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