Proving $e^z$ is holomorphic

Question

I read somewhere that $e^z$ is holomorphic function, but I can't think of an easy way to prove that. I thought of using Cauchy Riemann equations, but that's probably overkill. Is there a simple approach to show that this function is holomorphic?

Answer 1

Other answers assume you're defining the exponential in terms of power series. If you define $e^z = e^x\cos(y) + ie^x\sin(y)$, you see the following:

1. Because both coordinate functions are infinitely differentiable, $e^z$ is infinitely differentiable as a function of two variables
2. $\frac{\partial}{\partial x} e^z = e^z$
3. $\frac{\partial}{\partial y} e^z = ie^z$

Now it's just as simple as applying the Cauchy-Riemann equations.

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To address whether it's "overkill" -- first of all, if it works, it works. Worrying about overkill when you haven't even finished solving the problem yet is, well, overkill.

Second, as the course proceeds you should be accumulating a list of conditions that are equivalent to holomorphy ("complex-differentiable at every point in an open set", "satisfies the Cauchy-Riemann equations at every point in domain", "equals its power series at every point in an open set", ...).

Instead of trying to pick the "best" one off the bat, it would probably be more instructive to go through the list and see how each condition applies to $e^z$. That (a) give you practice with your whole toolset, and (b) set you up to better match tool to problem in the future.

Answer 2

It depends on your definition of $e^z$. The most standard definition is to define it via the power series $$ e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}, $$ which is easily seen to be convergent everywhere. Then you can just use the general theorem that a power series defines a holomorphic function everywhere inside its radius of convergence.

Answer 3

$e^{z}$ is defined as :

$$\sum_{n=0}^{\infty} \frac{z^{n}}{n!}.$$

The radius of convergence of this series is $\infty$. Thus it is an analytic function.