How do I interpret Euler's formula?

I like to think of Euler's formula as an algebraic result which can then be interpreted geometrically. The fact that the power series for $e^{ix}$, $\cos{x}$ and $\sin{x}$ match up has already been mentioned. Here is (I hope) some more convincing evidence:

1. if $f(x)=e^{ix}$ and $g(x)=\cos{x}+i\sin{x}$, $f(0)=1$ and $g(0)=1$.
2. $f'(x)=if(x)$ and $g'(x)=ig(x)$. (This is pretty easy to verify). Since $f$ and $g$ are both differentiable everywhere, they have the same value at $x=0$ and they are solutions to the same first order linear differential equation, they must be the same function.

So that's one way of looking at it, but it suffers from having to know about the existence and uniqueness of solutions to differential equations.

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Another way to look at it algebraically is using trig identities:

• $e^{i(a+b)}=e^{ia}e^{ib}$, or equivalently $f(a+b)=f(a)f(b)$

We can show that $g(x)$ has the same property by using the identities $\cos{(a+b)}=\cos{a}\cos{b}-\sin{a}\sin{b}$, and
$\sin{(a+b)}=\sin{a}\cos{b}+\cos{a}\sin{b}$.

• $g(a+b)=\cos{(a+b)}+i\sin{(a+b)}=\cos{a}\cos{b}-\sin{a}\sin{b}+i\sin{a}\cos{b}+i\cos{a}\sin{b}$
• $g(a)g(b)=\left(\cos{a}+i\sin{a}\right)\left(\cos{b}+i\sin{b}\right)$,

If we multiply this out it becomes the same thing as $g(a+b)$.

You can try this with any set of trig identities you like, and it will always work out that $f$ and $g$ have the same properties.

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If you buy all the algebraic evidence that $e^{ix}$ and $\cos{x}+i\sin{x}$ are really the same thing, then the geometric interpretation falls out for free.

• $\cos{\theta}+i\sin{\theta}$ is a point on the unit circle in the complex plane, at an $\theta$ counterclockwise from the positive real axis.
• By Euler's identity, $e^{i\theta}$ is the exact same thing, and can be interpreted in the exact same way.

Euler’s identity

$$e^{ix}=\cos x+i\sin x$$

is related to the following geometric interpretation

and can be proved by series

$$\begin{align} e^{i\theta} &{}= 1 + i\theta + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \frac{(i\theta)^5}{5!} + \frac{(i\theta)^6}{6!} + \frac{(i\theta)^7}{7!} + \frac{(i\theta)^8}{8!} + \cdots \\[8pt] &{}= 1 + i\theta - \frac{\theta^2}{2!} - \frac{i\theta^3}{3!} + \frac{\theta^4}{4!} + \frac{i\theta^5}{5!} - \frac{\theta^6}{6!} - \frac{i\theta^7}{7!} + \frac{\theta^8}{8!} + \cdots \\[8pt] &{}= \left( 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \frac{\theta^8}{8!} - \cdots \right) + i\left( \theta- \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \cdots \right) \\[8pt] &{}= \cos \theta + i\sin \theta . \end{align} $$

Maybe this is not answering your question, but I remember reading that Euler discovered some of the connection between $e$ and $i$ through this integral:

$$\int_{0}^{\infty}\frac{dx}{x^2+1}=\frac{\pi}{2}$$

We can also write $\frac{1}{x^2+1}=\frac{1}{2i}\left (\frac{1}{x-i}-\frac{1}{x+i}\right )$. This yields: $$\frac{1}{2i}\int_{0}^{\infty}\frac{1}{x-i}-\frac{1}{x+i}dx=\left [\frac{1}{2i}\ln\left (\frac{x-i}{x+i}\right)\right ]^{\infty}_{0}=\frac{1}{2i}\ln(-1)=\frac{\pi}{2}$$ So $\ln(-1)=\pi i$, or $$e^{\pi i}=-1$$