Euler's Formula

Just to be clear, this is the proof for the formula $e^{ix} = \cos{x}+i\sin{x}$, not the other weird equation with the same name involving polyhedrons. Let's start by considering the expression $\frac{1}{1+x^2}$. If you were asked to integrate this expression with respect to $x$, $tan^{-1}{x}+C$ is probably the first thing that comes to mind, but what if you wanted to be a rebel and integrate it another way? The only problem you would have is that annoying second degree polynomial in the denominator. Let's see if using imaginary numbers helps:

\begin{equation}
\frac{1}{1+x^2} = \frac{1}{2} \frac{1-ix+1+ix}{1+x^2}
\end{equation}
\begin{equation}
\frac{1}{1+x^2} = \frac{1}{2} \frac{1-ix+1+ix}{(1+ix)(1-ix)}
\end{equation}
\begin{equation}
\frac{1}{1+x^2} = \frac{1}{2} \left(\frac{1}{1+ix}+\frac{1}{1-ix}\right)
\end{equation}

And now, if we integrate both sides with respect to $x$:

\begin{equation}
\int\frac{1}{1+x^2}dx = \frac{1}{2} \int\left(\frac{1}{1+ix} + \frac{1}{1-ix}\right)dx
\end{equation}
\begin{equation}
\tan^{-1}{x} = \frac{-i}{2} [\text{Log}(1+ix)-\text{Log}(1-ix)] + C
\end{equation}

In case you didn't know, $\text{Log}{z}$ is the complex logarithm of $z$, defined by $\exp{\text{Log}z}=z$ where $z$ is any complex number. (To prove that $\frac{d}{dz}\text{Log}z = z^{-1}$, simply differentiate both sides of the statement, $\exp{\text{Log}z}=z$, using the chain rule and solve for $\frac{d}{dz}\text{Log}z$.)

\begin{equation}
2i\tan^{-1}{x}-2iC = \text{Log}(1+ix) - \text{Log}(1-ix)
\end{equation}
\begin{equation}
\exp\left(2i\tan^{-1}{x} - 2iC\right) = \exp[\text{Log}(1+ix)-\text{Log}(1-ix)]
\end{equation}
\begin{equation}
\exp\left(2i\tan^{-1}{x} - 2iC\right) = \frac{1+ix}{1-ix}
\end{equation}

That is an interesting relation, but can we simplify even further? Recall that $\cos{\tan^{-1}{x}} = 1/\sqrt{1+x^2}$ and $\sin{\tan^{-1}{x}} = x/\sqrt{1+x^2}$. (These trigonometric identities can be derived by considering a right triangle with an angle of $\tan^{-1}{x}$, an opposite leg length of $x$, an adjacent leg length of $1$ and a hypotenuse length of $\sqrt{1+x^2}$.) If we manipulate equation $(8)$ a little, these trigonometric identities can come in handy.

\begin{equation}
\exp\left(2i\tan^{-1}{x} - 2iC\right) = \frac{(1+ix)^2}{1+x^2}
\end{equation}

Let $A=\pm{e^{-iC}}$.

\begin{equation}
A\exp\left( i\tan^{-1}{x} \right) = \frac{1+ix}{\sqrt{1+x^2}}
\end{equation}
\begin{equation}
A\exp\left( i\tan^{-1}{x} \right) = \cos\tan^{-1}{x} + i\sin\tan^{-1}{x}
\end{equation}
\begin{equation}
Ae^{ix}=\cos{x}+i\sin{x} ,\qquad -\frac{\pi}{2} < x < \frac{\pi}{2} \end{equation}

If we substitute in $0$ for $x$, we find that $A=1$.

\begin{equation}
e^{ix}= \cos{x}+i\sin{x} ,\qquad -\frac{\pi}{2} < x < \frac{\pi}{2} \end{equation}

Now, to show that $e^{ix} = \cos{x}+i\sin{x}$ is true for all $x$, note that both $y=e^{ix}$ and $y = \cos{x}+i\sin{x}$ are solutions to the first-order differential equation $\frac{dy}{dx} = iy$ for all $x$, and both solutions share the point $(0,1)$. Therefore, the two solutions are equal for all $x$.

\begin{equation}
e^{ix} = \cos{x}+i\sin{x}
\end{equation}

And that's Euler's formula. This equation is quite important when dealing with stuff like Fourier Transforms and is an active ingredient in quantum mechanical wave functions, because it allows you to write complex wave forms as exponential functions.