The complex number with argument $\frac{5 \pi^c}{6}$ at a distance of 2 units from the origin is
The complex number with argument $\frac{5 \pi^c}{6}$ at a distance of 2 units from the origin is
$\sqrt{3}-\mathrm{i}$
$\sqrt{3}+\mathrm{i}$
$-\sqrt{3}-\mathrm{i}$
$-\sqrt{3}+\mathrm{i}$
Solution
Let $\mathrm{z}=\mathrm{a}+\mathrm{ib}$ and we have
$\frac{\mathrm{b}}{\mathrm{a}}=\tan \left(\frac{5 \pi}{6}\right)$ and $\sqrt{\mathrm{a}^2+\mathrm{b}^2}=2$
$\therefore \frac{\mathrm{b}}{\mathrm{a}}=\tan \left(\pi-\frac{\pi}{6}\right)=-\tan \frac{\pi}{6}=\frac{-1}{\sqrt{3}}$ and $\mathrm{a}^2+\mathrm{b}^2=4$
$\therefore \mathrm{b}=\frac{-\mathrm{a}}{\sqrt{3}}$ and $\mathrm{a}^2+\mathrm{b}^2+4$
$\therefore \mathrm{a}^2+\frac{\mathrm{a}^2}{3}=4 \Rightarrow 4 \mathrm{a}^2=12 \Rightarrow \mathrm{a}^2=3 \Rightarrow \mathrm{a}= \pm \sqrt{3}$
Also $\mathrm{b}=\frac{-\mathrm{a}}{\sqrt{3}}=\frac{ \pm \sqrt{3}}{\sqrt{3}}= \pm 1$
Since complex number lies in $2^{\text {nd }}$ quadrant, $\mathrm{a}=-\sqrt{3}$ and $\mathrm{b}=1 \Rightarrow \mathrm{z}=-\sqrt{3}+\mathrm{i}$