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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?2012-12-03

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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.

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    @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.2012-12-03