Question

The polar form of $\frac{-\sqrt{3}}{2}-\frac{i}{2}$ (where i = $\sqrt{-1}$) is

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Let Z = $\frac{-\sqrt{3}}{2}-i$.$\frac{1}{2}$

Sin $(\frac{-\sqrt{3}}{2},\frac{-1}{2})$ lies in 3rd quadrant

$\therefore$ Principal value of arg(z) = $-\pi +ta{n}^{-1}\left|\frac{\left(\frac{-1}{2}\right)}{\left(\frac{-\sqrt{3}}{2}\right)}\right|$

= $-\pi +ta{n}^{-1}\left(\frac{1}{\sqrt{3}}\right)$ = $-\pi +\frac{\pi }{6}$ = $\frac{-5\pi }{6}$

|z| = $\sqrt{[\frac{-\sqrt{3}}{2}{]}^2+(\frac{-1}{2}{)}^2}$ = $\sqrt{\frac{3}{4}+\frac{1}{4}}$ = 4

$\therefore$ polar from of z = |z|(cos (arg z)+i sin (arg z))

= cos $\left(\frac{-5\pi }{6}\right)+isin\left(\frac{-5\pi }{6}\right)$