Evaluate the following: $(1-i)^{1+i}$ My progression:
$(1-i)^{i+1} = e^{(i+1) * \ln(1-i)}$. I get stuck after this point.
 A: The expression $z^w$ where $z, w \in \mathbb{C}$ is not uniquely determined. In fact, we define
$$z^w = e^{w\log z}$$
where $\log z$ is any logarithm of $z$. There are infinitely many choices of $\log z$, and for most values of $z$ and $w$ there will be infinitely many possible values for $z^w$. 
To get something unique, you will have to specify a particular branch of the complex logarithm, but when you do so. $z^w$ won't be defined for all $z$ (or at the very least $z^w$ won't be continuous in $z$, depending on what your conventions with branches are).
In your particular case $\log(1-i) = \ln \sqrt 2 - \dfrac{i\pi}4 + 2\pi i k$ for some arbitrary integer $k$, and
\begin{align}
(1-i)^{1+i} &= e^{(1+i)\log(1-i)} \\
&= e^{(1+i)(\ln \sqrt 2 - \frac{i\pi}4 + 2\pi i k)} \\
&= e^{ \ln \sqrt 2 + \frac\pi4-2\pi k + i(\ln\sqrt 2 - i\frac{\pi}4 + 2\pi k)} \\
&= \sqrt 2 e^{\frac\pi4-2\pi k\pi}\cdot e^{i(\ln\sqrt2-\frac\pi4)} \\
&= \sqrt 2 e^{\frac\pi4-2\pi k\pi}\cdot \big( \cos (\ln\sqrt2-\frac\pi4) + i \sin(\ln\sqrt2-\frac\pi4) \big) \\
\end{align}
A: $$\ln\left((1-i)^{(1+i)}\right)=(1+i)\ln(1-i)=(1+i)\left[\frac{1}{2}\ln(2)-\frac{1}{4}i\pi\right]$$
So:
$$\ln[(1-i)^{(1+i)}]=F=\frac{1}{2}\ln(2)(1+i)+\frac{1}{4}\pi(1-i)$$
and:
$$(1-i)^{(1+i)}=\exp(F)$$
and then:
$$(1-i)^{(1+i)}=(1+i)\left(\sin\left(\frac{1}{2}\ln(2)\right)-i\cos\left(\frac{1}{2}\ln(2)\right)\right)\exp\left(\frac{1}{4}\pi\right)$$
A: $n \in {\mathbb Z}$.
\begin{align}
\left(1 - {\rm i}\right)^{1 + {\rm i}}
&=
\left[\sqrt{2\,}\,{\rm e}^{{\rm i}\left(-\pi/4 + 2n\pi\right)}\right]
^{1 + {\rm i}}
=
\left(\sqrt{2\,}\,\right)^{1 + {\rm i}}
{\rm e}^{{\rm i}\left(-\pi/4 + 2n\pi\right) - \left(-\pi/4 + 2n\pi\right)}
\\[3mm]&=
\sqrt{2\,}\,{\rm e}^{\pi/4 - 2n\pi}\,
\left(\sqrt{2\,}\,\right)^{\rm i}
{\rm e}^{{\rm i}\left(-\pi/4 + 2n\pi\right)}
=
\left(1 - {\rm i}\right){\rm e}^{\pi/4 - 2n\pi}\,
2^{{\rm i}/2}
\\[3mm]&=
{\rm e}^{\pi/4 - 2n\pi}\,
{\rm e}^{{\rm i}\ln\left(2\right)/2}\left(1 - {\rm i}\right)
=
{\rm e}^{\pi/4 - 2n\pi}\,
\left[%
\cos\left({1 \over 2}\ln\left(2\right)\right)
+
{\rm i}\sin\left({1 \over 2}\ln\left(2\right)\right)
\right]\left(1 - {\rm i}\right)
\\
----&------------------------------------
\end{align}
\begin{align}
&n \in {\mathbb Z}\,,
\\[5mm]
&\color{#ff0000}{\left(1 - {\rm i}\right)^{1 + {\rm i}}}
\\&=
\color{#ff0000}{{\rm e}^{\pi/4 - 2n\pi}\times
\left\{%
\left[
\cos\left({1 \over 2}\ln\left(2\right)\right)
+
\sin\left({1 \over 2}\ln\left(2\right)\right)\right]
+
{\rm i}\left[%
-\cos\left({1 \over 2}\ln\left(2\right)\right)
+
\sin\left({1 \over 2}\ln\left(2\right)\right)\right]
\right\}}
\end{align}
