The Cauchy-Riemann Theorem Examples 2

Recall from The Cauchy-Riemann Theorem page that if $A \subseteq \mathbb{C}$ is open, $f : A \to \mathbb{C}$ with $f = u + iv$, and $z_0 \in A$ then $f$ is analytic at $z_0$ if and only if there exists a neighbourhood $\mathcal N$ of $z_0$ with the following properties:

• 1) $\displaystyle{\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial v}{\partial x}, \frac{\partial v}{\partial y}}$ all exist and are continuous on $\mathcal N$.

• 2) The Cauchy-Riemann equations $\displaystyle{\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}}$ and $\displaystyle{\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}}$ on $\mathcal N$.

We also stated an important result that can be proved using the Cauchy-Riemann theorem called the complex Inverse Function theorem which says that if $f'(z_0) \neq 0$ then there exists open neighbourhoods $U$ of $z_0$ and $V$ of $f(z_0)$ such that $f : U \to V$ is a bijection and such that $\displaystyle{\frac{d}{dw} f^{-1}(w) = \frac{1}{f'(z)}}$ where $w = f(z)$.

We will now look at some more example problems in applying the Cauchy-Riemann theorem.

Example 1

Let $A \subseteq \mathbb{C}$ be open and $f : A \to \mathbb{C}$. Let $A^* = \{ z : \overline{z} \in A \} \subseteq A$ and define $g : A^* \to \mathbb{C}$ by $g(z) = \overline{f(\overline{z})}$. Prove that $g$ is then analytic on $A^*$.

Let $f = u + iv$. Then $f(x, y) = u(x, y) + iv(x, y)$. Then by the definition of $g$ we have:

(1)

\begin{align} \quad g(x, y) &= \overline{f(\overline(z)} \\ &= \overline{f(x, -y)} \\ &= \overline{u(x, -y) + iv(x, -y)} \\ &= u(x, -y) - iv(x, -y) \end{align}

We have that:

(2)

\begin{align} \quad \frac{\partial u(x, -y)}{\partial x} = \frac{\partial u}{\partial x} \biggr \lvert_{(x, -y)} = \frac{\partial v}{\partial y} \biggr \lvert_{(x, -y)} = \frac{\partial v(x, -y)}{\partial y} \end{align}

(3)

\begin{align} \quad \frac{\partial -v(x, -y)}{\partial y} = \frac{\partial v(x, -y)}{\partial y} \end{align}

And also:

(4)

\begin{align} \quad \frac{\partial u(x, -y)}{\partial y} = \frac{\partial u}{\partial y} \biggr \lvert_{(x, -y)} = -\frac{\partial v}{\partial x} \biggr \lvert_{(x, -y)} = - \frac{\partial v(x, -y)}{\partial x} \end{align}

(5)

\begin{align} \quad \frac{\partial -v(x, -y)}{\partial x} = -\frac{\partial v(x, -y)}{\partial x} \end{align}

So the Cauchy-Riemann equations are satisfied for $g$ which means that $g$ is analytic on $A^*$.

Example 2

Use the Cauchy-Riemann theorem to prove that $f(z) = \mid z \mid$ is not analytic.

We have that:

(6)

\begin{align} f(z) = f(x, y) = \sqrt{x^2 + y^2} + 0i \end{align}

So $u(x, y) = \sqrt{x^2 + y^2}$ and $v(x, y) = 0$. So the partial derivatives of these functions are:

(7)

\begin{align} \quad \frac{\partial u}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} \quad , \quad \frac{\partial u}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}} \quad , \quad \frac{\partial v}{\partial x} = 0 \quad , \quad \frac{\partial v}{\partial y} = 0 \end{align}

The Cauchy-Riemann equations hold nowhere for this function so $f$ is not analytic.

Example 3

Let $f$ be analytic on an open connected set $A$ and such that $\mathrm{Re} f(z) = C$ where $C \in \mathbb{R}$ is a constant. Prove that $f$ is constant on $A$.

We have that $\mathrm{Re} f(z) = u(x, y) = C$ and that $f$ is analytic on $A$, so by the Cauchy-Riemann equations we must have that:

(8)

\begin{align} \quad \frac{\partial u}{\partial x} = 0 = \frac{\partial v}{\partial y} \end{align}

And also:

(9)

\begin{align} \quad \frac{\partial u}{\partial y} = 0 = -\frac{\partial v}{\partial x} \end{align}

In particular, $\frac{\partial v}{\partial x} = 0$ and $\frac{\partial v}{\partial y} = 0$ on all of $A$. Thus $v$ must be constant on $A$ (since $A$ is open and connected), i.e., there exists a $D \in \mathbb{R}$ such that $v(x, y) = D$ on $A$. But $v(x, y) = \mathrm{Im} f(z) = D$, so on all of $A$:

(10)

\begin{align} \quad f(x, y) = C + Di \end{align}

In other words, $f$ is constant on $A$.