algebra precalculus - Exponential Form of Complex Numbers - Why e?

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Answer

If you remember series, notice that

$$e^{i x } = \sum_{n \geq 0} \frac{ i^n x^n }{n!}$$

Now, notice that $i^2 = -1$ but $i^{3} = -i$, and $i^4 = 1$ and $i^5 = i$ so on, and since

$$\sin x = \sum_{n \geq 0} \frac{ (-1)^n x^{2n+1 } }{(2n+1)!} \; \; \text{and} \; \;\cos x = \sum_{n \geq 0} \frac{ (-1)^n x^{2n } }{(2n)!}$$

after breaking the $n$ in the first summation for even cases and odd cases and seeing in the third line how the $i's$ alternate, one obtains the result