A second look at complex differentiation

What does it mean to differentiate a complex function?

If we have a complex function, $$f:\mathbb{C} \to \mathbb{C},$$ then we can try to define a derivative by analogy with the real version: we say that $f$ has derivative $f'(z)$ at $z$ if $$\lim_{h \to 0} \frac{f(z+h)-f(z)}{h} = f'(z).$$

The catch is that now $h$ is a complex number. For this to make sense, it must give the same answer no matter how $h \to 0$. The usual thing is then to insist that you get the same answer if $h$ is pure real or pure imaginary, and find some constraints that have to be satisfied: these are the Cauchy-Riemann equations.

If you're of a suspicious turn of mind, you might ask why it's enough for it to work for $h$ pure real or pure imaginary: that makes $h$ approach $0$ along one of two rather special lines. What if $h$ doesn't do that? Also, this argument shows that if $f$ is complex differentiable, it must satisfy the Cauchy-Riemann equations: but it doesn't tell us that if the Cauchy-Riemann equations are satisfied, then $f$ must actuallly be complex differentiable.

But there's another approach, which I think has a few advantages.

• It's based on a general definition of derivative, rather than guessing something that 'looks like' the real (variable, that is) derivative.
• It doesn't relay on $h$ approaching $0$ in any particular way.
• It shows that the Cauchy-Riemann equations are necessary and sufficient for complex differentiability.
• I just think it's prettier.

So, let's recall what the derivative really is: if you have a function $\pmb{f}:\mathbb{R}^n \to \mathbb{R}^m$, so$\pmb{f}$ is a vector with components $f^1 \ldots f^m$, then $\pmb{f}$ is differentiable at $\pmb{x}$ with derivative $D\pmb{f}(x)$ (an $m \times n$ matrix) if $$\pmb{f}(\pmb{x}+\pmb{h}) = \pmb{f}(\pmb{x})+D\pmb{f}(\pmb{x})\pmb{h}+o(\|\pmb{h}\|).$$

Then the $i,j$ element of $D\pmb{f}(x)$ is $\partial f^i/\partial x^j$.

In the case where $\pmb{f}:\mathbb{R}^2 \to \mathbb{R}^2$, we can be much more explicit than that. We write $$\pmb{f}(\pmb{x} ) = \left( \begin{array} {c} u(x,y)\\ v(x,y) \end{array} \right)$$ where $$\pmb{x} = \left( \begin{array} {c} x\\ y \end{array} \right)$$ and then $$D\pmb{f} = \left( \begin{array}{cc} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{array} \right)$$ has the property that $$\begin{split} \left( \begin{array}{c} u(x+h_x,y+h_y)\\ v(x+h_x,y+h_y) \end{array} \right) &= \left( \begin{array}{c} u(x,y)\\ v(x,y) \end{array} \right) + \left( \begin{array}{cc} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{array} \right) \left( \begin{array}{c} h_x\\ h_y \end{array} \right) +o(\sqrt{h_x^2+h_y^2})\\ &= \left( \begin{array}{c} u(x,y)\\ v(x,y) \end{array} \right) + \left( \begin{array}{c} \frac{\partial u}{\partial x}h_x + \frac{\partial u}{\partial y} h_y\\ \frac{\partial v}{\partial x}h_x + \frac{\partial v}{\partial y} h_y \end{array} \right) +o(\sqrt{h_x^2+h_y^2}). \end{split}$$

But now we can define the complex number $z=x+iy$, and the complex function $f:\mathbb{C} \to \mathbb{C}$ by $$f(x+iy) = u(x,y)+iv(x,y).$$ So $f$ will be differentiable at $z$ with derivative $f'(z)$ if $$f(z+h)=f(z)+f'(z)h+o(|h|) = f(z)+f'(z)h + o(\sqrt{h_x^2+h_y^2})$$ where $h=h_x+ih_y$.

So what is this $f'(z)$?

Let's call it $d_x+id_y$, so that $d_x$ and $d_y$ are the real and imaginary parts of $f'(z)$ respectively. Then we have $$\begin{split} f'(z)h &= (d_x+idy)(h_x+ih_y) \\ &= d_x h_x-d_y h_y + i(d_y h_x + d_x h_y). \end{split}$$

So now we have two different ways of describing the same function: once as a function $\mathbb{R}^2 \to \mathbb{R^2}$, and once as a function $\mathbb{C} \to \mathbb{C}$. But it's the same function, both times, so the derivative must really be the same. In other words, since the expressions must match for all possible $h_x$ and $h_y$, $\pmb{f}$ can be thought of as a complex differentiable function if and only if $$\begin{split} \frac{\partial u}{\partial x}h_x + \frac{\partial u}{\partial y} h_y &= d_x h_x-d_y h_y \\ \frac{\partial v}{\partial x}h_x + \frac{\partial v}{\partial y} h_y &= d_y h_x + d_x h_y. \end{split}$$

Matching these up, then, we have the conditions $$\begin{split} d_x &= \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\\ d_y &= \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y}. \end{split}$$

And putting it all together, we see that the real function given by $(x,y) \to (u(x,y),v(x,y))$ can be thought of as a complex differentiable function $x+iy \to u(x,y)+iv(x,y)$ if and only if $u$ and $v$ satisfy the Cauchy-Riemann equations $$\begin{split} \frac{\partial u}{\partial x} &= \frac{\partial v}{\partial y}\\ \frac{\partial v}{\partial x} &= -\frac{\partial u}{\partial y}. \end{split}$$

I, at least, find this more satisfying.