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Fix a domain $D \subset \C$ and $f : D \to \C$ a function. We saw in Week 1 how to write \[ f(x+iy) = u(x,y) + i v(x,y) \] where $u,v : D \to \R$ are real-valued functions. Here we will explore what differentiability of $f$ as a complex function tells us about the functions $u$ and $v$. The result will be the Cauchy-Riemann equations that give two relationships between the partial derivatives of $u$ and $v$.
Fix a real-valued function $g : D \to \R$. Define \[ \begin{aligned} (\partial_1 g)(x,y) &= \lim_{h \to 0} \frac{ g(x+h,y)-g(x,y)}{h} \\ (\partial_2 g)(x,y) & = \lim_{k \to 0} \frac{ g(x,y+k)-g(x,y)}{k} \end{aligned} \] for any $(x,y) \in D$ at which the limits exist. We call $\partial_1 g$ and $\partial_2 g$ the partial derivatives of $g$.
Thus, to calculate $\partial_1 g$ we treat $y$ as a constant and differentiate with respect to $x$ using the usual rules, and to calculate $\partial_2 g$ we treat $x$ as a constant and differentiate with respect to $y$. Note that one often sees \[ \frac{\partial g}{\partial x} \qquad \frac{\partial g}{\partial y} \] written for $\partial_1 g$ and $\partial_2 g$ respectively.
Calculate $\partial_1 g$ and $\partial_2 g$ for $g(x,y) = xy^2 + (x^2 + y^2)^2$.
When calculating $\partial_1 g$ we use the usual rules of differentiation with $x$ as the variable, treating $y$ as a constant. For $\partial_2 g$ we treat $y$ as the variable and $x$ as the constant. Thus \[\begin{aligned} (\partial_1 g)(x,y) &= y^2 + 4x(x^2 + y^2) \\ (\partial_2 g)(x,y) &= 2xy + 4y(x^2 + y^2) \end{aligned}\]
Let $f : D \to \C$ and write $f(x+iy) = u(x,y)+iv(x,y)$. Suppose that $f$ is differentiable at $z_0 =x_0+iy_0$. Then the partial derivatives \[ (\partial_1 u)(x_0,y_0) \qquad (\partial_2 u)(x_0,y_0) \qquad (\partial_1 v)(x_0,y_0) \qquad (\partial_2 v)(x_0,y_0) \] all exist and the relations \[ \begin{aligned} (\partial_1 u)(x_0,y_0) &= (\partial_2 v)(x_0,y_0) \\ (\partial_2 u)(x_0,y_0) &= - (\partial_1 v)(x_0,y_0) \end{aligned} \] hold.
Recall that to calculate $f'(z_0)$ we take a limit as $h \to 0$. The idea of this proof is to calculate the limit in in two different ways: by taking $h \to 0$ along the real, and then by taking $h \to 0$ along the imaginary axis.
Let $k$ be real. Taking $h = k$ in the definition of differentiability, we have \[ \begin{aligned} f'(z_0) & = \lim_{k \to 0} \frac{ f(z_0+k)-f(z_0)}{k} \\ & = \lim_{k \to 0} \frac{ u(x_0+k,y_0) + iv(x_0+k,y_0) - u(x_0,y_0) -iv(x_0,y_0)}{k} \\ & = \lim_{k \to 0} \frac{u(x_0+k,y_0)-u(x_0,y_0)}{h} + i \lim_{k \to 0} \frac{v(x_0+k,y_0)-v(x_0,y_0)}{k} \\ & = (\partial_1 u)(x_0,y_0) + i (\partial_1 v)(x_0,y_0) \end{aligned} \] because if a limit exists then the limits along the real and imaginary parts exist. Taking next $k$ real and $h = ik$ we get \[ \begin{aligned} f'(z_0) & = \lim_{k \to 0} \frac{ f(z_0+ik)-f(z_0)}{ik} \\ & = \lim_{k \to 0} \frac{ u(x_0,y_0+k) + iv(x_0,y_0+k) - u(x_0,y_0) -iv(x_0,y_0)}{ik} \\ & = \lim_{k \to 0} - i \frac{u(x_0,y_0+k)-u(x_0,y_0)}{h} + \frac{v(x_0,y_0+k)-v(x_0,y_0)}{ik} \\ & = - i (\partial_2 u)(x_0,y_0) + (\partial_2 v)(x_0,y_0) \end{aligned} \] by similar reasoning and the fact that $1/i = -i$. We have two expressions for $f'(z_0)$. Comparing their real and imaginary parts yields the claimed relationships. $\square$
The relations in the conclusion of the theorem are called the Cauchy-Riemann equations.
