Euler's Formula

Proof 1: By Taylor series

Here are the Maclaurin expansions ¹ for $\cos(x), \sin(x),$ and $e^y$.

$$\begin{aligned} \cos(x) &= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots\\ \sin(x) &= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots\\ e^y &= \sum_{n=0}^{\infty} \frac{y^n}{n!} = 1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \frac{y^4}{4!} + \cdots\\ \end{aligned}$$

Letting $y = ix$, we get ²

$$\begin{aligned} e^{ix} = \sum_{n=0}^{\infty} \frac{(ix)^n}{n!} = \sum_{n=0}^{\infty}i^n\frac{x^n}{n!} &= 1 + ix - \frac{x^2}{2!} - i \frac{x^3}{3!} + \frac{x^4}{4!} + i \frac{x^5}{5!} - \frac{x^6}{6!} + \cdots\\ &= \left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots\right) + i\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots\right)\\ &= \cos(x) + i\sin(x) \quad \blacksquare \end{aligned}$$

Lemma 1: The magnitude of $e^{ix}$ is $1 \quad \forall x \in \mathbb{R}$

Proof. We have shown that $e^{ix}$ is a complex number $a+bi$, where $a=\cos x$ and $b=\sin x$. The magnitude of a complex number is its distance, on the complex plane, from $(0, 0)$. We find this distance using the distance formula. $$|e^{ix}| = |\cos(x) + i\sin(x)\:| = \sqrt{\cos^2(x) + \sin^2(x)} = \sqrt{1} = 1 \quad \blacksquare$$

The significance of this result is that on the complex plane ³, $e^{i\theta}$ represents a point that is a distance of one unit away from the origin $(0, 0)$. Varying $\theta$ from $0$ to $2 \pi$ results in the unit circle.

The $\theta$ parameter measures the counterclockwise angle the point makes with the positive x axis. At this point, since we’re determining a point’s position by its angle, we should wonder how this could be connected with polar coordinates.

A polar coordinate requires an extra parameter, $r$. Otherwise, we are stuck with only the unit circle and can’t represent points a distance of, for example, 2 units from the origin. We simply add that parameter $r$ by multiplying by $r$.

Lemma 2 (polar coordinates): For all points $x+iy$ on the complex plane, there exist $r \in \mathbb{R}^{\geq 0}$ and $\theta \in [0, 2\pi)$ such that $re^{i\theta} = r(\cos \theta + i \sin \theta) = r\cos \theta + i(r\sin \theta) = x+iy$

Proof 2: By differential equation

Consider the following differential equation: $$\frac{df(x)}{dx} = if(x)$$

One solution is $f(x) = e^{ix}$ because $$\frac{d}{dx}e^{ix} = ie^{ix}$$

Another solution is $f(x) = \cos(x)+i\sin(x)$ because$$\frac{d}{dx}\left(\cos(x)+i\sin(x)\right) = -\sin(x)+i\cos(x) = i\left(\cos(x)+i\sin(x)\right)$$

and both functions satisfy $f(0) = 1$: $$f(0) = e^{i(0)} = \cos(0) + i\sin(0) = 1$$

Hence, the functions are identical; that is, $e^{ix} = \cos(x)+i\sin(x) \quad \blacksquare$

Lemma 3 (Euler’s identity): $e^{\pi i} = -1$

Proof.

$$ \begin{aligned} e^{\pi i} &= \cos(\pi) + i\sin(\pi) \\ &= -1 + i(0)\\ &= -1\\ & \blacksquare \end{aligned} $$

Lemma 4: $\ln(-1)=i \pi$

Proof. Take the natural log of both sides of Euler’s identity: $$ e^{\pi i} = -1 \implies \pi i = \ln(-1) \quad \blacksquare$$

So then $\ln(-23) = \ln(-1) + \ln(23) = \pi i + \ln(23)$.