Question

If w=-1+ root 3i/2, then arg(w^2) is

Answer 1

$$\begin{gathered} \mathrm{HI}, \\ \mathrm{\omega }=\frac{-1+\sqrt{3}\mathrm{i}}{2} \\ {\mathrm{\omega }}^2=\frac{1+3{\mathrm{i}}^2-2\sqrt{3}\mathrm{i}}{4}=\frac{1-3-2\sqrt{3}\mathrm{i}}{4} \\ =\frac{-1-\sqrt{3}\mathrm{i}}{2} \\ =\frac{-1}{2}-\frac{\sqrt{3}}{2}\mathrm{i} \\ \mathrm{so}\arg \left({\mathrm{\omega }}^2\right)={\tan }^{-1}\left(\frac{-\frac{\sqrt{3}}{2}}{\frac{-1}{2}}\right)={\tan }^{-1}\sqrt{3}=60\mathrm{degree} \end{gathered}$$