# Holomorphic functions

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 ${U\subseteq{\mathbb C}}$ be open, and let ${f:U\rightarrow{\mathbb C}}$. If ${f'}$ is defined on all of ${U}$ (i.e., ${f}$ is differentiable), we say that ${f}$ is holomorphic on ${U}$. We denote by ${{\mathcal H}(U)}$ the set of holomorphic functions.

It’s quite common to think about ${{\mathbb C}}$ in terms of its real and imaginary parts, so let’s do that. Write ${f=u+iv}$, where ${u}$ and ${v}$ are functions from ${U}$ to ${{\mathbb R}}$, so we have ${f(z)=u(z)+iv(z)}$. But in fact, if ${z=x+iy}$, we can think of it as the point ${(x,y)}$ in ${{\mathbb R}^2}$. We can write this as

$\displaystyle f(x+iy)=u(x,y)+iv(x,y).$

I’ll go back and forth between writing ${u(z)}$ and ${u(x,y)}$ (and similarly for ${v}$), without mentioning it. If you like, you can think of the comma as shorthand for “${+i\cdot}$

For the functions ${u}$ and ${v}$, let ${u_x}$, ${u_y}$, ${v_x}$, and ${v_y}$ denote the partial derivatives in the ${x}$ (real) and ${y}$ (imaginary) directions.

Lemma 1

Let ${f:U\rightarrow{\mathbb C}}$ be continuous and ${f=u+iv}$. Then ${f\in{\mathcal H}(U)}$ if and only if ${u_x,u_y,v_x,v_y}$ all exist and satisfy the Cauchy-Riemann equations:

$\displaystyle u_x=v_y\mbox{ and }u_y=-v_x.$

Proof: First assume ${f}$ is holomorphic. Then certainly all of the partials exist. We can see, for instance, that ${u_x={\mbox{Re}}(f')}$. Let ${z_0=x_0+iy_0\in U}$. If we take the limit as ${z\rightarrow z_0}$ by taking ${z=z_0+t}$ for ${t\in{\mathbb R}}$ going to zero, then

$\displaystyle \begin{array}{rcl} f'(z_0) &=& \displaystyle\lim_{t\rightarrow0}\frac{f(z_0+t)-f(z_0)}{t}\\ &=& \displaystyle\lim_{t\rightarrow0}\frac{u(x_0+t,y_0)-u(x_0,y_0)}{t}+i\cdot\left(\lim_{t\rightarrow0}\frac{v(x_0+t,y_0)-v(x_0,y_0)}{t}\right)\\ &=& u_x(z_0)+iv_x(z_0) \end{array}$

But we could also take take ${z}$ to ${z_0}$ by approaching from another direction. I could approach along the imaginary axis, taking ${z=z_0+it}$, and letting ${t\in{\mathbb R}}$ go to zero. From this direction, we get that

$\displaystyle \begin{array}{rcl} f'(z_0) &=& \displaystyle\lim_{t\rightarrow0}\frac{f(z_0+it)-f(z_0)}{it}\\ &=& \displaystyle\lim_{t\rightarrow0}\frac{u(x_0,y_0+t)-u(x_0,y_0+t)}{it}+i\cdot\left(\lim_{t\rightarrow0}\frac{v(x_0,y_0+t)-v(x_0,y_0)}{it}\right)\\ &=& -iu_y(z_0)+v_y(z_0) \end{array}$

These give two different expressions for ${f'}$, and so we can equate them. Taking the real parts of each tells us that ${u_x=v_y}$. Taking the imaginary parts of each tells us that ${v_x=-u_y}$.

In the other direction, let ${w=a+bi}$. From the existence of the partials, we know

$\displaystyle \frac{u(z_0+w)-u(z_0)-u_x(z_0)a-u_y(z_0)b}{w}\rightarrow0$

as ${w\rightarrow0}$. Similarly, for ${v}$, we have that

$\displaystyle \frac{v(z_0+w)-v(z_0)-v_x(z_0)a-v_y(z_0)b}{w}\rightarrow0$

as ${w\rightarrow 0}$. Now what we want is to find some ${L}$ for which ${\displaystyle\lim_{w\rightarrow0}\frac{f(z_0+w)-f(z_0)}w=L}$. Let’s try ${L=u_x(z_0)+iv_x(z_0)}$, because we used it in the first half of the proof. If anything should work, that should. We wish to show that

$\displaystyle \displaystyle\frac{f(z_0+w)-f(z_0)-L\cdot w}w\rightarrow0$

as ${w\rightarrow0}$. So consider just the real part of the numerator. From the Cauchy-Riemann equations, ${v_x(z_0)=-u_y(z_0)}$. Then, from the existence of the partials, we have

$\displaystyle \begin{array}{rcl} \displaystyle\frac{{\mbox{Re}}(f(z_0+w)-f(z)-w\cdot L)}w &=& \displaystyle\frac{u(z_0+w)-u(z_0)-u_x(z_0)a+v_x(z_0)b}{w}\\ &=& \displaystyle\frac{u(z_0+w)-u(z_0)-u_x(z_0)a-u_y(z_0)b}w\rightarrow0. \end{array}$

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

$\displaystyle \begin{array}{rcl} \displaystyle\frac{{\mbox{Im}}(f(z_0+w)-f(z)-w\cdot L)}w &=& \displaystyle\frac{v(z_0+w)-v(z_0)-u_y(z_0)a-v_y(z_0)b}{w}\\ &=& \displaystyle\frac{v(z_0+w)-v(z_0)-v_x(z_0)a-v_y(z_0)b}w\rightarrow0. \end{array}$

Putting these together gives the desired result. $\Box$

### 10 Responses to Holomorphic functions

1. BenX says:

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.

• soffer801 says:

Well, there’s a typo which is now fixed. I forgot to subtract $u(z_0)$.

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 $|w|$, then as $w$ gets small, the whole thing does. I didn’t put the absolute value bars in because $|x|$ goes to zero if and only if $x$ 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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