The Cauchy-Riemann Theorem Examples 1

# The Cauchy-Riemann Theorem Examples 1

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 example problems in applying the Cauchy-Riemann theorem.

## Example 1

Determine whether the function $f(z) = \overline{z}$ is analytic or not.

Let $f(z) = f(x + yi) = x - yi = \overline{z}$. Then $u(x, y) = x$ and $v(x, y) = -y$. The first order partial derivatives of $u$ and $v$ clearly exist and are continuous. They are:

(1)

So the first condition to the Cauchy-Riemann theorem is satisfied. However note that $\displaystyle{1 = \frac{\partial u}{\partial x} \neq \frac{\partial v}{\partial y} = -1}$ ANYWHERE. So one of the Cauchy-Riemann equations is not satisfied anywhere and so $f(z) = \overline{z}$ is analytic nowhere.

## Example 2

Determine whether the function $f(z) = e^{z^2}$ is analytic or not using the Cauchy-Riemann theorem.

Let:

(2)
\begin{align} \quad f(z) = f(x + yi) = e^{(x + yi)^2} = e^{(x^2 - y^2) + 2xyi} = e^{x^2 - y^2} e^{2xyi} = e^{x^2 - y^2} \cos (2xy) + e^{x^2 - y^2} \sin (2xy) i \end{align}

Then $u(x, y) = e^{x^2 - y^2} \cos (2xy)$ and $v(x, y) = e^{x^2 - y^2} \sin (2xy)$. The partial derivatives of these functions exist and are continuous. They are given by:

(3)
\begin{align} \quad \frac{\partial u}{\partial x} = 2x e^{x^2 - y^2} \cos (2xy) - 2y e^{x^2 - y^2} \sin (2xy) = e^{x^2 - y^2} [2x \cos (2xy) - 2y \sin (2xy)] \end{align}
(4)
\begin{align} \quad \frac{\partial v}{\partial y} = -2ye^{x^2 - y^2} \sin(2xy) + 2x e^{x^2 - y^2} \cos (2xy) = e^{x^2 - y^2}[2x \cos (2xy) - 2y \sin (2xy)] \end{align}

So $\displaystyle{\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}}$ everywhere. Also:

(5)
\begin{align} \quad \frac{\partial u}{\partial y} =-2ye^{x^2 - y^2} \cos (2xy) - 2x e^{x^2 - y^2} \sin (2xy) = -e^{x^2 - y^2}[2x \sin (2xy) + 2y \cos (2xy)] \end{align}
(6)
\begin{align} \quad \frac{\partial v}{\partial x} = 2xe^{x^2 - y^2}\sin(2xy) + 2ye^{x^2 - y^2}\cos(2xy) = e^{x^2 - y^2}[2x \sin (2xy) + 2y \cos(2xy)] \end{align}

So $\displaystyle{\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}}$ everywhere as well.

Thus by the Cauchy-Riemann theorem, $f(z) = e^{z^2}$ is analytic everywhere. This should intuitively be clear since $f$ is a composition of two analytic functions.

## Example 3

Prove that if $f$ is analytic at then $\displaystyle{\mid f'(z) \mid^2 = \left (\frac{\partial u}{\partial x} \right )^2 + \left ( \frac{\partial v}{\partial x} \right )^2}$ and $\displaystyle{\mid f'(z) \mid^2 = \left (\frac{\partial u}{\partial y} \right )^2 + \left ( \frac{\partial v}{\partial y} \right )^2}$.

Suppose that $f$ is analytic. Then from the proof of the Cauchy-Riemann theorem we have that:

(7)
\begin{align} \quad f'(z) = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} \end{align}

Therefore:

(8)
\begin{align} \quad \mid f'(z) \mid = \sqrt{ \left( \frac{\partial u}{\partial x} \right )^2 + \left ( \frac{\partial v}{\partial x} \right )^2} \end{align}

Hence:

(9)
\begin{align} \quad \mid f'(z) \mid^2 = \left( \frac{\partial u}{\partial x} \right )^2 + \left ( \frac{\partial v}{\partial x} \right )^2 \end{align}

The other formula can be derived by using the Cauchy-Riemann equations or by the fact that in the proof of the Cauchy-Riemann theorem we also have that:

(10)
\begin{align} \quad f'(z) = \frac{\partial v}{\partial y} -i\frac{\partial u}{\partial y} \end{align}