# 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.