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?

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You should see Needham's book. – J. M. Jul 13 '12 at 15:23
I think it would help if you asked yourself: what does it mean that "Taylor's series hold ..."? A series is just a series; it doesn't "hold", but it may or may not converge. – Marc van Leeuwen Jul 13 '12 at 17:11
It seems like the OP might wonder if there is another way to understand complex exponents -- whether this can be done without Taylor Series at all. Is this idea so compelling that other intelligent, non-human civilizations would have this same formula or is it somewhat arbitrary, coming perhaps from the way we thing about complex numbers? – Jeff Apr 29 at 17:56

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....

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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.

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