Can someone give intuition behind understanding $i^i = e^{\frac{-\pi}{2}}$ and more so on complex powers? I have no problem accepting that $$i^i = e^{\frac{-\pi}{2}} $$ Here im assuming that $-\pi\leq \theta \leq \pi$
But what I am wondering is that, can someone give me some intuition behind understanding taking powers of a complex number where the power is also complex ? Like I am curious what you used to get an understanding of this property ? I chose $i^i$ because it seems like an easy example to give intuition behind but I am also looking at this from the point of view of also trying to understand say: $$(4+5i)^{(2-3i)}$$ Like how do you get an intuitive understanding of this ?  
 A: To calculate $z^u=(4+5i)^{2-3i}$ you have to use three formulas


*

*convert $z=(4+5i)$ from rectangular form to polar form $z=re^{i\theta}$

*calculate $\ln(z)=\ln(r)+i\theta+2ik\pi\qquad k\in\mathbb Z$

*calculate $\displaystyle z^u=e^{u\ln(z)}$

Let's go:
$r^2=|z|^2=4^2+5^2=41\implies r=\sqrt{41}$ 
Since $z$ is in the first quadrant then $\theta=\tan^{-1}(\frac 54)$
$\begin{array}{ll}z^u &=\exp\bigg((2-3i)\left(\ln(r)+i\theta+2ik\pi\right)\bigg)\\&=\exp\bigg(\left(2\ln(r)+3\theta+6k\pi\right)+i\left(-3\ln(r)+2\theta+4k\pi\right)\bigg)\\ &=r^2\ e^{3\theta}\ e^{6k\pi}\bigg(\cos\big(2\theta-3\ln(r)\big)+i\sin\big(2\theta-3\ln(r)\big)\bigg) \end{array}$
And from there you can calculate a numerical value by replacing $r,\theta$ by their respective value:

$(4+5i)^{2-3i}\approx e^{6k\pi}\left(-484.77 + 358.42\,i\right)$


Note that $\ln(z)$ is multivalued since it is the reciprocal function of exponential which is $2i\pi$ periodic.
So in the end you get a factor $e^{6k\pi}$ and a principal value $[z^u]_{k=0}=-484.77+358.42i$

You need to keep this factor for things like $z^u\times z^v=z^{u+v}$ to stay true.
If you only consider the principal value then $[z^u]_{k=0}\times[z^v]_{k=0}=[z^{u+v}]_{k=0}$ may be false.
You have to see it like $\exists (k_1,k_2,k_3)\in\mathbb Z^3\mid\  [z^u]_{k=k_1}\times [z^v]_{k=k_2}=[z^{u+v}]_{k=k_3}$

Now that you understood the principle, let's do it for $i^i$ :
$\displaystyle i^i=\exp\bigg(i\ln(e^{i\frac{\pi}2})\bigg)=\exp\bigg(i(\ln(1)+i\frac{\pi}2+2ik\pi)\bigg)=\exp\bigg(-\frac{\pi}2-2k\pi\bigg)=e^{-\frac{\pi}2}e^{-2k\pi}$
So the principal value is $[i^i]_{k=0}=e^{-\frac{\pi}2}$ and the factor is $e^{2k\pi}$ 
[note that since $k$ is arbitrary, $2k\pi$ or $-2k\pi$ does not matter].

To conclude let's calculate $i^2$. Should it also have multiple values ?
$\displaystyle i^2=\exp\bigg(2\ln(e^{i\frac{\pi}2})\bigg)=\exp\bigg(2(\ln(1)+i\frac{\pi}2+2ik\pi)\bigg)=\exp\bigg(i\pi-2ik\pi\bigg)=\underbrace{e^{i\pi}}_{-1}\underbrace{e^{2ik\pi}}_{1}=-1$
This time it has only one value as expected $i^2=-1$, because the factor $e^{2ik\pi}=1$ for any value of $k$.
A: I am afraid that the  intuition you ask for gets lost somewhere on the way. 
We may start by defining $$ \tag 1a^b:=\underbrace{a\cdot \ldots \cdot a}_b$$
for $b\in \Bbb N$ (or maybe define $a^b$ as the cardinality of the set of maps from a set of cardinality $b$ to a set of cardinality $a$, which allows us to use some infinite $b$-values, but poses a restriction on the $a$'s used; for this discussion. I'll stick with $(1)$ as a starting point) and then extend the definition step by step to larger domains, based on a permanence principle: For $(1)$ we prove rules such as $$\tag2a^{b+c}=a^ba^c$$ and extend the definition in  a way that (among others) these rules remain valid. For example, this forces us to define $a^{-b}=\frac1{a^b}$ when extending to $b\in\Bbb Z$ (and which also shows that we cannot define $0^b$ for $b<0$).
Sooner or late, the only way of extending the domain is via the exponential function
$$\exp(x)=\sum_{k=0}^\infty\frac{x^k}{k!} $$
which is fortunately convergent for all $c\in\Bbb C$ (and even in many situations where $x$ is not even a number, thus allowing us to define exponentiation even in some at the first glance strange areas).
Among the many fascinating properties of $\exp$ is
$$\exp(x+y)=\exp( x)\exp (y),$$
just as needed for $(2)$.
But that requires us to use the inverse function $\ln$ and set
$$\tag3a^b=\exp(b\ln a).$$
This is all fine, but $\ln a$ cannot be defined at for at $a=0$ (thus we better still use $(1)$ instead of $(3)$ in that case) and $\ln$ can only be defined "consistently" in a simply connected domain (not containing $0$). The most common convention(!) to pick  such a domain as large as possible is to remove the non-positive reals from $\Bbb C$. That is, for all other $z$, $\ln z$ is a complex number $w$ with $\exp(w)=z$ and $-\pi<\arg w<\pi$. As $\exp$ is its own derivative, we quickly find that $\exp''(ix)+\exp(ix)=0$ and by comparing the initial conditions, find $\exp(ix)=\cos(x)+i\sin(x)$. Hence $\exp(i\frac\pi2)=i$, so that our chosen convention tells us $\ln(i)=i\frac\pi2$ and ultimately $$ i^i=\exp(i\cdot i\tfrac\pi2)=\exp(-\tfrac\pi2).$$
I think it is the forced use of some convention when picking a good domain in $\Bbb C$ which ultimately kills the intuition you are looking for.
Note that I consequently wrote $\exp(x)$, not $e^x$ in order to avoid the idea that the exponential function "is" taking some number dubbed $e$ to some power.
