## Complex Differentiability, Part 1: Complex Continuity, and Where do the Cauchy-Riemann Equations Come From?

### September 11, 2010

When we do differential calculus, we work in the real numbers more times than not — we’re used to them, and they’re easy to manipulate.  A good number of applications of calculus also can naturally be worked on using the real numbers.

But there are some obvious drawbacks to using the real numbers — the most obvious being that not every polynomial actually has a root in the real numbers.  In fact, some polynomials will have all of their roots in the complexes.  This is kind of crappy if we’d like to know everything there is to know about a certain function, and so we often will extend our field to become the complex numbers.

I’m not going to run through what complex numbers are, or what the field topology looks like, or anything like that.  I’m going to assume that you know this.  The first point that we’re going to go through is just what it means for a function to be continuous — and, really, if a function is not continuous, what chance does it have of being differentiable?

## Continuity in the Complexes.

There’s not much to say here.  Continuity in the complexes is really just about the same as continuity in the reals, with one (drastic!) change.  Consider $f:{\mathbb C}\rightarrow {\mathbb C}$.  Then we have that $f$ is continuous at $z_{0}$ if

$\displaystyle \lim_{z\rightarrow z_{0}} f(z) = f(z_{0})$

Does this look familiar?  It should.  It’s the same definition as in the reals.  (In fact, this holds true for any decent space whose topology satisfies a countable local basis!)  So what’s the problem here?

Well, in the reals, it was kind of an easy deal.

As the picture above describes, we can really only go in from the “left” (red dots) or from the “right” (blue dots) to get to our limit point (green dot).  We can alternate back and forth between left and right, but ultimately there are just two directions that we can go in to get to this point.  This makes it fairly easy to have a notion of continuity — you just need to sort of check a little to the left, and a little to the right.

As this slightly less clear graph tries to show, our sequence (the red dots) can really come in from anywhere.  They can be above the point we want to get to (green dot), below, to the left, to the right, and really along any line extending outwards from our limit point.  So, we’ve gone from checking two directions to checking an uncountable infinite number of directions.  Kind of a drastic change.  Nonetheless, usually this only requires a bit more cleverness on our part.

In order to prove something is not continuous, simply find two lines going through the desired limit point and show that, along these two lines, the limit you get is different.  For example, let

$f(z) = f(x + yi) = f(x,y) = \frac{xy}{x^2 + y^2}$

and we’ll show this function is not continuous at the point $z = 0$.  So, first, take the limit along the line $y = 0$.  This means that $y = 0$ and $x\rightarrow 0$.

$\displaystyle \lim_{x\rightarrow 0} \frac{xy}{x^2 + y^2} = \lim_{x\rightarrow 0} 0 = 0$

But now take the limit along the line $y = x$.

$\displaystyle \lim_{x\rightarrow 0} \frac{x^2}{2x^2} = \lim_{x\rightarrow 0} \frac{1}{2} = \frac{1}{2}$.

Since $\frac{1}{2} \neq 0$, we have that this limit doesn’t exist, and that $f(z)$ is not continuous at $z= 0$.  Sad.

On the other hand, it’s much harder to prove that something is continuous.  We rely mainly on the $\epsilon -\delta$ definition, which is pretty irritating, and which I will not give an example of, because it’s in pretty much every basic calculus and complex analysis book, and it’s really annoying to write out.

## Complex Differentiability.

So, if we know something is continuous, is it also differentiable?  Well, how did we do this in real space?

$\displaystyle \lim_{h\rightarrow 0}\frac{f(x+h) - f(x)}{h} = f'(x)$.

If such a limit existed, in other words, it would be the derivative.  So, as we saw, real limits are more or less easy — just two directions to check.  But complex ones?  Ugh.  Worst.  It’s a huge pain to prove that something is sufficient for a complex derivative to exist, but it’s really rather easy to prove that something is necessary for it to exist; in other words, we assume that it exists, and then we derive some basic consequence that we require every function which is complex differentiable at a point to have at that point.  Enter the Cauchy-Riemann equations.

### Cauchy-Riemann Equations.

Let’s first suppose that $f:{\mathbb C}\rightarrow {\mathbb C}$ and suppose we have that $\displaystyle \lim_{h\rightarrow 0}\frac{f(z_{0}+h) - f(z_{0})}{h} = f'(z_{0})$; or, in other words, suppose the complex derivative actually exists at the point $z_{0} = x_{0} + iy_{0}$.

Letting $y = y_{0}$, and going along the “real axis”, we find

$\displaystyle \lim_{h\rightarrow 0}\frac{f(x_{0} + h, y_{0}) - f(x_{0}, y_{0})}{h} = \frac{\partial f}{\partial x}(z_{0}) = f_{x}(z_{0})$

Letting $x = x_{0}$ and going along the “imaginary axis”, we find

$\displaystyle \lim_{h\rightarrow 0}\frac{f(x_{0}, y_{0}+h) - f(x_{0}, y_{0})}{ih} = -i\frac{\partial f}{\partial y}(z_{0}) = -if_{y}(z_{0})$

Since we can write $f(x,y) = u(x,y) + iv(x,y)$, then

$f_{x}(x_{0}, y_{0}) = u_{x}(x_{0},y_{0}) + iv_{x}(x_{0},y_{0})$

$-if_{y}(x_{0}, y_{0}) = -iu_{y}(x_{0},y_{0}) + v_{y}(x_{0},y_{0})$

Since differentiability holds, we have that $f_{x} = if_{y}$ and we can set the real and imaginary parts of these derivatives equal to each other.

$u_{x}(x_{0},y_{0}) = v_{y}(x_{0},y_{0})$

$u_{y}(x_{0},y_{0}) = -v_{x}(x_{0},y_{0})$

or, more concisely,

$u_{x} = v_{y}$

$u_{y} = -v_{x}$

These equations are called the Cauchy-Riemann equations, most likely after the mathematicians of the same names.  Every single function which is complex differentiable must satisfy these equations.

But, if a function satisfies these equations, does that mean that it is complex differentiable?  Upsettingly, no.  In part 2, we’ll explore what exactly makes a complex function differentiable.  It’s a slightly stronger condition than just satisfying the CR-equations above, but it does take some proving.