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