Euler's Identity from first principles

Why does$${𝑒}^{𝑖𝑡}=\mathrm{<mtext>cos</mtext>}\left(𝑡\right)+𝑖\mathrm{<mtext>sin</mtext>}\left(𝑡\right)?$$

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 $\mathrm{<mtext>cos</mtext>}$ and $\mathrm{<mtext>sin</mtext>}$, 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 ${𝑒}^{𝑖𝑡}$ might be, but it will surely be a complex number of some sort, so we can separate it into real and imaginary parts:$$𝑓\left(𝑡\right)≔{𝑒}^{𝑖𝑡}=𝑥\left(𝑡\right)+𝑖𝑦\left(𝑡\right)$$where $𝑥\left(\right)$ and $𝑦\left(\right)$ are some yet-unknown functions.

Whatever this new exponential is, it should satisfy$$\frac{𝑑}{𝑑𝑡}{𝑒}^{𝑖𝑡}=𝑖{𝑒}^{𝑖𝑡}$$Otherwise it wouldn’t really have the right to call itself ${𝑒}^{𝑖𝑡}$, since this rate of change equation is basically the core of what $𝑒$ is all about.

Rewriting this a bit, we can see that$${𝑓}^{\prime }\left(𝑡\right)=𝑖𝑓\left(𝑡\right)=𝑖𝑥\left(𝑡\right)-𝑦\left(𝑡\right).$$and therefore$$\begin{array}{cc}{𝑦}^{\prime }\left(𝑡\right)& =𝑥\left(𝑡\right),\\ {𝑥}^{\prime }\left(𝑡\right)& =-𝑦\left(𝑡\right).\end{array}$$

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 ${𝑒}^{𝑖𝑡}$ from the origin:$$𝑔\left(𝑡\right)≔{\left|𝑓\left(𝑡\right)\right|}^2={𝑥\left(𝑡\right)}^2+{𝑦\left(𝑡\right)}^2$$Its derivative is$$\begin{array}{cc}{𝑔}^{\prime }\left(𝑡\right)& =2𝑥\left(𝑡\right){𝑥}^{\prime }\left(𝑡\right)+2𝑦\left(𝑡\right){𝑦}^{\prime }\left(𝑡\right)\\ & =-2𝑥\left(𝑡\right)𝑦\left(𝑡\right)+2𝑦\left(𝑡\right)𝑥\left(𝑡\right)\\ & =0.\end{array}$$Together with $𝑔\left(0\right)={\left|𝑓\left(0\right)\right|}^2={\left|1\right|}^2=1$, this tells us that$$\left|{𝑒}^{𝑖𝑡}\right|\equiv 1,$$so ${𝑒}^{𝑖𝑡}$ always lies on the unit circle.

To find out where on the circle ${𝑒}^{𝑖𝑡}$ is, we calculate the distance travelled from $𝑡=0$ to $𝑡={𝑡}_1$:$$𝐷={\int }_0^{{𝑡}_1}\left|{𝑓}^{\prime }\left(𝑡\right)\right|𝑑𝑡.$$Note that$$\left|{𝑓}^{\prime }\left(𝑡\right)\right|=\left|𝑖{𝑒}^{𝑖𝑡}\right|=\left|{𝑒}^{𝑖𝑡}\right|=1,$$so we just get$$𝐷={\int }_0^{{𝑡}_1}𝑑𝑡={𝑡}_1.$$Together with the fact that ${𝑓}^{\prime }\left(0\right)=𝑖$, we can thus deduce that $𝑓\left(𝑡\right)$ is a length of $𝑡$ units counterclockwise around the circle, starting at $𝑓\left(0\right)=1$.

With this, we’re done. We now know exactly what $𝑓\left(𝑡\right)$ 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 ${𝑒}^{𝑖𝜏}$ means going a length of $𝜏$ around the circle, and since $𝜏$ is the full circumference, we have$${𝑒}^{𝑖𝜏}=1,$$so we even get the periodicity for free.

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