Question

Find the principal argument of ${\left(1+i\sqrt{3}\right)}^2$.

Answer 1

$$\begin{gathered} z={\left(1+i\sqrt{3}\right)}^2 \\ =1+3i^2+2\sqrt{3}i \\ =1-3+2\sqrt{3}i \\ =-2+2\sqrt{3}i \end{gathered}$$

$$\begin{gathered} \mathrm{Let}\beta \mathrm{be}\mathrm{an}\mathrm{acute}\mathrm{angle}\mathrm{given}\mathrm{by}\tan \beta =\frac{\left|\mathrm{Im}\left(z\right)\right|}{\left|\mathrm{Re}\left(z\right)\right|}.\mathrm{Then}, \\ \tan \beta =\frac{\left|{\displaystyle 2\sqrt{3}}\right|}{\left|{\displaystyle 2}\right|}=\left|\sqrt{3}\right| \\ \Rightarrow \tan \beta =\left|\tan \frac{\mathrm{\pi }}{3}\right| \\ \Rightarrow \beta =\frac{\mathrm{\pi }}{3} \\ \mathrm{Clearly},z\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{second}\mathrm{quadrant}.\mathrm{Therefore},\arg \left(z\right)=\mathrm{\pi }-\frac{\mathrm{\pi }}{3}=\frac{2\mathrm{\pi }}{3}. \\ \mathrm{Hence},\mathrm{the}\mathrm{principal}\mathrm{argument}\mathrm{of}z\mathrm{is}\frac{2\mathrm{\pi }}{3}. \end{gathered}$$