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? 
 A: There are several reasons. Even without going into the technical details of why it's correct, here is a small list of reasons for why might be a good idea:


*

*It's easy to use that form to read off the length and angle of your complex number

*It's easy to recognize the form and see that it is indeed meant to convey length and angle as opposed to, for instance, the width and height we see in the $a+bi$ form.

*The rules for complex multiplication means that hijacking the exponential notation and (mis)use the intuition you have from real exponentiation gives the correct results

A: In this answer, it is shown that
$$
\lim_{n\to\infty}\left(1+\frac{i\theta}n\right)^n=\cos(\theta)+i\sin(\theta)\tag1
$$
Therefore, we can say
$$
e^{i\theta}=\cos(\theta)+i\sin(\theta)\tag2
$$
We can also use the power series for $e^x$, $\cos(x)$, and $\sin(x)$ to derive $(2)$.
In any case, once we have $(2)$, any point on the unit circle in $\mathbb{C}$ can be represented as $e^{i\theta}$ for some $\theta\in\mathbb{R}/2\pi\mathbb{Z}$; $\theta$ is the argument of that point.
Furthermore, using $(2)$, we can write any point in $\mathbb{C}$ as
$$
\begin{align}
x+iy
&=\overbrace{r\cos(\theta)}^x+i\,\overbrace{r\sin(\theta)}^y\\
&=re^{i\theta}\tag3
\end{align}
$$
One important identity is
$$
\begin{align}
e^{i\theta}e^{i\phi}
&=(\cos(\theta)+i\sin(\theta))(\cos(\phi)+i\sin(\phi))\\
&=(\cos(\theta)\cos(\phi)-\sin(\theta)\sin(\phi))+i(\sin(\theta)\cos(\phi)+\cos(\theta)\sin(\phi))\\
&=\cos(\theta+\phi)+i\sin(\theta+\phi)\\
&=e^{i(\theta+\phi)}\tag4
\end{align}
$$
Equation $(4)$ tells us that we can combine imaginary exponents like we do real ones.
A: If $z=0$ it is clear that we can take $r=0$ and any value for $\theta$.
Note that any number on the complex unit circle can be written as $\cos t + i \sin t$ for some $t$.
Note that any non zero complex number $z$ can be written as $z= |z| {z \over |z|}$ and ${z \over |z|}$ lies on the complex unit circle.
If we let $r=|z|$, then we see that there is some $t$ such that $z = r (\cos t + i \sin t)$.
As to why $e^{it} = \cos t + i \sin t$, let 
$\phi(t) = e^{it} - ( \cos t + i \sin t ) $, note that $\phi(0) = 0$ and $\phi'(t) = i\phi(t)$. Hence $e^{-it} \phi(t)$ is a constant from which it follows that $\phi(t) = 0$ for all $t$.
A: 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 $\cos x + i\sin x$
A: Here's a rather elegant proof.
The function $f : t\mapsto \cos t+i\sin t$ is differentiable and satisfies 
\begin{align*}
f'(t) &= i\,f(t)  \\
f(0) &= 1
\end{align*}
Now let's solve it.
We have $f(0) = 1$ and 
$$f'(t) = (\cos t+i\sin t)' = -\sin t+ i\cos t = i(\cos t+i\sin t) =if(t) $$
Now let us solve this differential equation
$$f'(t) = if(t)\Longleftrightarrow e^{-it}f'(t) -ie^{-it} f(t)=0 \Longleftrightarrow \frac{d}{dt}\left(e^{-it} f(t)\right)  = 0$$
That is 
$$e^{-it} f(t) = c\Longleftrightarrow  f(t) = ce^{it}$$ 
But $f(0)=1 $ i.e $c=1$. Hence $f(t)=e^{it}$.

A: It's a definition. 
As to why it's a good definition, the answer comes from the fact that the Taylor series for exp(x) stays the same when we let $x \in \mathbb{C}$. 