In particular, we conclude from the proof of the theorem that \[f'(x+iy) = (\partial_1 u)(x,y) + i (\partial_1 v)(x,y) = (\partial_2 v)(x,y) - i (\partial_2 u)(x,y)\] wherever $f$ is differentiable.
We can use the Cauchy-Riemann equations to examine whether the function $f(z)=\bar{z}$ is differentiable on $\CC$. Note that writing $z=x+iy$ allows us to write $f(z) = \bar{z} = x-iy$. Hence $f(z) = u(x,y)+iv(x,y)$ with $u(x,y)=x$ and $v(x,y)=-y$. Now \[ \begin{gather*} (\partial_1 u)(x,y) = 1 & \qquad (\partial_2 u)(x,y) = 0 \\ (\partial_1 v)(x,y) = 0 & \qquad (\partial_2 v)(x,y) = -1 \end{gather*} \] and there are no points $(x,y)$ at which \[(\partial_1 u)(x,y) = (\partial_2 v)(x,y)\] holds. We conclude that $f(z) = \bar{z}$ is not differentiable at any point in $\CC$.
Notice however that $f(z)=\bar{z}$ is continuous at every point in $\CC$. Hence $f(z)=\bar{z}$ is an example of an everywhere continuous but nowhere differentiable function. Such functions also exist in real analysis, but they are much harder to write down and considerably harder to study. One of the simplest is known as Weierstrass' function \[w(x) = \sum_{n=0}^{\infty} \frac{1}{2^{n\alpha}}\cos (2 \pi b^n x)\] for some $\alpha \in (0,1)$ and some $b \geq 2$; such functions are still of interest in current research.
The Cauchy-Riemann equations hint at what is special about differentiability for a function of a complex variable. Writing \[f(x+iy) = u(x,y) + iv(x,y)\] again, we can think of $f$ as a function $D \to \R^2$. As with any such function, its real derivative at a point $(x,y) \in D$ is the matrix \[ (\mathsf{D} f)(x,y) = \begin{bmatrix} (\partial_1 u)(x,y) & (\partial_2 u)(x,y) \\ (\partial_1 v)(x,y) & (\partial_2 v)(x,y) \end{bmatrix} \] provided all partial derivative exist. The derivative of a function $f : D \to \R^2$ can be any $2 \times 2$ matrix \[ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \] i.e. $a,b,c,d \in \R$ could be any real number. Compared to this general situation, derivatives of holomorphic functions thought of as functions $D \to \R^2$ are very special: the Cauchy-Riemann equations force the structure \[ \begin{bmatrix} a & -c \\ c & a \end{bmatrix} \] on the real derivative. In particular, if the real derivative of a holomorphic function $f : D \to \C$ is non-zero then it is invertible, consisting of a scaling followed by a rotation.
The Cauchy-Riemann equations also tell us that the level curves of $u$ and $v$ interact in a special way when $f$ is holomorphic, as the following theorem explains.
Let $f : D \to \C$ be holomorphic and fix $z \in D$. Write $z = x + iy$ and $f(z) = r + is$. If $f'(z) \ne 0$ then the level curves $u(x,y) = r$ and $v(x,y) = s$ are orthogonal at $z$.
It suffices to prove that the gradients \[ (\nabla u)(x,y) = \begin{bmatrix} (\partial_1 u)(x,y) \\ (\partial_2 u)(x,y) \end{bmatrix} \qquad (\nabla v)(x,y) = \begin{bmatrix} (\partial_1 v)(x,y) \\ (\partial_2 v)(x,y) \end{bmatrix} \] are orthogonal. But \[ \begin{align*} & (\nabla u)(x,y) \cdot (\nabla v)(x,y) \\ ={} & (\partial_1 u)(x,y)(\partial_2 u)(x,y) + (\partial_1 v)(x,y)(\partial_2 v)(x,y) \\ ={} & 0 \end{align*} \] by the Cauchy-Riemann equations. $\square$