# Solve the following

Question:

If $z=\cos \frac{\pi}{4}+i \sin \frac{\pi}{6}$, then

(a) $|z|=1, \arg (z)=\frac{\pi}{4}$

(b) $|z|=1, \arg (z)=\frac{\pi}{6}$

(c) $|z|=\frac{\sqrt{3}}{2}, \arg (z)=\frac{5 \pi}{24}$

(d) $|z|=\frac{\sqrt{3}}{2}, \arg (z)=\tan ^{-1} \frac{1}{\sqrt{2}}$

Solution:

(d) $|z|=\frac{\sqrt{3}}{2}, \arg (z)=\tan ^{-1} \frac{1}{\sqrt{2}}$

$z=\cos \frac{\pi}{4}+i \sin \frac{\pi}{6}$

$\Rightarrow z=\frac{1}{\sqrt{2}}+\frac{1}{2} i$

$\Rightarrow|z|=\sqrt{\left(\frac{1}{\sqrt{2}}\right)^{2}+\frac{1}{4}}$

$\Rightarrow|z|=\sqrt{\frac{1}{2}+\frac{1}{4}}$

$\Rightarrow|z|=\sqrt{\frac{3}{4}}$

$\Rightarrow|z|=\frac{\sqrt{3}}{2}$

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

$=\frac{1}{\sqrt{2}}$

$\Rightarrow \alpha=\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)$

Since, the point $z$ lies in the first quadrant.

Therefore, $\arg (z)=\alpha=\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)$