The complex number with argument \(\frac{5 \pi^c}{6}\) at a distance of 2 units from the origin is

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 \(\mathrm{b}=1 \Rightarrow \mathrm{z}=-\sqrt{3}+\mathrm{i}\)