All Values of a Complex expression

Question

I am asked to find all values to $$\left(\frac{1-i}{\sqrt2}\right)^{1+i}$$

I do not know how to approach a power with complex part. Any help would be appreciated.

Answer 1

Hints: $(1-i) = \sqrt(2)e^{-i\frac{\pi}{4}}$. Now let $ln(k) = i\cdot ln(1-i)$. Can you complete from here?

Answer 2

Since $$ (1-i) = \sqrt{2}\,e^{-\pi i/4} $$ we have: $$ (1-i)^i = e^{\frac{i}{2}\log 2+2k\pi} e^{\pi/4} = (\cos(\log\sqrt{2})+i\sin(\log\sqrt{2}))\,e^{2k\pi+\pi/4}$$ with $k\in\mathbb{Z}$.

Answer 3

Using $1 - i = \sqrt{2} e^{-\pi i/4}$ it is seen that \begin{align} \frac{(1-i)^{1+i}}{\sqrt{2}} &= \left( \frac{1-i}{\sqrt{2}} \right) (1-i)^{i} \\ &= e^{-\pi i/4} \cdot (\sqrt{2})^{i} e^{\pi/4} \\ &= e^{\pi/4} \cdot e^{-\pi i/4} e^{i \ln(\sqrt{2})} \\ &= e^{\pi/4} \cdot e^{-i (\pi/4 - \ln(2)/2)} \\ &= e^{\pi/4} \left[ \cos\left(\frac{\pi - 2 \ln(2)}{4}\right) - i \sin\left(\frac{\pi - 2 \ln(2)}{4}\right) \right]. \end{align}