Calculating complex numbers What is $(1-i)^{1+i}$ ?
I did:
$(1-i)^{1+i} = \left(e^{ln(1-i)}\right)^{1+i} = e^{(1+i)\cdot ln(1-i)} = e^{(1+i)\cdot [ ln\sqrt{2} + i\cdot(-\pi/4) + 2ki\pi ]}$
$ = e^{(ln\sqrt{2} + i\cdot(-\pi/4) + 2ki\pi)} \cdot e^{i\cdot(ln\sqrt{2} - \pi/4 + 2ki\pi)}$
This is it?
 A: HINT :
$$1-i = \sqrt{2}e^{-i\pi/4}$$
Then the calculations depend under which form your want it but here's another hint for later :
$$\sqrt{2}^{1+i} = \sqrt{2}\cdot2^{i/2} = \sqrt{2}\cdot e^{i\log(2)/2}$$
A: When you work with logarithms in the complex plane you have to be very careful. The logarithm cannot be defined in a good way on the whole complex plane (even if you remove the $0$). You have to chose a branch of log, that means choosing a particular definition of logarithm. I would take the principal branch (i.e. the one with imaginary part lying in $(-\pi,\pi)$ and which is undefined on $\mathbb{R}_{\le 0}$) and do it as follows:
$$(1-i) = \sqrt{2}\ e^{-i\frac{\pi}{4}}$$
(you can see why this is true e.g. identifying $\mathbb{C}$ with $\mathbb{R}^2$ and drawing the position of $(1-i)$ as a "vector" in this space). Then you get:
$$(1-i)^{1+i}=\left(\sqrt{2}\ e^{-i\frac{\pi}{4}}\right)^{1+i} = 2^{\frac{1+i}{2}}e^{-i\frac{\pi}{4}(1+i)}$$
This gives you:
$$2^{\frac{1+i}{2}} = e^{\frac{i}{2}\ln(2)}\sqrt{2}$$
$$e^{-i\frac{\pi}{4}(1+i)} = e^{-i\frac{\pi}{4}}e^{\frac{\pi}{4}}$$
and thus:
$$(1-i)^{1+i}=e^{-i\frac{\pi}{4}}e^{\frac{\pi}{4}}e^{\frac{i}{2}\ln(2)}\sqrt{2}=$$
$$(1-i)e^{\frac{\pi}{4}}\left(\cos\left(\frac{\ln(2)}{2}\right)+i\sin\left(\frac{\ln(2)}{2}\right)\right)$$
For more details on branches of log, try to give a look at this page.
