Rigourous proof of Euler's identity. [closed]

Question

We wish to prove that $$e^{i\pi}+1=0.$$

The standard approach is to use Euler's formula (immediate, for example, from the series definition of the exponential, sine and cosine) and then to use the facts that, $$\sin(\pi)=0\text{, and }\cos(\pi)=-1$$ to draw the necessary conclusion.

But, starting with the series definitions of sine and cosine, I struggle to conclude that $$\sin(\pi)=0\text{, and }\cos(\pi)=-1.$$

The goal is to show that $$\sum_{n=0}^\infty\frac{(2\pi i)^{2n+1}}{(2n+1)!}=0,$$ and that $$\sum_{n=0}^\infty\frac{(2\pi i)^{2n}}{(2n)!}=-1.$$

Attempts by me to prove these relations always end up back where I started. My feeling is that it must somehow come down to some fundamental geometric consideration of the polar description of $\mathbb{C}$ - after all, what is $\pi$ but the circumference of a circle divided by its diameter? Somehow this must be mentioned in any valid proof.

I feel that this issue is not always taken into consideration when a proof of Euler's identity is presented.

Something to gawk at; a Mathematica notebook which plots the partial sums of $e^{ix}$ in the complex plane: ExponentialPartialSumsPlotted.nb

Answer 1

Denote by $f(x)$ the point on the unit circle with angle $x$. A simple geometric consideration shows that $f'(x)=if(x)$ which expresses the fact that any tangent line to the unit circle is perpendicular to the corresponding radius. Now $f^{(n)}(x)=i^nf(x)$ by induction, hence $f^{(n)}(0)=i^n$, and Taylor's formula with remainder term shows that $f(x)=\sum_{n=0}^\infty (ix)^n/n!$. This means that if we define the complex exponential function with the usual Taylor series, then $f(x)=e^{ix}$. Now $e^{i\pi}=-1$ is immediate, once we define $\pi$ as the half-perimeter of the unit circle.