Exponential Form of Complex Numbers - Why e?

Question

You know how $\sin(\theta) \cdot i + \cos \theta$ is in exponential form $e^{i\theta}$, why exactly is $\sin(\theta) \cdot i + \cos \theta = e^{i\theta}$?

Answer 1

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