The Cauchy-Riemann Theorem

# The Cauchy-Riemann Theorem

Recall from the Analytic/Holomorphic Complex Functions page that if $A \subseteq \mathbb{C}$ is open,

 Theorem 1 (The Cauchy-Riemann Theorem): Let $A \subseteq \mathbb{C}$ be an open set, $z_0 \in A$, and let $f : A \to \mathbb{C}$ where $f = u + iv$ ($u, v : \mathbb{R}^2 \to \mathbb{R}$). Then $f$ is analytic at $z_0$ if and only there is a neighbourhood $\mathcal N \subseteq A$ of $z_0$ for wihch: 1) $\displaystyle{\frac{\partial u}{\partial x}}$, $\displaystyle{\frac{\partial u}{\partial y}}$, $\displaystyle{\frac{\partial v}{\partial x}}$, and $\displaystyle{\frac{\partial v}{\partial y}}$ all exist and are continuous on $\mathcal N$ 2) The equations $\displaystyle{\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}}$ and $\displaystyle{\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}}$, called the Cauchy-Riemann Equations, are satisfied on all of $\mathcal N$. Moreover we have that $\displaystyle{f'(x + iy) = \frac{\partial u}{\partial x} (x, y) + i \frac{\partial v}{\partial x} (x, y) = \frac{\partial v}{\partial y} (x, y) - i \frac{\partial u}{\partial y}(x, y)}$.
• Partial Proof: $\Rightarrow$ Suppose that $f$ is analytic at $z_0$. Then there exists a neighbourhood $\mathcal N$ of $z_0$ such that $f$ is complex differentiable on $\mathcal N$. In particular, the limit $\displaystyle{f'(z_0) = \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0}}$ exists. Since this limit exists, the limit also exists along any path as $z \to z_0$ and so the existence of these partial derivatives on $\mathcal N$ is given.
• If $z_0 = x_0 + iy_0$ and $z = x + iy_0$ then consider this limit as $x \to x_0$ (so that $z \to z_0$):
(1)
\begin{align} \quad f'(z_0) &= \lim_{(x, y) \to (x_0, y_0)} \frac{u(x, y) + iv(x, y) - [u(x_0, y_0) + iv(x_0, y_0)]}{[x + iy] - [[x_0 - iy_0]} \\ &= \lim_{(x, y) \to (x_0, y_0)} \left (\frac{u(x, y) - u(x_0, y_0)}{[x - x_0] + i[y - y_0]} + i \frac{v(x, y) - v(x_0, y_0)}{[x - x_0] + i[y - y_0]} \right ) \\ &= \lim_{x \to x_0, y = y_0} \left (\frac{u(x, y_0) - u(x_0, y_0)}{x - x_0} + i \frac{v(x, y_0) - v(x_0, y_0)}{x - x_0} \right ) \\ &= \lim_{x \to x_0} \frac{u(x, y_0) - u(x_0, y_0)}{x - x_0} + i \lim_{x \to x_0} \frac{v(x, y_0) - v(x_0, y_0)}{x - x_0} \\ &= \frac{\partial u}{\partial x} (z_0) + i \frac{\partial v}{\partial x} (z_0) \end{align}
• And more generally, $\displaystyle{f'(z) = \frac{\partial u}{\partial x} (z) + i \frac{\partial v}{\partial x} (z)}$ $(*)$ for all $z \in \mathcal N$ since $f$ is analytic at $z_0$.
• Similarly we consider the limit as $z = x_0 + iy$ where $y \to y_0$:
(2)
\begin{align} \quad f'(z_0) &= \lim_{(x, y) \to (x_0, y_0)} \frac{u(x, y) + iv(x, y) - [u(x_0, y_0) + iv(x_0, y_0)]}{[x + iy] - [x_0 - iy_0]} \\ &= \lim_{(x, y) \to (x_0, y_0)} \left (\frac{u(x, y) - u(x_0, y_0)}{[x - x_0] + i[y - y_0]} + i \frac{v(x, y) - v(x_0, y_0)}{[x - x_0] + i[y - y_0]} \right ) \\ &= \lim_{x = x_0, y \to y_0} \left ( \frac{1}{i} \frac{u(x_0, y) - u(x_0, y_0)}{y - y_0} + \frac{v(x_0, y) - v(x_0, y_0)}{y - y_0} \right ) \\ &= -i \lim_{y \to y_0} \frac{u(x_0, y) - u(x_0, y_0)}{y - y_0} + \lim_{y \to y_0} \frac{v(x_0, y) - v(x_0, y_0)}{y - y_0} \\ &= -i \frac{\partial u}{\partial y} (z_0) + \frac{\partial v}{\partial y} (z_0) \end{align}
• And more generally, $\displaystyle{f'(z) = -i \frac{\partial u}{\partial y} (z) + \frac{\partial v}{\partial y} (z)}$ $(**)$ for all $z \in \mathcal N$ since $f$ is analytic at $z_0$.
• Comparing $(*)$ and $(**)$ gives us the following equality which holds for all $z \in \mathcal N$:
(3)
\begin{align} \quad \frac{\partial u}{\partial x}(z) + i \frac{\partial v}{\partial x}(z) = \frac{\partial v}{\partial y}(z) + i \left ( - \frac{\partial u}{\partial y}(z) \right ) \end{align}
• Therefore we have that $\displaystyle{\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}}$ and $\displaystyle{\frac{\partial v}{\partial x} = - \frac{\partial u}{\partial y}}$, (the Cauchy-Riemann equations) are satisfied on $\mathcal N$
• All that remains to show in this direction is that the partial derivatives of $u$ and $v$ are continuous on $\mathcal N$ which we currently do not have the tools to prove, so we will come back to this later on. $\square$

## The Complex Inverse Function Theorem

The extremely important inverse function theorem that is often taught in advanced calculus courses appears in many different forms. One of such forms arises for complex functions. We will state (but not prove) this theorem as it is significant nonetheless.

 Theorem 2 (The Complex Inverse Function Theorem): Let $A \subseteq \mathbb{C}$ be open and let $f : A \to \mathbb{C}$ and $z_0 \in A$. If $f$ is analytic at $z_0$ and $f'(z_0) \neq 0$ then there exists open neighbourhoods $U$ of $z_0$ and $V$ of $f(z_0)$ for which $f : U \to V$ is a bijection and the inverse function (which exists since $f$ restricted to $U$ is bijective), $f^{-1} : V \to U$ is analytic at $f(z_0)$ with $\displaystyle{\frac{d}{dw} f^{-1} (w) = \frac{1}{f'(z)}}$ where $w = f(z)$.

It is important to note that the complex inverse function tells us that provided that $f'(z_0) \neq 0$ then there exists a LOCAL analytic inverse function of $f$. It may be that no global inverse function exists even if $f'(z_0) \neq 0$ for every $z_0 \in \mathbb{C}$.

## Example 1

Determine whether or not the function $f(z) = \overline{z}$ is analytic on a subset $A$ of $\mathbb{C}$.

Let $z = x + yi$. Then if $f = u + iv$ then $f(z) = f(x + yi) = x - yi$ and so $u(x, y) = x$ and $v(x, y) = -y$. Notice that for all $z \in \mathbb{C}$ that:

(4)
So $\displaystyle{\frac{\partial u}{\partial x} \neq \frac{\partial v}{\partial y}}$, i.e., the Cauchy-Riemann equations are not satisfied for all $z \in \mathbb{C}$ and so $f$ is analytic nowhere.