How Does Euler's Formula Lead to e^(iπ) = -1?

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Euler's formula establishes that e^(ix) = cos(x) + i sin(x), which is crucial for understanding e^(iπ) = -1. By substituting π into Euler's formula, it simplifies to e^(iπ) = cos(π) + i sin(π), yielding -1 as cos(π) = -1 and sin(π) = 0. The discussion highlights the connection between the exponential function and trigonometric functions through their power series expansions. Participants express curiosity about deriving this relationship without numerical approximation, emphasizing the importance of recognizing the series for cos(x), sin(x), and e^(ix). Ultimately, the thread illustrates the elegance of Euler's identity in connecting complex numbers and trigonometry.
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e^{i\pi}=-1

I was wondering how on Earth this was possible. I know that:

<br /> e^z = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!}+...+\frac{z^n}{n!}<br />

So

<br /> e^{i\pi}=1+i\pi+\frac{-\pi^2}{2!}+\frac{-\pi^3i}{3!}+\frac{\pi^4}{4!}...<br />

I was wondering if there is any way to solve this other than punching out actual numbers and seeing about where they converge to?
 
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e^ix = cos x + i sin x
 
thanks, I didn't know about that equation
 
If you look at the power series for cos(x), sin(x) and eix, the relationship will be obvious.
 
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