# Holomorphic functions

July 5, 2012 10 Comments

Last time we said that we would prove differentiable functions, infinitely differentiable functions, and analytic functions were all the same. So let’s give a name to one of these classes of functions: Let be open, and let . If is defined on all of (i.e., is differentiable), we say that is **holomorphic** on . We denote by the set of holomorphic functions.

It’s quite common to think about in terms of its real and imaginary parts, so let’s do that. Write , where and are functions from to , so we have . But in fact, if , we can think of it as the point in . We can write this as

I’ll go back and forth between writing and (and similarly for ), without mentioning it. If you like, you can think of the comma as shorthand for “”

For the functions and , let , , , and denote the partial derivatives in the (real) and (imaginary) directions.

Lemma 1Let be continuous and . Then if and only if all exist and satisfy the

Cauchy-Riemann equations:

*Proof:* First assume is holomorphic. Then certainly all of the partials exist. We can see, for instance, that . Let . If we take the limit as by taking for going to zero, then

But we could also take take to by approaching from another direction. I could approach along the imaginary axis, taking , and letting go to zero. From this direction, we get that

These give two different expressions for , and so we can equate them. Taking the real parts of each tells us that . Taking the imaginary parts of each tells us that .

In the other direction, let . From the existence of the partials, we know

as . Similarly, for , we have that

as . Now what we want is to find some for which . Let’s try , because we used it in the first half of the proof. If anything should work, that should. We wish to show that

as . So consider just the real part of the numerator. From the Cauchy-Riemann equations, . Then, from the existence of the partials, we have

As one might expect, something nearly identical happens when we take the just the imaginary part of the numerator.

Putting these together gives the desired result.

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Would you mind clarifying how we know from the existence of partials that those two quotients (from the beginning of “the other direction”) go to 0 as w goes to 0? I know it’s a rewriting of a limit but I can’t see which one.

Well, there’s a typo which is now fixed. I forgot to subtract .

Roughly speaking, what’s what it means to have the partials. It means that small changes are approximated by a linear factor given by the partials, and that this approximation is better than linear.

By better than linear, I mean that If I divide the whole thing by the distance traveled, which in this case is , then as gets small, the whole thing does. I didn’t put the absolute value bars in because goes to zero if and only if does. This way I could avoid extra symbols.

That probably wasn’t particularly helpful, so you can also take a look at a reasonably similar approach towards the bottom of page 31 of Professor Mario Bonk’s notes: http://www.math.ucla.edu/~mbonk/246c.1.12s/complana.pdf

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