Question

If $z=\frac{1+i\sqrt{3}}{2i(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})},$ then

• A. $|z|=1$
• B. $|z|=\frac{1}{2}$
• C. $\text{arg(z)}=\frac{5\pi }{6}$
• D. $\text{arg(z)}=-\frac{\pi }{2}$

Answer 1

Answer: (A), (D)

The correct options are
A $|z|=1$
D $\text{arg(z)}=-\frac{\pi }{2}$

Given,
$z=\frac{1+i\sqrt{3}}{2i(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})}\Rightarrow z=\frac{1+i\sqrt{3}}{2i(\frac{1}{2}+i\frac{\sqrt{3}}{2})}\Rightarrow z=\frac{1+i\sqrt{3}}{-\sqrt{3}+i}$
$\Rightarrow z=\frac{i(1+i\sqrt{3})}{-1-i\sqrt{3}}=-i$
$\therefore |z|=1,\text{arg(z)}=-\frac{\pi }{2}$