Question

Convert $\frac{-16}{1+i\sqrt{3}}$ into polar form.

• A. $8\left(cos\,\frac{2\pi}{3}+i\, sin \frac{2\pi}{3}\right)$
• B. $4\left(cos\frac{\pi}{3}+i\,sin\frac{\pi}{3}\right)$
• C. $cos \frac{2\pi}{3}+i\, sin \frac{2\pi}{3}$
• D. None of these

Answer 1

Answer: (A)

Let $z=\frac{-16}{1+i\sqrt{3}}$
$=\frac{-16}{1+i\sqrt{3}}\times\frac{1-i\sqrt{3}}{1-i \sqrt{3}}$
$=\frac{-16\left(1-i\sqrt{3}\right)}{1+3}$
$=-4\left(1-i\sqrt{3}\right)$
$=-4+i 4\sqrt{3}$
Let $-4 = rcos\theta, 4\sqrt{3}=r\, sin\,\theta$
By squaring and adding, we get
$16+48=r^{2}\left(cos^{2}\theta+sin^{2}\theta\right)$
$\Rightarrow \, r^{2}=64$
$\Rightarrow \, r=8$
Hence, cos $\theta=\frac{-1}{2}$,
sin $\theta=\frac{\sqrt{3}}{2}$
$\Rightarrow \, \theta=\pi-\frac{\pi}{3}=\frac{2\pi}{3}$
Thus, the required polar form is $8 \left(cos\frac{2\pi}{3}+i\,sin \frac{2\pi}{3}\right)$