# Category Archives: Complex Analysis

## Euler’s famous formula

A hand waving proof of Euler’s formula $e^{i\theta}=\cos\theta +i\sin\theta$ starts from the following Taylor series:

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots$

$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots$

$e^{x}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$

Plugging in $i\theta$ in this series and separating the real and imaginary parts yields the formula.

Here is a sketch for a more formal proof:

1. Consider the initial value problem $\frac{dw}{dz}=w$; $w(0)=1$. We solve it by formally setting

$w=a_0+a_1z+a_2z^2+\cdots$

$\frac{dw}{dz}=a_1+2a_2z+\cdots$

Evidently to satisfy the differential equation we must have $a_{n-1}=na_n$ and the initial condition gives $a_0=1$. It follows by induction that $a_n=1/n!$.

2. The solution $w$ is now redesignated as $e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots$.

3. We note that this power series has an infinite radius of convergence as $(n!)^{1/n}\rightarrow\infty$. By the existence and uniqueness theorem of differential equations, it is certain that this is the only solution of the given IVP.

4. The trigonometric functions are now defined by $\cos z=\frac{e^{iz}+e^{-iz}}{2}$ and $\sin z=\frac{e^{iz}-e^{-iz}}{2i}$ and it immediately follows that $e^{iz}=\cos z+i\sin z$. We thus have Euler’s formula.