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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?

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    $\begingroup$ There no ±in the argument of $i$. $\endgroup$ – Bernard Oct 21 '16 at 0:24
  • $\begingroup$ In full generality you don't have an identity $(-x)^y = -(x^y)$. But having that toggled sign is making things more complicated than they need to be. $\endgroup$ – Excluded and Offended Oct 21 '16 at 0:24
  • $\begingroup$ math.stackexchange.com/questions/301929/… $\endgroup$ – rlartiga Oct 21 '16 at 0:25
  • $\begingroup$ Okay, so if it were just $i^i = e^{-2\pi n-\pi /2}$ how would I got about plotting it? $\endgroup$ – Spuds Oct 21 '16 at 0:26
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    $\begingroup$ @SSimple Art: in a very weak sense — per chance (as though two computational errors yielded the right answer). It lacks rigour, from my point of view. $\endgroup$ – Bernard Oct 21 '16 at 1:13
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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, $\exp(-\frac{\pi}{2}).$

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$i^i$ 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 $x\in \Bbb Z$

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