What is the shortest way to prove:\[ e^{ix}=\cos(x)+i\sin(x) \]

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What is the shortest way to prove:\[ e^{ix}=\cos(x)+i\sin(x) \]

Mathematics
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I know one way is to use the series expansion for cos(x) and sin(x) and show that it matches the series expansion for e^(ix) - but is there a shorter proof?
Do you know what is the polar form of a complex number ?
yes

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I read once a short proof by using differential equations.
do you recall that proof @Mr.Math?
Look here http://en.wikipedia.org/wiki/Euler%27s_formula#Proofs
@FoolForMath - do you mean this: |dw:1325522941156:dw|
thanks @Mr.Math - that is what I was looking for.
You're welcome!
asnaseer take a look at this thread: http://math.stackexchange.com/questions/3510/how-to-prove-eulers-formula-expi-t-costi-sint
The complex number approach has been explained there too.
i don't think there is a shorter way than series expansion
@asnaseer: Glad to help :)
Using the uniqueness theorem with differentials seems to be the shortest method.
I guess it all depends on how you define "shortest"
...and elementary. Without looking at all the alternatives in a lot of detail, I'd hypothesize the series proof is the most mathematically elementary.
I still find the proof using derivatives much simpler.
Yes, I agree with you asnaseer.
it look more "elegant" as well.
*looks
yea but inaccessible without knowledge of calculus.
Returning for a moment to the idea of being elementary, what the differentiation proofs assume is that e^ix is differentiable. That's not obvious before the fact.
btw asnaseer, what's your need?
no need really - I was just wondering if there were any other ways to prove this apart from the series expansion.

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