Q:

How do we know Taylor's Series works with complex numbers?

Euler famously used the Taylor's Series of $\exp$:
$$\exp (x)=\sum_{n=0}^\infty \frac{x^n}{n!}$$
and made the substitution $x=i\theta$ to find
$$\exp(i\theta) = \cos (\theta)+i\sin (\theta)$$
How do we know that Taylor's series even hold for complex numbers?  How can we justify the substitution of a complex number into the series?

A:

The series $f(z)=\sum_{n=0}^{\infty} \frac{z^n}{n!}$ is convergent on $C$, and thus it defines an Analytic function.
Now there are few different ways to convince yourself that this has to be $e^z$. 
For once, it is the only Analytic continuation of $e^x$ to the complex plane...
Or, alternately, you can prove that $f(z_1+z_2)=f(z_1)f(z_2)$ and $f(1)=e$. Also, you can show that it is the only differentiable function satisfying these two relations.
If you prefer differential equations, $f'(z)=f(z)$ and $f(1)=e$ uniquely determine a solution, and bot $e^z$ and $f(z)$ are solutions.... 

A:

You define a complex function by the formula
$f(z)=\sum_{n=0}^{\infty} \frac{z^n}{n!}  $. You prove that it converges everywhere and defines a holomorphic function of $z$.  Then you prove that for $z=x \in \mathbb{R}$ it agrees with the usuall exponential.

