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