How to raise a complex number to the power of another complex number? How do I calculate the outcome of taking one complex number to the power of another, ie  $\displaystyle {(a + bi)}^{(c + di)}$?
 A: Well, assuming principal values of the complex logarithm (otherwise much craziness ensues):
$$(a+bi)^{c+di}=\exp((c+di)\ln(a+bi))$$
$$=\exp((c+di)(\ln|a+bi|+i\arg(a+bi)))$$
$$=\exp((c\ln|a+bi|-d\arg(a+bi))+i(c\arg(a+bi)+d\ln|a+bi|))$$
$$=\exp((c\ln|a+bi|-d\arg(a+bi)))\exp(i(c\arg(a+bi)+d\ln|a+bi|))$$
and I'll let you finish it at this point, using the fact that $\exp(ix)=\cos\;x+i\sin\;x$
A: I transcribe part of my answer to this question.
The complex exponential $e^z$ for complex $z=x+iy$ preserves the law of exponents of the real exponential and satisfies $e^0=1$. 
By definition 
$$e^z=e^{x+iy}=e^xe^{iy}=e^x(\cos y+\sin y)$$ 
which agrees with the real exponential function when $y=0$. The principal logarithm of $z$ is the complex number 
$$w=\text{Log }z=\log |z|+i\arg z$$
so that $e^w=z$, where $\arg z$ (the principal argument of $z$) is the real number in $-\pi\lt \arg z\le \pi$, with $x=|z|\cos (\arg z)$ and $y=|z|\sin (\arg z)$. 
The complex power is 
$$z^w=e^{w\text{ Log} z}.$$ 
In your case you have: $z=a+bi,w=c+di$ 
$$\begin{eqnarray*}
\left( a+bi\right) ^{c+di} &=&e^{(c+di)\text{Log }(a+bi)} \\
&=&e^{(c+di)\left( \ln |a+bi|+i\arg (a+bi)\right) } \\
&=&e^{c\ln \left\vert a+ib\right\vert -d\arg \left( a+ib\right) +i\left(
c\arg \left( a+ib\right) +d\ln \left\vert a+ib\right\vert \right) } \\
&=&e^{c\ln \left\vert a+ib\right\vert -d\arg(a+bi)}\times  \\
&&\times \left( \cos \left( c\arg \left( a+ib\right) +d\ln \left\vert
a+ib\right\vert \right) +i\sin \left( c\arg \left( a+ib\right) +d\ln
\left\vert a+ib\right\vert \right) \right). 
\end{eqnarray*}$$    
A: First you need to realize that this is a multi-valued function.
Let us choose the principal logarithm for convenience. So our argument $\theta$ will lie between $(-\pi,\pi]$.
We then write $a+ib = r e^{i \theta} = e^{(ln(r) + i \theta)}$.
Now, we get $(a+ib)^{c+id} = e^{(ln(r) + i \theta)(c+id)}$. Do the necessary algebraic manipulations in the exponent to get $e^{(cln(r) - d \theta) + i(d ln(r) + c \theta)}$. You might also want to take a look at the previous question asked on a similar topic.
