Analytic Complex Functions Examples 1

Analytic Complex Functions Examples 1

Recall from the Analytic Complex Functions page that if $A \subseteq \mathbb{C}$ is open, $z_0 \in A$, and $f : A \to \mathbb{C}$ then $f$ is said to be analytic at $z_0$ if there exists an $r > 0$ for which $f$ is complex differentiable on the open disk $D(z_0, r)$, and $f$ is said to be analytic on $A$ if $f$ is analytic at each $z_0 \in A$.

We will now look at some example problems involving the concept of analyticity.

Example 1

Show that the function $f(z) = z^2$ is analytic on all of $\mathbb{C}$.

To show that $f$ is analytic everywhere we will show that $f$ is differentiable everywhere.

Let $z_0 \in \mathbb{C}$. Then:

(1)
\begin{align} \quad f'(z_0) &= \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0} \\ &= \lim_{z \to z_0} \frac{z^2 - z_0^2}{z - z_0} \\ &= \lim_{z \to z_0} \frac{(z + z_0)(z - z_0)}{z - z_0} \\ &= \lim_{z \to z_0} (z + z_0) \\ &= 2z_0 \end{align}

So $f$ is differentiable on all of $\mathbb{C}$ and in fact $f'(z) = 2z$.

Example 2

Show that the function $f(z) = \mid z \mid^2$ is differentiable only at the point $z_0 = 0$ and deduce that $f$ is analytic nowhere.

Let $z_0 \in \mathbb{C}$ and consider:

(2)
\begin{align} \quad f'(z_0) &= \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0} \\ &= \lim_{z \to z_0} \frac{\mid z \mid^2 - \mid z_0 \mid^2}{z - z_0} \\ &= \lim_{z \to z_0} \frac{z \cdot \overline{z} - z_0 \cdot \overline{z_0}}{z - z_0} \\ &= \lim_{z \to z_0} \frac{z \cdot \overline{z} - z \cdot \overline{z_0} + z \cdot \overline{z_0} - z_0 \cdot \overline{z_0}}{z - z_0} \\ &= \lim_{z \to z_0} \frac{z(\overline{z} - \overline{z_0} + \overline{z_0} (z - z_0)}{z - z_0} \\ &= \lim_{z \to z_0} \left ( z \cdot \frac{\overline{z} - \overline{z_0}}{z - z_0} + z_0 \right ) \end{align}

Suppose that $z_0 = 0$. Then:

(3)
\begin{align} \quad f'(0) = \lim_{z \to 0} z \cdot \frac{\overline{z}}{z} = \lim_{z \to 0} \overline{z} = 0 \end{align}

So $f$ is differentiable at $0$ and $f'(0) = 0$.

Now suppose that $z_0 \neq 0$. We will show that the following limit does not exist:

(4)
\begin{align} \quad \lim_{z \to z_0} \frac{\overline{z} - \overline{z_0}}{z - z_0} \end{align}

We will look at the limit along the vertical and horizontal lines passing through these points. We have that:

(5)
\begin{align} \quad \lim_{z \to z_0, y = y_0} \frac{\overline{z} - \overline{z_0}}{z - z_0} &= \lim_{x \to x_0, y = y_0} \frac{(x - yi) - (x_0 - y_0i)}{(x - x_0) + (y - y_0)i} \\ &= \lim_{x \to x_0} \frac{x - x_0}{x - x_0} \\ &= 1 \end{align}

We also have that:

(6)
\begin{align} \quad \lim_{z \to z_0, x = x_0} \frac{\overline{z} - \overline{z_0}}{z - z_0} &= \lim_{x = x_0, y \to y_0} \frac{(x - yi) - (x_0 - y_0i)}{(x - x_0) + (y - y_0)i} \\ &= \lim_{y \to y_0} \frac{-(y - y_0)i}{(y - y_0)i} \\ &= -1 \end{align}

Since the limit as $z \to z_0$ on $f$ approaches two different values along different paths we must have that the general limit, $\displaystyle{\lim_{z \to z_0} \frac{\overline{z} - \overline{z_0}}{z - z_0}}$ does not exist. Therefore $f'(z_0)$ does not exist whenever $z_0 \neq 0$.

Thus the only point for which $f$ is differentiable is at $z_0 = 0$ and so $f$ cannot be analytic anywhere since there exists no open subset $A$ of $\mathbb{C}$ for which $f$ is differentiable on all of $A$.

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