# Proving a complex function is continuous.

I've recently started complex analysis but I have very little background in complex numbers and to make sure I don't fall behind I'm doing some extra exercises one of which is

Show $f$ is continuous on $\mathbb C$

$${f(z)=\begin{cases} z^{2} / |z| \qquad &\mbox{when } z\neq0,\\ 0 \qquad &\mbox{when }z=0. \end{cases}}$$

I know the epsilon delta proof for limits quite well and the limit definition of continuous but I don't know how to apply them to this function. Can anybody help or at least get me started?

## 2 Answers

I'm sure you will agree with me that for $z \not= 0$, $f(z)$ is continuous.

The only point of concern is the continuity at $z=0$.

Now consider

$$|f(z)|=|\frac{z^2}{|z|}|=\frac{|z|^2}{|z|}=|z|$$

and we can use a result:

$|f|$ has a limit $0$ as $z \rightarrow c$ $\iff$ $f$ has a limit $0$ as $z \rightarrow c$

And we have

$$\lim_{z \to 0}|z|=0$$ So $$\lim_{z \to 0}f(z)=0=f(0)$$

and it follows that $f$ is continuous at $z=0$.

• Take the modulus of f(z)! Thank you. I had your first statement written and had gotten down to zero being the point of concern. Thanks again very easy to follow. – Padraic Feb 18 '14 at 12:53

$$\lim_{z \to 0} \ \ \left | \frac{ \ \ z^{2}}{|z|} \right | = \lim_{z \to 0} \ \ \left |z \cdot \frac{ z}{|z|} \right | = \lim_{z \to 0} \ \ |z| \cdot\left | \frac{ z^{}}{|z|} \right | = \lim_{z \to 0} \ \ |z| = 0$$

But $0$ is the only complex number with modulus $0$ , so $$\lim_{z \to 0} \ \ \frac{ \ \ z^{2}}{|z|} = 0$$ and then your function is continous