## Euler’s formula e^ix = cos x + i sin x: a geometric approach

Today I mentioned the famous Euler’s formula briefly in my calculus class (when discussing hyperbolic functions, lecture notes here):

$\displaystyle \begin{array}{rl} \displaystyle \boxed{e^{ix}=\cos x+ i \sin x} \ \ \ \ \ (1)\end{array}$

where ${i\in \mathbb C}$ is a solution to ${z^2=-1}$ (usually denoted by “${i=\sqrt{-1}}$”, but indeed there is no single-valued square root for complex numbers, or even negative real numbers).

One of the usual ways to derive this formula is by comparing the power series of the exponential function and the trigonometric functions ${\sin}$ and ${\cos}$:

$\displaystyle \begin{array}{rl} \displaystyle \begin{cases} e^z=& \displaystyle 1+\frac{z^1}{1!}+\frac{z^2}{2!} +\frac{z^3}{3!} +\frac{z^4}{4!}+\cdots\\ \sin x=& \displaystyle \frac{x^1}{1!} -\frac{x^3}{3!} +\frac{x^5}{5!}-\cdots\\ \cos x=& \displaystyle 1 -\frac{x^2}{2!} +\frac{x^4}{4!}-\cdots \end{cases} \end{array}$

Putting ${z=ix}$ in the first expansion and comparing with the remaining two, it’s easy to see that

$\displaystyle \begin{array}{rl} \displaystyle e^{ix} =& \displaystyle \left(1-\frac{x^2}{2!} +\frac{x^4}{4!}-\cdots\right) +i\left(\frac{x^1}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right)\\ =& \displaystyle \cos x+i\sin x. \end{array}$

Here, I am going to give another approach which does not require any knowledge of series (thus avoid the problem of convergence), but only basic knowledge in complex number (complex addition and multiplication). I am sure this approach must has been taken before but I couldn’t find a suitable reference, especially an online one.

Let us agree that we define the Euler’s number to be

$\displaystyle \begin{array}{rl} \displaystyle e:=\lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n. \end{array}$

From this it is easy to see that

$\displaystyle \begin{array}{rl} \displaystyle e^x=\lim_{n\rightarrow \infty}\left(1+\frac{x}{n}\right)^n\textrm{ for }x\in \mathbb R. \end{array}$

It is then natural to define the complex exponential function by

$\displaystyle \begin{array}{rl} \displaystyle e^z :=\lim_{n\rightarrow \infty}\left(1+\frac{z}{n}\right)^n\textrm{ for }z\in \mathbb C. \end{array}$

Here I am cheating a little bit because I have implicitly assumed that this limit exists.

Now recall the geometry of the complex plane. We can identify a complex number ${x+iy}$ with the point ${(x, y)}$ on the plane. We can write a complex number in its polar form ${z=r(\cos \theta+i \sin\theta)}$, which is identified with ${(r, \theta)}$ in polar coordinates. We call ${r=|z|}$ and ${\theta=\mathrm{arg}(z)}$ the modulus (or length) and the argument (or angle) of ${z}$ respectively.

The complex plane

The complex multiplication of ${z_1=r_1(\cos \theta_1+i\sin \theta_1)}$ and ${z_2=r_2(\cos \theta_2+i\sin \theta_2)}$ is then

$\displaystyle \begin{array}{rl} \displaystyle z_1z_2 =r_1r_2\left(\cos (\theta_1+\theta_2)+i\sin (\theta_1+\theta_2)\right). \end{array}$

i.e. the modulus of ${z_1z_2}$ is the product of the two moduli and the argument of ${z_1z_2}$ is the sum of the two arguments.

So now, let’s fix ${x\in \mathbb R}$ and compute

$\displaystyle \begin{array}{rl} \displaystyle \lim_{n\rightarrow \infty}\left(1+\frac{ix}{n}\right)^n, \end{array}$

which by definition would be ${e^{ix}}$. We will argue that its length is ${1}$ and its argument is ${x}$, i.e. (1) holds:

$\displaystyle \begin{array}{rl} \displaystyle e^{ix}=\cos x+i \sin x. \ \ \ \ \ (2)\end{array}$

The geometry of the powers of 1+yi

To see this, let ${z_n=1+\frac{ix}{n}}$. Then ${|z_n|=\left(1+\frac{x^2}{n^2}\right)^{\frac{1}{2}}}$ and so

$\displaystyle \begin{array}{rl} \displaystyle |z_n|^n=\left(1+\frac{x^2}{n^2}\right)^{\frac{n}{2}} \end{array}$

From this we have

$\displaystyle \begin{array}{rl} \displaystyle |({z_n})^n| =|z_n|^n =\left(1+\frac{x^2}{n^2}\right)^{\frac{n}{2}} =& \displaystyle \left[\left(1+\frac{x^2}{n^2}\right)^{n^2}\right]^{\frac{1}{2n}}\\ \rightarrow& \displaystyle \left(e^{x^2}\right)^{0}=1 \ \ \ \ \ (3)\end{array}$

as ${n\rightarrow \infty}$. On the other hand, the argument of ${z_n}$ (which is well-defined up to a multiple of ${2\pi}$) can be chosen to be

$\displaystyle \begin{array}{rl} \displaystyle \arg(z_n)=\theta_n=\tan^{-1}\left(\frac{x}{n}\right). \end{array}$

Then by the L’Hôpital’s rule,

$\displaystyle \begin{array}{rl} \displaystyle \lim_{n\rightarrow \infty} n\theta_n= \lim_{n\rightarrow \infty} n\tan^{-1}\left(\frac{x}{n}\right) =\lim_{t\rightarrow 0^+} \frac{\tan^{-1}(tx)}{t} =\lim_{t\rightarrow 0^+} \frac{x}{1+t^2x^2}=x. \end{array}$

So we have

$\displaystyle \begin{array}{rl} \displaystyle \lim_{n\rightarrow \infty} \arg [(z_n)^n ]= \lim_{n\rightarrow \infty} n\arg(z_n)= \lim_{n\rightarrow \infty} n\theta_n=x. \ \ \ \ \ (4)\end{array}$

Combining (3) and (4), we have

$\displaystyle \begin{array}{rl} \displaystyle e^{ix}:=\lim_{n\rightarrow \infty}\left(1+\frac{ix}{n}\right)^n=\lim_{n\rightarrow \infty}(z_n)^n= \cos x+i\sin x. \end{array}$

I found a video explaining $e^{i\pi}=-1$ (but without giving the full mathematical details) in the above approach: