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

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  3. 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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