TITLE: Solving for m in $\left ( 1+i \right )^{m}= \left ( 1-i \right )^{m}, m\in \mathbb{Z}$
QUESTION [6 upvotes]: Taking the log of both sides , I wound up with an equality that looks like this:
$$m\cdot i\left ( \ln \left ( \frac{\pi }{4} +2k\pi\right )-\ln \left ( \frac{7\pi}{4} +2k\pi\right ) \right )=0$$
which to my knowledge is satisfied only when $m=0$. However, wolfram alpha gives multiple solutions:
http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427edlc1a076ou

REPLY [4 votes]: $$\left(\frac{1+i}{1-i}\right)^m=1\iff i^m=1\iff m\equiv 0\pmod{4}$$