I have an exercise that I like to solve.

Find the branch cut of $\displaystyle (i-1)^{2+2i}$.

I can simplify it s bit but I don't know what a branch cut is.

$\displaystyle \displaystyle e^{(2+2i)Ln(1-i)} = e^{2(1+i)(log_{e}\sqrt{2}+iArg(1-i))} = e^{(log_{e}{2}+i log_{e}{2}+i\frac{\pi}{4}-\frac{\pi}{4})} = e^{(log_{e}{2}-\frac{\pi}{4} +i ( log_{e}{2}+\frac{\pi}{4}))}$

Now I don't know quite what to do, or what conclusions to make..?

Or is it nicer to write:

$\displaystyle \displaystyle e^{(1+i)log_{e}{2}+\frac{\pi}{4}(i-1)} = 2^{1+i} e^{\frac{\pi}{4}(i-1)}$

Ah well... Pointers are welcomed!