<s>[INST] Why can complex numbers be written in exponential form? $z=r(\cos \theta+i\sin \theta)$ is $z=re^{i\theta}$.     Why can complex numbers be written in exponential form? $z=r(\cos \theta+i\sin \theta)$ is $z=re^{i\theta}$.

I have studied that the exponential form of a complex number $z=r(\cos \theta+i\sin \theta)$ is $z=re^{i\theta}$.
Can someone explain why? 
 [/INST] Lets consider a function from $\mathbb R\to \mathbb C$
$z(\theta) = \cos \theta + i\sin \theta\\
z(\theta)z(\phi) = (\cos \theta + i\sin \theta)(\cos \phi + i\sin \phi) = \cos(\theta + \phi) + i\sin (\theta+\phi) = z(\theta + \phi)$
That is a property of an exponential function.  We do not know the base.
For some base:
$\exp (iy) = z(y) =\cos y + i\sin y$
and:
$\exp (x + iy) = \exp(x)\exp(iy) =\exp(x) (\cos y + i\sin y)$
And then you can define $e$ to be the required base.  In much of complex analysis, it does not matter that it is the same $e$ as you have learned to be Euler's constant.
However, if you have taken calculus, you should recognize these Taylor expansions.
$e^x = \sum_\limits{n=0}^{\infty} \frac {x^n}{n!}\\
\cos x = \sum_\limits{n=0}^{\infty} \frac {(-1)^nx^{2n}}{(2n)!}\\
\sin x = \sum_\limits{n=0}^{\infty} \frac {(-1)^nx^{2n+1}}{(2n+1)!}$
what is
$e^{ix}$ ? 
$e^{ix} = \sum_\limits{n=0}^{\infty} \frac {{ix}^n}{n!}\\
1 + ix + \frac {(ix)^2}{2} + \frac {(ix)^3}{3!}+ \frac {(ix)^4}{4!} \cdots\\
1 + ix + \frac {-x^2}{2} + \frac {-ix^3}{3!} + \frac {x^4}{4!} \cdots$
collect the real terms and the imaginary terms
$(1 - \frac {x^2}{2} + \frac {x^4}{4!}\cdots )+ i( x  - \frac {x^3}{3!} + \frac {x^5}{5!}  \cdots)\\
e^{ix} = \cos x + i\sin x$
Without calculus.
we can define 
$e = \lim_\limits{n\to\infty}(1+\frac {1}{n})^n\\
e^x =\lim_\limits{n\to\infty} (1+\frac {1}{n})^{nx} $
Make a substitution $m = nx$
$e^x =\lim_\limits{m\to\infty} (1+\frac {x}{m})^m $
Then look at what happens as $m = 1, 2,3, etc.$
We have already shown that multiplication of complex numbers multiplies the lengths and adds the angles.

As $m$ increases hopefully you can see how that sequence of line segments begins to lie on the curve of the circle.
and when $m$ is very large comes to rest on $\cos x + i\sin x$
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