$\frac{1+7 i}{(2-i)^2}=\frac{(1+7 i)}{(3-4 i)} \frac{(3+4 i)}{(3+4 i)}=\frac{-25+25 i}{25}=-1+i$
Let $z=x+i y=-1+i$
$\therefore r \cos \theta=-1$ and $r \sin \theta=1 \quad \therefore \theta=\frac{3 \pi}{4}$ and $r=\sqrt{2}$ Thus $\frac{1+7 i}{(2-i)^2}=\sqrt{2}\left[\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right]$
Aliter :$\left|\frac{1+7 i}{(2-i)^2}\right|=\left|\frac{1+7 i}{3-4 i}\right|=\sqrt{2}$
and $\arg \left(\frac{1+7 i}{3-4 i}\right)=\tan ^{-1} 7-\tan ^{-1}\left(-\frac{4}{3}\right)$
$=\tan ^{-1} 7+\tan ^{-1} \frac{4}{3}=\frac{3 \pi}{4}$
$\therefore \frac{1+7 i}{(2-i)^2}=\sqrt{2}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)$