How can $i^i = e^{-\pi/2}$ !! I was asked a homework question: find $i^i$. The solution provided was as follows:
Let $A = i^i$.
$\log A = i \log i$.
Now, $\log i = \log e^{i\pi/2} = \frac{i\pi}{2}$.
So, $\log A = -\frac{\pi}{2}$
Thus, $i^i = e^{-\pi/2}$.
I understood how the result was obtained, but it is illogical. I understand that multiplying by $i$ is equivalent to rotating the position vector of the complex number in Argand Plane by $90$ degrees anti-clockwise. How can rotating $i$ anti-clockwise $i$ number of times give $e^{-\pi/2}$?
So can somebody explain to me graphically or more intuitively, how $i^i = e^{-\pi/2}$ ? 
 A: The homework question is wrong, and so is the provided answer. Exponentiation $a^b$ is well defined when either $b$ is integer (and $a$ is invertible in case $b<0$) or when $a\in\Bbb R_{>0}$; in the former case the "repeated multiplication" definition of exponentiation applies, and in the latter case the definition $a^b=\exp(b\ln a)$ where the functions $\exp:\Bbb C\to\Bbb C$ and $\ln:\Bbb R_{>0}\to\Bbb R$ are the usual well defined ones. In the case of $\def\ii{{\bf i}}\ii^{\ii}$ however neither of these cases applies, so the expression is not well defined.
Many will try to nevertheless use the formula $\exp(b\ln a)$ to give a value to$~a^b$, as is done (somewhat indirectly) in the answer presented in the question. However, this overlooks that fact that the justification for $a^b=\exp(b\ln a)$, namely $$a^b =(\exp(\ln a))^b =\exp((\ln a)b),$$ uses a rule, namely $(\exp y)^z=\exp(yz)$ (or maybe even more generally $(a^y)^z=a^{yz}$ for $a\in\Bbb R_{>0}$), that simply does not hold for all $y,z\in\Bbb C$ (although it does hold for $y\in\Bbb R$ and $z\in\Bbb C$). For a simple example where the rule fails, take $y=2\pi\ii$ and $z=\pi$, then
$$
  (\exp2\pi\ii)^\pi=1^\pi=1\neq
  \exp(2\pi^2\ii)\approx 0.629681725+0.77685322\ii
  .
$$
An alternative form of the rule is $\ln(x^y)=y\ln(x)$ that also fails in general when $y\notin\Bbb R$, for instance when $y=2\pi\ii$ and $x=e$, where it would give $0=2\pi\ii$. Your "answer" uses this latter rule at the very beginning with $y=\ii$, which is outside of the range where the rule is valid.
See also this answer.
A: Like DHMO says in his comment, the complex map $\ln$ is multivalued, so complex exponentiation is a multivalued operation. Accordingly, 
$$
i^i=\exp(i\ln(i))=\exp(i\cdot (\pi/2+2k\pi)i),k\in\mathbb{Z}
$$
Then if you want, you may consider the principal branch of the above for $k=0$, which gives the desired answer. $i^i$ gives a set equality and not a single number.
A: By definition $$e^{ix} = \cos x+i\sin x = cisx$$
This definition can be proved by observing the Taylor expansions of both the RHS and LHS. You will find they are both identical. 
Letting $x= \frac{\pi}{2}$ we get this: 
$$i = e^{\frac{i\pi}{2}}$$
Then just play with the powers. 
$$i^i = e^{\frac{i^2\pi}{2}}$$
$$i^2=-1$$
Hence, 
$$i^i = e^{-\frac{\pi}{2}}$$
Of course $i^i$ attains an infinite elemental set of real values (due to the periodic nature of $cisx$) but since your question was to prove $i^i = e^{-\frac{\pi}{2}}$, I have done so accordingly. 
A: If you would like, there is also another solution to this problem. Since we cannot count "i" times, the solution is to find a formula for doing exponents, namely, Euler's formula. When you plug in $e^{xi}$, the output will be $\cos(x)+i\sin(x)$ (in radians), also notated as the $\operatorname{cis}(x)$.  Now, we can define $i$ as $e^{π/2i}$. So therefore our goal is $(e^{π/2i})^i$. Luckily, with the way that exponents work, we can simplify this to $e^{π/2i\cdot i}$, which of course, by definition, is simplified to
$e^{-π/2}$.
