Why is e^(i * theta) = cos(theta) + i * sin(theta)

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Stallmp

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Why is this specific equation true? This is applied all the time in for example polar coordinates, where re^(itheta) is equal to r(costheta+isintheta).
 
Stallmp said:
Why is this specific equation true? This is applied all the time in for example polar coordinates, where \(\displaystyle re^{(i\theta)}\) is equal to \(\displaystyle r(cos\theta+isin\theta)\).
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To answer that question, one needs to understand the following series:
\(\displaystyle {e^x} = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} ,~~\sin (x) = \sum\limits_{k = 0}^\infty {\frac{{{{( - 1)}^k}{x^{2k + 1}}}}{{(2k + 1)!}}} ,~~\&~~\cos (x) = \sum\limits_{k = 0}^\infty {\frac{{{{( - 1)}^k}{x^{2k}}}}{{(2k)!}}}\)

Using those series and letting \(\displaystyle x=i\theta\) we get \(\displaystyle e^{i\theta}=\cos(\theta)+i\sin(\theta)\)
 
But then again, you would have to understand why those specific series are true, so tackling this problem with even a more difficult problem is beyond my knowledge I'm afraid.
 
Stallmp said:
But then again, you would have to understand why those specific series are true, so tackling this problem with even a more difficult problem is beyond my knowledge I'm afraid.
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Your reply begs the question: "why don't you know those fundamental series?"
You did ask for the reason for the formula. I post the derivation.
 
pka said:
Your reply begs the question: "why don't you know those fundamental series?"
You did ask for the reason for the formula. I post the derivation.
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I get your point, but based on my question, you can probably already tell that I don't have too much specialised knowledge regarding this specific topic. Thanks for your help though, don't get me wrong. I really appreciate it. It's just that derivations are often really hard and probably not how Euler discovered that theorem. I was wondering if there was an intuitive proof for it, rather than proving by series or a formula.
 
I couldn't edit that quickly enough. Here's what I meant to say:

Yes, but, you see, we don't know what mathematics you do know so we don't know what responses you would understand. Have you taken a Calculus course? Do you know what "Taylor's series" are? Do you know the Taylor's series for $e^x$? For sin(x) and cos(x)? If you do, replace "x" with "ix", then separate the even powers (which will have no "i" since \(\displaystyle i^2= -1\) so \(\displaystyle i^{2n}= (-1)^n\) from the odd powers (which will have a single "i" since \(\displaystyle i^{2n+ 1}= i^{2n}i= (-1)^ni\) and compare those to the Taylor's series for sine and cosine. If you do not know what "Taylor's series" are, you probably would not understand any explanation![/tex][/tex]
 
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