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\begin{document}
\noindent
{\sc Notes on Imaginary Exponents}\\
Eric Rasmusen, erasmuse61@gmail.com\\
http:$\backslash\backslash$rasmusen.org$\backslash$papers$\backslash$imaginary\_exponents.pdf\\
February 26, 2020.
What is $10^i$? We know that $10^3 = 10 \cdot 10 \cdot 10$, and after some effort we can figure out that logic requires $10^{-3} = \frac{1}{10^3}$ and $10^{1/3} = \sqrt[3]{10}$. But how do we even get started with an imaginary number $i$ in place of 3?
Let's take a first step which makes things a bit more complicated but will help in the end. Note that $10^i = e^{log(10) i}, $ where $e$ is Euler's number, special because if $f(x) = e^x$, $f'(x) = e^x$ also: the function $f$'s slope is the same as its height. So if we can figure out what $e^{i log (10)}$ should be, we've figured out $10^i$ too.
So let's see if we can calculate what $e^{im}$ is, for any number $m$.
We can write the answer as a complex number $ e^{im} = g(m) + i h(m) $, where maybe $h=0$ or $g=0$, and this makes our goal to find the functions $g(m)$ and $h(m)$.
We want this to be consistent with our existing mathematics. One part of our existing mathematics is that if $f(m) = e^{a m}$, then $f’(m) = a e^{am}$. Let $a= i$, and we have $f(m) = e^{i m}$ and $f’(m) = i e^{im}$.
Using our $g$ and $h$ functions, we also have $ e^{im} = g(m) + i h(m) $ so
\begin{equation}
f’(m) = i e^{im} = i g(m) + i^2 h(m) = ig(m) - h(m).
\end{equation}
But we could also write the derivative as
\begin{equation}
f'(m)= g'(m) + i h'(m)
\end{equation}
Putting together these last two ways to write $f'(m)$, we see that $g'(m) = - h(m)$ and $ ih'(m) = i g(m)$, so $g''(m) = -h''(m) = -g(m)$. There is only one pair of functions whose derivatives flip around like that: $g(m) = cos(m)$ (with $ g'(m) = -sin(m)$) and $h(m) = sin(m)$ (with $h'(m) = cos(m)$). Going back to $ e^{im} = g(m) + i h(m), $ we get Euler's Formula,
\begin{equation}
e^{im} = cos(m) + sin (m) i.
\end{equation}
This works out quite nicely, because we also, for consistency, want $e^{0} = 1$, and indeed, $e^{0 i} = cos (0) + sin(0) i = 1 + (0) i = 1.$
Our formula for imaginary exponents thus tells us that $e^{im}$ starts with $e^{i \cdot 0 } =1$ and then starts going counterclockwise around the unit circle in the imaginary plane with real numbers on the $x-axis$ and imaginary numbers on the vertical axis. It's on the unit circle because the coefficients are $(x, y) = (cos(m), sin(m))$ and from trigonometry we know $cos^2(m) +sin^2(m) = 1$, which means $x^2+y^2=1$, the equation for the unit circle. As $m$ gets bigger, $e^{im}$ just keeps going around the circle, over and over.
Speaking of going around the ciricle, one thing remains. We need to define units for $m$, which is a measure of angle size. If there are 360 degrees to the circle, then $e^ {180i} = cos (180) + sin (180) i = -1 + (0) i = -1.$ If, instead, there are 720 degrees to the circle, then $e^ {180i} = cos (180) + sin (180) i = 0 + (1) i = i.$ The convention is to use $2 \pi$ radians to the circle, so $e^ { \pi i} = cos (\pi ) + sin (\pi) i = -1$, which is Euler’s Identity. Note that we could start off and make any real number we like to produce a simple exponent with $i$. We could, for example, start with $20^i = -1$. If we do that, we are in effect assuming that $e^{log(20) i } = -1$, so we are making the very ugly assumption that there are $ 2*log(20)$ degrees in the circle. We would end up with $40^i =-1$ too, though, and $60^i =-1$, and so forth.
