Question

If $z=\cos \frac{\mathrm{\pi }}{4}+i\sin \frac{\mathrm{\pi }}{6}$, then
(a) $\left|z\right|=1,arg\left(z\right)=\frac{\mathrm{\pi }}{4}$
(b) $\left|z\right|=1,arg\left(z\right)=\frac{\mathrm{\pi }}{6}$
(c) $\left|z\right|=\frac{\sqrt{3}}{2},arg\left(z\right)=\frac{5\mathrm{\pi }}{24}$
(d) $\left|z\right|=\frac{\sqrt{3}}{2},arg\left(z\right)={\tan }^{-1}\frac{1}{\sqrt{2}}$

Answer 1

(d) $\left|z\right|=\frac{\sqrt{3}}{2},\arg \left(z\right)={\tan }^{-1}\frac{1}{\sqrt{2}}$
$$\begin{gathered} z=\cos \frac{\mathrm{\pi }}{4}+i\sin \frac{\mathrm{\pi }}{6} \\ \Rightarrow z=\frac{1}{\sqrt{2}}+\frac{1}{2}i \\ \Rightarrow \left|z\right|=\sqrt{{\left(\frac{1}{\sqrt{2}}\right)}^2+\frac{1}{4}} \\ \Rightarrow \left|z\right|=\sqrt{\frac{1}{2}+\frac{1}{4}} \\ \Rightarrow \left|z\right|=\sqrt{\frac{3}{4}} \\ \Rightarrow \left|z\right|=\frac{\sqrt{3}}{2} \\ \tan \alpha =\left|\frac{\mathrm{Im}\left(z\right)}{\mathrm{Re}\left(z\right)}\right| \\ =\frac{1}{\sqrt{2}} \\ \Rightarrow \alpha ={\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right) \\ \mathrm{Since},\mathrm{the}\mathrm{point}z\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{quadrant}. \\ \mathrm{Therefore},\arg \left(z\right)=\alpha ={\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right) \end{gathered}$$