Answer
$\frac{cos(2ln(\pi))}{\pi}+ \frac{isin(2ln(\pi))}{\pi}$
Work Step by Step
$\pi^{-1+2i} = \frac{1}{\pi}\pi^{2i}$
Applying Euler's formula to $\pi^{2i}$ yields $cos(2ln(\pi))+ isin(2ln(\pi))$
$\frac{cos(2ln(\pi))}{\pi}+ \frac{isin(2ln(\pi))}{\pi}$