I just want to make sure I got the right calculation.
$\log[(1+i)^{2i}]=\log[e^{i\ln2-\pi/2-4k\pi}]=i\ln2-\pi/2-4k\pi=i\ln2-\pi/2.$
I just want to make sure I got the right calculation.
$\log[(1+i)^{2i}]=\log[e^{i\ln2-\pi/2-4k\pi}]=i\ln2-\pi/2-4k\pi=i\ln2-\pi/2.$
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.