e^ix = cos x + i sin x

By Alexander (p195) on April 14, 1998 :

How to prove that the CIS equasion is right -
e^(i*x) =cos x - isin x

Alexander (p195)

By Rup on April 14, 1998 :

I've never heard it called the CIS equation, but I suppose that makes sense; I always thought of it as a theorem -- probably Euler's. And I'm afraid that you've got a sign error... it's

e^(ix) = cos x + i sin x

The easiest way to see this is from the Maclaurin expansions of exponential, cos and sin;
e(ix) = 1 + ix + (1/2)x² + (i/6)x³ + (1/24)x⁴ +...
cos x = 1 + (1/2)x² + (1/24)x⁴ +...
i sin x = ix + (i/6)x³ + +...

Writing (summing over n from 0 to infinity):

$e^{i x} = \sum [(i x)^{n} / n!]$
the next step is to break this sum into real and imaginary parts; let's make this easy on ourselves by splitting into even and odd powers by renumbering n=2m and now summing from m=0 to infinity:
$= \sum [(i x)^{2 m} / (2 m)! + (i x)^{2 m + 1} / (2 m + 1)!]$

$= \sum [(- 1)^{m} x^{2 m} / (2 m)!] + i \sum [(- 1)^{m} x^{2 m + 1} / (2 m + 1)!]$

$= \cos x + i \sin x$
I don't know if this seems rigorous enough to you or seems perhaps a little vague -- I'm not really sure quite what else can be said. This is the only proof I've been presented with, and I can't find anything else in the books at hand.

Hope this helps,
Rup.

[Editor: The following alternative suggestion was made by Brice Yokem. Define f(x)=e^ix and g(x)=cos(x)+isin(x) then f'(x)=if(x) and g'(x)=ig(x). That is, the complex differentiable functions f and g satisfy the same differential equation. It is easily verified that f(0)=g(0) (initial conditions) therefore the two functions f and g must be equal.]