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: Magnitude of exp(ix)

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 \(re^{i\theta} = r(\cos \theta + i \sin \theta) = r\cos \theta + i(r\sin \theta) = x+iy\).

Essentially, \(re^{i\theta}\) represents a polar coordinate with radius \(r\) and angle \(\theta\) on the complex plane.

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)

\[\ln(-1)=i \pi\]

Proof. Take the natural log of both sides of Euler’s identity:

\[e^{\pi i} = -1 \implies \pi i = \ln(-1) \quad \blacksquare\]

Then, for example, \(\ln(-23) = \ln(-1) + \ln(23) = \pi i + \ln(23)\).