Show that $Log((1+i)^2) = 2Log(1+i),$ while $Log((-1+i)^2) \not= 2Log(-1+i)$
Solution:
I am using $[0 , 2\pi]$ for the principle argument.
The first part worked out fine, I got:
$Log((1+i)^2) = log(2) + i \frac{\pi}{2}$
$2Log(1+i)= log(2) + i \frac{\pi}{2}$
But the second part is not working out because I found that the results are equal:
$Log((-1+i)^2) = log(2) + i \frac{3\pi}{2}$
$2Log((-1+i) = log(2) + i \frac{3\pi}{2}$