Question

If $z=\frac{1+7i}{(2-i{)}^2}$, then
(a) $\left|z\right|=2$
(b) $\left|z\right|=\frac{1}{2}$
(c) amp (z) = $\frac{\mathrm{\pi }}{4}$
(d) amp (z) = $\frac{3\mathrm{\pi }}{4}$

Answer 1

(d) amp (z) = $\frac{3\mathrm{\pi }}{4}$

$$\begin{gathered} z=\frac{1+7i}{{\left(2-i\right)}^2} \\ \Rightarrow z=\frac{1+7i}{4+i^2-4i} \\ \Rightarrow z=\frac{1+7i}{4-1-4i}\left[\because i^2=-1\right] \\ \Rightarrow z=\frac{1+7i}{3-4i} \\ \Rightarrow z=\frac{1+7i}{3-4i}\times \frac{3+4i}{3+4i} \\ \Rightarrow z=\frac{3+4i+21i+28i^2}{9-16i^2} \\ \Rightarrow z=\frac{3-28+25i}{9+16} \\ \Rightarrow z=\frac{-25+25i}{25} \\ \Rightarrow z=-1+i \\ \tan \alpha =\left|\frac{\mathrm{Im}\left(z\right)}{\mathrm{Re}\left(z\right)}\right| \\ =1 \\ \Rightarrow \alpha =\frac{\mathrm{\pi }}{4} \\ \mathrm{Since},z\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{second}\mathrm{quadrant}. \\ \mathrm{Therefore},\mathrm{amp}\left(z\right)=\mathrm{\pi }-\mathrm{\alpha } \\ =\mathrm{\pi }-\frac{\mathrm{\pi }}{4} \\ =\frac{3\mathrm{\pi }}{4} \end{gathered}$$