Can complex exponential e^{ix} be proven without using Taylor series?

[eddo]

e^{ix}=cosx + isinx

I know this can be easily proven using the Taylor series, but I recall seeing a proof which doesn't use the Taylor series. I'm pretty sure it has something to do with derivatives, but the problem is I don't remember how it went and I can't find it anywhere. So if anyone knows it or has any idea of where to start could you let me know? Thanks.

 

[Hurkyl]

Once you want to ask that question, you must first ask yourself how you wish to define things like e^z and cos z.

 

[Tide]

You might consider that if w = e^{iz} then

\frac {d^2w}{dz^2} + w = 0[/itex]<br /> <br /> from which <br /> <br /> w = A \cos z + B \sin z<br /> <br /> Does that help?

 

[devious_]

z=\cos x + i\sin x

\frac{dz}{dx}=-\sin x + i\cos x=i^2\sin x + i\cos x=i(\cos x + i\sin x )=i.z

\int\frac{1}{z}\;dz=i\int dx

\ln|z|=ix+c

When x=0, z=1 => c=0.

\ln|z|=ix

e^{ix}=z

e^{ix}=\cos x + i\sin x


Is that what you're talking about?