\newpage
Figure 1 illustrates this. It show the imaginary plane, and various exponents, with $m$ as an angle, the real part of the x-axis, and the imaginary part on the y-axis. Figure 2 shows the same information, but with $m$ on the horizontal axis instead of as an angle, and $m^i$ on the vertical axis, where the vertical axis has two curves, for the real part and the imaginary part of $m^i$.
\begin{center}
{\sc Figure 1:
$f(m) = e^{im}$, in real-imaginary space} \label{circle.png}
\includegraphics[width=4in]{circle.png}
\end{center}
%Send to David Joyce.
\begin{center}
{\sc Figure 2:
$f(m) = e^{im} $, in m-Real and m-Imaginary Space} \label{circle.png}
\includegraphics[width=3in]{sinewave.png}
\end{center}
A second way to derive imaginary exponents is to start with a Taylor Series for $e^{x}$ around 0, adapt it to $e^{ix}$, and then show that the resulting series is the sum of Taylor series's for $sin(x)$ and $cos(x)$. In general, a Taylor Series around $x=0$ is
\begin{equation}
\begin{array}{ll}
f(x) & = f(0) + \frac{f'(0)}{1!} (x-0) + \frac{f''(0)}{2!} (x-0)^2 + \frac{f'''(0)}{3!} (x-0)^3 + \frac{f''''(0)}{4!} (x-0)^4 +\ldots\\
\end{array}
\end{equation}
If $f(x) = e^x$, then $f'x) = e^x$, $f''(x) = e^x$, and so forth. Since $e^0 = 1$, this means $f'(0) =1 , f''(0) =1, f'''(0) =1$, and so forth. Thus,
\begin{equation}
\begin{array}{ll}
e^x & = e^0 + \frac{e^0}{1!} x+ \frac{e^0}{2!} x^2 + \frac{e^0}{3!} x^3 + \frac{e^0}{4!}x^4 + \frac{e^0}{5!} x^5+\ldots\\
& \\
&= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} +\frac{x^5}{5!} + \ldots\\
\end{array}
\end{equation}
Put in $ix$ instead of $x$, and we get
\begin{equation}
\begin{array}{ll}
e^{ix}& = 1 + ix + \frac{i^2 x^2}{2!} + \frac{i^3x^3}{3!} + \frac{i^4x^4}{4!} + \frac{i^5x^4}{5!} + \ldots\\
&\\
&= 1 + ix - \frac{ x^2}{2!} - \frac{ i x^3}{3!} + \frac{ x^4}{4!}+ \frac{i x^4}{5!} + \ldots \\
&\\
&= ( 1 - \frac{ x^2}{2!} + \frac{ x^4}{4!} + \ldots) + i (x - \frac{ x^3}{3!} + \frac{ x^4}{5!}
+ \ldots ) \\
\end{array}
\end{equation}
That last grouping of the real terms and the imaginary terms is important because it corresponds to Taylor Series around 0, for $cos(x)$ and for $sin(x)$. First, noting that $cos(0) =1$ and $sin(0) = 0$, and $\frac{d}{dx} cos (x) = - sin(x)$ and $\frac{d}{dx} sin (x) = cos(x)$,
\begin{equation}
\begin{array}{ll}
cos (x) & = cos(0) - sin (0) x - \frac{ cos(0) x^2}{2!} + \frac{ sin (0) x^3}{3!} + \frac{cos(0) x^4}{4!} - \frac{ sin (0) x^5}{5!} + \ldots\\
&\\
& = 1 - (0)x - \frac{ (1) x^2}{2!} + \frac{(0) x^3}{3!} + \frac{ (1) x^4}{4!} - \frac{ (0) x^5}{5!} + \ldots\\
&\\
& = 1 - \frac{ x^2}{2!} + \frac{ x^4}{4!} + \ldots\\
\end{array}
\end{equation}
Thus, $cos(x)$ is the first group in our $e^{ix}$ Taylor Series. Doing the Taylor Series for $sin(x)$, we get
\begin{equation}
\begin{array}{ll}
sin(x) & = sin(0) + cos (0) x - \frac{ sin(0) x^2}{2!} - \frac{ cos (0) x^3}{3!} + \frac{sin(0) x^4}{4!} + \frac{ cos(0) x^5}{5!} + \ldots\\
&\\
& = 0 +(1)x - \frac{ (0) x^2}{2!} - \frac{(1) x^3}{3!} + \frac{ (0) x^4}{4!} +\frac{ (1) x^5}{5!} + \ldots\\
&\\
& = x - \frac{ x^3}{3!} +\frac{ x^5}{5!} + \ldots\\
\end{array}
\end{equation}
Thus, $ i sin(x)$ is the second group in our $e^{ix}$ Taylor Series. We can conclude that
\begin{equation}
e^{ix} = cos(x) + i sin(x),
\end{equation}
which is Euler's formula.
This second approach is more complicated, because we not only need to use the facts that $f''(x) = f(x)$ for $f(x) = e^x$ and that $f''(x) = -f(x)$ for $f(x) = sin(x)$ and $f(x) =cos(x)$, but also the idea of the Taylor series.
\end{document}