# Rigourous proof of Euler's identity. [closed]

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

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## closed as off topic by Andres Caicedo, Andy Putman, Wadim Zudilin, Pete L. Clark, AngeloJan 6 '11 at 7:53

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You could take pi to be the smallest positive solution of sin(x) = 0, where sin(x) is defined by a power series. With this definition I believe Spivak, in his book Calculus, derives a formula for pi which shows it is 3.14... and perhaps Rudin's Real and Complex Analysis (near the beginning) also develops matters from sin and cos as power series and pi as a minimal positive zero of the sine series. –  KConrad Jan 6 '11 at 6:18
Sure, I can take Euler's identity to be true and use it to calculate a numerical approximation to \pi to arbitrary precision, but that is not really a proof - it's a definition. Maybe it should be Euler's definition of \pi; Let \pi be the real number such that e^(i\pi)+1=0. To call something an identity for my part would require that \pi has been used as an input (coming from some separate definition) somehow. –  Stephen Mc Ateer Jan 6 '11 at 21:57
@Stephen: I think you misunderstood. In Rudin's "Principles of Mathematical Analysis 3E" around theorem 8.7 they define $sin(x)$ to be its power series, define $2\pi$ to be the period, and then derive that $2\pi r$ is the circumference of the circle radius $r$. This is simply doing things in the other direction. –  Eric Naslund Feb 25 '11 at 4:40

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.