Question

If $z=\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}$, then

• A. $\left|z\right|=1,\arg z=\frac{\pi }{4}$
• B. $\left|z\right|=1,\arg z=\frac{\pi }{6}$
• C. $\left|z\right|=\frac{\sqrt{3}}{2},\arg z=\frac{5\pi }{24}$
• D. $\left|z\right|=\frac{\sqrt{3}}{2},\arg z={\tan }^{-1}\frac{1}{\sqrt{2}}$

Answer 1

The correct option is B $\left|z\right|=1,\arg z=\frac{\pi }{6}$

if $z=\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right)$ then.

As $z=|z|e^{iarq\left(z\right)}$

$\therefore z=\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right)=e^{i\left(\pi /6\right)}[\because e^{i\theta }=\cos \theta +i\sin \theta ]$

$\Rightarrow |z|=1,arq=\frac{\pi }{6}$