Question:
If $|z|=4$ and $\arg (z)=\frac{5 \pi}{6}$, then $z=$ ___________________
Solution:
If $|z|=4$ and $\arg z=\frac{5 \pi}{6}=\frac{\pi-\pi}{6}$
$\therefore z=|z|\left[\cos \left(\frac{5 \pi}{6}\right)+i \sin \left(\frac{5 \pi}{6}\right)\right]$
$=4\left[\cos \left(\pi-\frac{\pi}{6}\right)+i \sin \left(\pi-\frac{\pi}{6}\right)\right]$
[Since $\cos (\pi-\theta)=-\cos \theta$ and $\sin (\pi-\theta)=\sin \theta]$
$\therefore z=4\left[-\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right]$
Since $\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}$
$\sin \frac{\pi}{6}=\frac{1}{2}$
$\Rightarrow z=4\left[-\frac{\sqrt{3}}{2}+\frac{i}{2}\right]=2(-\sqrt{3}+i)$
Hence, $z=-2 \sqrt{3}+2 i$