Evaluate and find the principal value of $(-1+i)^ {2-i}$

Question

Can anyone please help me evaluate and find the principal value of

$(-1+i)^{2-i}$

I got up to

$=e^{2-i}(ln(-1+i))$ $=e^{(2-i)(1/2 ln(2)+i(3pi/4))}$

Answer 1

\begin{align} (-1+i)^{2-i} &= \left[ e^{\pi i} \, \sqrt{2} \, e^{- \pi i/4} \right]^{2 - i} \\ &= e^{2 \pi i + \pi} \, e^{(2-i) \ln(\sqrt{2})} \, e^{- \pi i/2 - \pi/4} \\ &= 2 \, e^{ 3 \pi/4} \left( e^{i(3 \pi + \ln(2))/2} \right)\\ &= 2 \, e^{ 3 \pi/4} \left[ \cos\left( \frac{3 \pi + \ln(2)}{2} \right) + i \, \sin\left( \frac{3 \pi + \ln(2)}{2} \right) \right] \end{align} The principle value is \begin{align} 2 \, e^{ 3 \pi/4} \, \cos\left( \frac{3 \pi + \ln(2)}{2} \right) \end{align} and the argument is \begin{align} 2 \, e^{ 3 \pi/4} \, \sin\left( \frac{3 \pi + \ln(2)}{2} \right) \end{align}