Euler's Identity from first principles

Why does

𝑒^{𝑖 𝑡} = cos (𝑡) + 𝑖 sin (𝑡) ?

This question has puzzled many math learners and the usual proof based on Taylor series explains nothing. There’s a nicer proof at https://en.wikipedia.org/wiki/Euler%27s_formula#Using_polar_coordinates, but that also kind of feels wrong since it already assumes the existence of $cos$ and $sin$ , which I don’t want to do here. But my proof is based on that, so I had to mention it.

Anyway, imagine we’re defining complex exponentials for the first time.

We don’t know what $𝑒^{𝑖 𝑡}$ might be, but it will surely be a complex number of some sort, so we can separate it into real and imaginary parts:

𝑓 (𝑡) ≔ 𝑒^{𝑖 𝑡} = 𝑥 (𝑡) + 𝑖 𝑦 (𝑡)

where $𝑥 ()$ and $𝑦 ()$ are some yet-unknown functions.

Whatever this new exponential is, it should satisfy

\frac{𝑑}{𝑑 𝑡} 𝑒^{𝑖 𝑡} = 𝑖 𝑒^{𝑖 𝑡}

Otherwise it wouldn’t really have the right to call itself $𝑒^{𝑖 𝑡}$ , since this rate of change equation is basically the core of what $𝑒$ is all about.

Rewriting this a bit, we can see that

𝑓^{'} (𝑡) = 𝑖 𝑓 (𝑡) = 𝑖 𝑥 (𝑡) - 𝑦 (𝑡) .

and therefore

\begin{aligned} 𝑦^{'} (𝑡) & = 𝑥 (𝑡), \\ 𝑥^{'} (𝑡) & = - 𝑦 (𝑡) . \end{aligned}

Now I’ll admit that this next step is a bit unmotivated; In a real discovery scenario, you might arrive at this after trying several other things.

Let’s look at the distance of $𝑒^{𝑖 𝑡}$ from the origin:

𝑔 (𝑡) ≔ {| 𝑓 (𝑡) |}^{2} = {𝑥 (𝑡)}^{2} + {𝑦 (𝑡)}^{2}

Its derivative is

\begin{aligned} 𝑔^{'} (𝑡) & = 2 𝑥 (𝑡) 𝑥^{'} (𝑡) + 2 𝑦 (𝑡) 𝑦^{'} (𝑡) \\ = - 2 𝑥 (𝑡) 𝑦 (𝑡) + 2 𝑦 (𝑡) 𝑥 (𝑡) \\ = 0 . \end{aligned}

Together with $𝑔 (0) = {| 𝑓 (0) |}^{2} = {| 1 |}^{2} = 1$ , this tells us that

| 𝑒^{𝑖 𝑡} | \equiv 1,

so $𝑒^{𝑖 𝑡}$ always lies on the unit circle.

To find out where on the circle $𝑒^{𝑖 𝑡}$ is, we calculate the distance travelled from $𝑡 = 0$ to $𝑡 = 𝑡_{1}$ :

𝐷 = \int_{0}^{𝑡_{1}} | 𝑓^{'} (𝑡) | 𝑑 𝑡 .

Note that

| 𝑓^{'} (𝑡) | = | 𝑖 𝑒^{𝑖 𝑡} | = | 𝑒^{𝑖 𝑡} | = 1,

so we just get

𝐷 = \int_{0}^{𝑡_{1}} 𝑑 𝑡 = 𝑡_{1} .

Together with the fact that $𝑓^{'} (0) = 𝑖$ , we can thus deduce that $𝑓 (𝑡)$ is a length of $𝑡$ units counterclockwise around the circle, starting at $𝑓 (0) = 1$ .

With this, we’re done. We now know exactly what $𝑓 (𝑡)$ is. For example, $𝑖$ is on the unit circle and the arc from 1 to it has length $\frac{𝜏}{4}$ . This means that

𝑒^{𝑖 \frac{𝜏}{4}} = 𝑖 .

Similarly, $- 1$ is also on the unit circle and the arc from 1 to it has length $\frac{𝜏}{2}$ , which gives us the famous¹

𝑒^{𝑖 \frac{𝜏}{2}} = - 1 .

Of course, the way we’ve derived it here also tells us that $𝑒^{𝑖 𝜏}$ means going a length of $𝜏$ around the circle, and since $𝜏$ is the full circumference, we have

𝑒^{𝑖 𝜏} = 1,

so we even get the periodicity for free.

To conclude, we just define $cos (𝑡) ≔ 𝑥 (𝑡)$ and $sin (𝑡) ≔ 𝑦 (𝑡)$ , which can be justified for $𝑡 \in [0, \frac{𝜏}{4}]$ by looking at the triangle made from $(0, 0)$ , $(𝑥 (𝑡), 0)$ and $(𝑥 (𝑡), 𝑦 (𝑡))$ , but for other $𝑡$ this is really the definition of $cos$ and $sin$ and thus explains why they’re periodic.

Tho for some historical reason people seem to prefer to use $𝜋 = \frac{𝜏}{2}$ here. ↩︎