# 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)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)Suppose that $z_0 = 0$. Then:

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

(5)We also have that:

(6)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 $A$.