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.
