# A simple computation of log function

I just want to make sure I got the right calculation.

$$\log[(1+i)^{2i}]=\log[e^{i\ln2-\pi/2-4k\pi}]=i\ln2-\pi/2-4k\pi=i\ln2-\pi/2.$$

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What do you mean by $\log$? All possible values or a specific branch? –  mrf Dec 3 '12 at 10:55

Let's see, $$1+i=\sqrt2e^{\pi i/4}$$ so $$\log((1+i)^{2i})=2i\log(1+i)=2i(\log\sqrt2+{\pi i\over4})=i\log2-(\pi/2)$$ Looks OK to me.
How do you justify the first step? $\log a^b$ is not necessarily the same as $b\log a$ for complex numbers, at least not if you work with a particular branch of $\log$ (as you seem to be doing). –  mrf Dec 3 '12 at 10:56
@mrf, you have a point, but OP didn't seem to be worried about multiples of $2\pi$, so I chose not to worry about them, either. –  Gerry Myerson Dec 3 '12 at 11:41