Intuition for Euler's Formula

| math

As in my post about the exponential power series, I set out from the fact that the derivative of the exponential function is the exponential function itself.

An educated guess

The trigonometric functions sine and cosine are continuous and periodic and if you cycle through their derivatives, they eventually re-appear. This seems promising. Maybe they could be used to represent the exponential function? $$ \begin{align*} (\cos \theta)' &= -\sin \theta \\ (\sin \theta)' &= \cos \theta \end{align*} $$

Take a look at the unit circle. For any angle \(\theta\), \(\cos \theta\) is the x-value and \(\sin \theta\) the y-value of a point on this circle. Now take the derivative of both functions and stay on the same axis such that you get a new point with \(-\sin \theta\) as x-value and \(\cos \theta\) as y-value. To get to the original point, all you need is a clockwise rotation of \(\pi / 2\) around the origin.

The complex plane

This idea can be represented nicely in the complex plane, where the x-axis represents the real part and the y-axis the imaginary part of a complex number. To get to a point \(z\) on the circle, add the real and imaginary part \((z = \cos \theta + i \sin \theta).\) Now take the derivative of this sum. $$ \frac{d}{d \theta} \left( \cos \theta + i \sin \theta \right) = -\sin \theta + i \cos \theta $$

Multiplication with \(i\) must be a counterclockwise rotation of \(\pi / 2\) around the origin since it takes the point \((1, 0)\) to \((0, i).\) So I multiply both sides by \(1 / i\) to rotate clockwise. $$ \begin{align*} \frac{1}{i} \cdot \frac{d}{d \theta} \left( \cos \theta + i \sin \theta \right) &= \frac{-\sin \theta + i \cos \theta}{i}\\ &= \frac{-\sin \theta}{i} + \cos \theta \\ &= \cos \theta - \frac{\sin \theta (-i)}{i (-i)} \\ &= \cos \theta + i \sin \theta \end{align*} $$

Finding the exponential via the differential equation

Now I'm looking for another function \(f\) such that the following equation holds. $$ \frac{1}{i} \cdot \frac{d f}{d \theta} = f $$

If you know the chain rule, at this point it becomes quite clear that \(f(\theta) = e^{i \theta}\) is such a function because \(\frac{d}{d \theta} e^{i \theta} = e^{i \theta} i.\) So we have two functions that satisfy the same differential equation. $$ \begin{align*} f' &= i e^{i \theta} &= i f\\ g' &= -\sin \theta + i \cos \theta &= i g \end{align*} $$

Proving equality

But are they actually the same? Using the quotient rule to compute the derivative of \(f\) divided by \(g\) , we find that the rate of change of their ratio is always zero. (Dividing by \(g = \cos \theta + i \sin \theta\) is fine since it describes the points on the unit circle, so it is never zero.) $$ \begin{align*} \left( \frac{f}{g} \right)' &= \frac{f' g - f g'}{g^2}\\ &= \frac{ifg - fig}{g^2}\\ &= 0 \end{align*} $$

This means that \(\frac{f}{g}\) must be constant and we can compute this constant by evaluating the ratio at any point. Zero is easy. $$ \begin{align*} \left( \frac{f}{g} \right) (0) &= \frac{e^{i \cdot 0}}{\cos 0 + i \sin 0}\\ &= \frac{1}{1 + 0}\\ &= 1 \end{align*} $$ The ratio is \(1\) for any \(\theta\), so \(f\) and \(g\) must be equal and we end up with Euler's famous formula. $$ e^{i \theta} = \cos \theta + i \sin \theta $$

Update: I added the last section to turn this into a proper derivation.