+4\sqrt{3}i$

$\Rightarrow |z|=\sqrt{(-4{)}^2+(4\sqrt{3}{)}^2}$

$\Rightarrow |z|=\sqrt{16+48}$

$\Rightarrow |z|=8\dots \left(2\right)$

Let $\alpha$ be acute angle such that-

$\tan \alpha =\frac{|\text{Im}\left(z\right)|}{|\text{Re}\left(z\right)|}$

$\Rightarrow \tan \alpha =\frac{|4\sqrt{3}|}{|-4|}$

$\Rightarrow \tan \alpha =\sqrt{3}$

$\Rightarrow \alpha =\frac{\pi }{3}$

Since, $z$ lies on the $2^{nd}$ quadrant,

$\therefore arg\left(z\right)=\pi -\alpha$

$\Rightarrow arg\left(z\right)=\pi -\frac{\pi }{3}$

$\Rightarrow arg\left(z\right)=\frac{2\pi }{3}$

Hence, Polar form of $z$

$z=|z|\left[\cos arg\right(z)+i\sin arg(z\left)\right]$

$\Rightarrow z=8(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3})$