# Imaginary Powers

Imaginary numbers ($i = \sqrt{-1}$) are used to relate Cartesian coordinates to polar coordinates as shown in the figure below. The argument $\varphi$ is the angle from the real axis along the unit circle, and can be expressed as two terms: a real and an imaginary, giving the Cartesian (x,y) position.
$e^{i\varphi} = \text{cos}(\varphi) + i \text{sin}(\varphi) = x + i y$
(If you want to expand beyond the unit circle just multiply by the desired radius: $e^{i\varphi} \rightarrow re^{i\varphi}$)

This is known as the Euler relation, and it gives us some neat results, like:
$e^{i\pi} = -1$
$e^{i \pi /2} = i$
But what happens when you exponentiate $i$ with itself? $i^i$.

From the Euler relation we know that $i=e^{i \pi /2}$, so it’s really a matter of substitution:
$i^i = e^{i \pi / 2 i} = e^{-\pi / 2} = 0.20788...$
So $i^i$ is a real number!

But wait a second, this isn’t the whole answer. We can add or subtract $2\pi$ from any angle and wind up in the same place on the circle. So $i = e^{i \pi /2 + i 2 \pi n}$, where $n$ is any integer. So we end up with a set of values:
$i^i = e^{- \pi /2 - 2 \pi n}$
which are all real!

The curve above goes through the set of values for $i^i$, where $n$ is an integer.
$i^i = \{...,111.32,0.20788,0.00038820,...\}$
But 0.20788… will be looked at as the primary value for now since it is the result when $n=0$, and having one value keeps things simple.

So what happens when we repeat this exponentiation? $i^{i^i}$
First, this can be written in a simpler way:
$^3i = i^{i^i}$
This operation is called “tetration” or “power tower”.
You start from the top and work your way down. So $^3i = i^{i^i} = i^{0.20788}$
Now use the Euler relation again (ignoring $2 \pi n$ terms in the exponential):
$i^{0.20788} = e^{i \pi/2 \times 0.20788} = e^{i 0.32653}$
So the result is complex:
$^3i = 0.94715+0.32076i$

We can keep going using results from the last iteration:
$^4i = i^{0.94715+0.32076 i} = 0.05009+0.60211 i$

Eventually you find that with $^mi$ as $m$ increases, the results start to converge on one number. This implies there’s a value for $^\infty i$. After plugging and chugging we can eventually conclude that:
$^\infty i = 0.43828 + 0.36059 i$
Convergence can be seen in both the real and imaginary components, around $m \approx 70$ if we require the difference between iterations to be less than 0.1%. (See figures below).

Each result has a different radius associated with it, since the imaginary component of the previous result puts a real term in the exponential of the next. This real term can be expressed as a factor $r$:
$e^{i\varphi} = a + bi \rightarrow i^{a+bi} = e^{i \pi/2 (a + bi)} = e^{-b\pi/2}e^{i a \pi/2} = r e^{i a \pi/2}$
The radius can be found by multiplying the result by its complex conjugate and then find the square root (i.e. the Pythagorean theorem).

We can then plot radius of $^mi$ as a function of $m$ (see plot below).

The python code for making these plots is:

import os
import sys
import math
import numpy
import pylab

REAL = []
IMAG = []
ITERA = []

pp = 1j
for n in range(100):
ppL = pp #save previous value for cut off
pp = (1j)**pp #interate power tower
RA = 0.001 #ratio for cut off
if (abs(pp.real - ppL.real)/(ppL.real + 0.00001) < RA) and (abs(pp.imag - ppL.imag)/(ppL.imag + 0.00001) < RA): #establish a cut off
print n
print pp
break
else:

REAL.append(pp.real) #save real component
IMAG.append(pp.imag) #save imaginary component
ITERA.append(n+2) #save iteration "m"

ITERA = [1]+ITERA
REAL = [0]+REAL
IMAG = [1]+IMAG

pylab.figure(1)
pylab.xlabel("m")

# pylab.show()

pylab.figure(2)
pylab.xlabel("m")
pylab.ylabel("imaginary component")
pylab.plot(ITERA,IMAG,'k')

pylab.savefig('Imfig.png')

pylab.figure(3)
pylab.xlabel("m")
pylab.ylabel("real component")
pylab.plot(ITERA,REAL,'k')

pylab.savefig('Refig.png')