Plotting $i^i$ in the complex plane I'm asked to find all the values of $i^i$ and then plot them in the complex plane.
First I used Euler's identity to show that
$i = \pm e^{i2\pi n\pm \pi i/2}$,
so then
$i^i = \pm e^{-2\pi n \mp \pi/2}$
Now that I have this, I'm not sure how to plot it. Does it go in the complex z plane of the complex w plane?
 A: In complex numbers you have to take care with power functions.
In complex analysis you have infinity logarithms. A logarithm function $l$ in a region D is a holomorphic function in $D$ such that $\exp(l(z))=z$ for all $z \in D$. As $\exp$ is not injective in $\mathbb{C}$, a complex number can have infinite logarithms. But you also know that the period of complex exponential function is $2\pi i \mathbb{Z}$, so if $\bar l$ is a logarithm function in $D$, then other logarithms functions in $D$ are given by $l = \bar l + 2 \pi i k$ for some $k \in \mathbb{Z}$.
The principal branch of logarithm is defined in $\mathbb{C}$ without the real negative axis, by $l(z)= \log|z|+i \theta$, where $\theta$ is the principal argument of $z$, and $\log$ is the usual real logarithm.
Now that you fix a logarithm function $l$, you can define $z^\sigma = \exp(\sigma l(z))$. With the principal branch, $l(i)=\log|1|+i\frac{\pi}{2}=i\frac{\pi}{2}$, so $i^i = \exp(i l(i))=\exp(i^2\frac{\pi}{2})=\exp(-\frac{\pi}{2})$.
So all values of $i^i$ are given by $\exp(-\frac{\pi}{2}+2\pi ik)$ with $k \in \mathbb{Z}$.
Notice that $\exp(2\pi ik)=1$ for all $k \in \mathbb{Z}$ so your plot its just one point, $\exp(-\frac{\pi}{2}).$
A: $i^i$ is Real for every interpretation. In short,
$$i^i=e^{\frac{-\pi}{2}+2n\pi}, n \in \Bbb Z$$
Remember, n itself CAN be negative, so there is no need for $\pm$.
Note that $2n\pi-\frac{\pi}{2}=\frac{4n-1}{2}\pi$. 
Thus I would plot $$y=e^{\frac{4x-1}{2}\pi}$$ 
where $i^i$ is every point where $x\in \Bbb Z$
