Around 1740, Leonard Euler discovered a formula that connected functions of complex arguments to trigonometric functions, effectively forming a link between analytic functions and geometric functions which eventually extended to topology, differential equations, and mathematical physics. All of this began with one simple formula, lauded by *Richard Feynman* as “the most remarkable formula in mathematics,” and it is

$$e^{ix}=\cos(x)+i\sin(x), \text{ with } x\in\mathbb{R}. $$

Here, we aim to prove the above formula. In 1702, *Johann Bernoulli* discovered the relationship

$\displaystyle \frac{1}{1+x^2} = \frac{1}{(1+ix)(1-ix)} =\frac{1}{2}\left(\frac{1}{1+ix}+\frac{1}{1-ix}\right),$

which was instrumental for Gauss as mathematicians at the time were unable to solve the integral

$\displaystyle \int^x\frac{dx}{1+x^2}.$

With Bernoulli’s relationship, the integral can be rewritten as

$\displaystyle \int^x\frac{dx}{1+x^2}=\frac{1}{2}\left(\int^x\frac{dx}{1+ix}+\int^x\frac{dx}{1-ix}\right).$

Now, these integrals can be evaluated by a change of variables to obtain the solution as a combination of complex logarithms. Clearly, we have

$\displaystyle \frac{1}{2}\left(\int^x\frac{dx}{1+ix}+\int^x\frac{dx}{1-ix}\right)=\frac{1}{2i}\text{Log}(1+ix)-\frac{1}{2i}\text{Log}(1-ix).$

Definition $-$ The complex principle logarithm is defined as

$\displaystyle \text{Log}(z)=\ln\|z\|+i\theta,\text{ where } -\frac{\pi}{2} \le \theta \doteq \text{Arg}(z) < \frac{\pi}{2}.$

Now, we switch to a geometric interpretation to determine the form of $\|z\|$ and $\text{Arg}(z)$. In fact, $\|z\|$ is the hypotenuse of the right triangle shown below, while $\text{Arg}(z)$ is the angle between the horizontal and the hypotenuse and can be obtained by considering $\arctan(b/a)$ since $z=a+ib.$

The geometric intrepretation for $z=1+ix$ may be seen here. The quantities of interest are $||z||=1+x^2$ and $\theta=\arctan(x/1).$ Returning to the logarithms, we have that

$$\displaystyle \begin{align*} \frac{1}{2i}\text{Log}(1+ix)-\frac{1}{2i}\text{Log}(1-ix) &= \frac{1}{2i} (\ln(1+x^2)+i\arctan(x)) \\ & -\frac{1}{2i} (\ln(1+x^2)+i\arctan(-x)).\end{align*}$$

Since $\arctan(-x)=-\arctan(x)$, we have that

$\displaystyle \frac{1}{2i}\text{Log}(1+ix)-\frac{1}{2i}\text{Log}(1-ix)$

$\displaystyle =\frac{1}{2i}(\ln(1+x^2)-\ln(1+x^2)+i\arctan(x)+i\arctan(x)).$

By the properties of the Log, we can collapse the left hand side

$\displaystyle \frac{1}{2i} \text{Log}\left(\frac{1+ix}{1-ix}\right)=\arctan(x).$

Collecting the complex parts to the right hand side and exponentiating, we obtain

$\displaystyle \frac{1+ix}{1-ix}=\exp(2i\arctan(x)).$

Here, we have to be a bit more clever to realize that

$\displaystyle \cos(\arctan(x))=\frac{1}{\sqrt{1+x^2}}\text{ and } \sin(\arctan(x)) = \frac{x}{\sqrt{1+x^2}}.$

Then, we can write

$\displaystyle \exp(2i\arctan(x))=\frac{1+ix}{1-ix}=\frac{(1+ix)^2}{1+x^2}=\cos^2(\arctan(x))(1+ix)^2.$

When we take the square root of both sides, we take the positive principle branch cut, as dictated by the principle logarithm, in order to obtain

$\exp(i\arctan(x))=\cos(\arctan(x))(1+ix)= \cos(\arctan(x)) + ix \cos(\arctan(x)).$

With the simple relation, $x\cos(\arctan(x))=\sin(\arctan(x))$, and our definition, $\theta=\arctan(x)$, we arrive at our desired result:

$\exp(i\theta)=\cos(\theta)+i\sin(\theta).$

A beautiful consequence of this is $\displaystyle \exp(i\theta)=-1.$

Photo Credit: The unit circle constructed by Euler’s formula is shown here, clearly forming a link between geometry and analyticity. Credit to “Euler’s formula” by gunther for the image. Licensed under CC BY-SA 3.0 via Wikimedia Commons.