3
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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 $z=0$.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Padraic Feb 18 '14 at 12:53
2
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.