# Proof that a complex function is continuous at $z=0$

Given the function $f\colon \mathbb{C}\to\mathbb{C}$ by $f(z)=\begin{cases} \frac{xy(x+iy)}{x^2+y^2} & \text{if } z\neq 0\\ 0 & \text{if } z=0 \end{cases}$ with $z=x+iy$.

How do I proof that $f$ is continuous in $z=0$?

I've tried using the $\epsilon$-$\delta$ -definition, giving me to proof $|x+iy|<\delta \Rightarrow \left|\frac{xy(x+iy)}{x^2+y^2}\right|<\epsilon$.

I've tried rewriting the right component, but I just can't seem to figure it out. I just don't 'see' it. Can someone give me a hint to the solution? Thanks.

$$\left|\frac{xy(x+iy)}{x^2+y^2}\right| = |z| \frac{|xy|}{x^2+y^2} \leq \frac{1}{2}|z|$$ because $|xy| \leq \frac{1}{2}(x^2+y^2)$.
• So that would give me a solution, by using $\epsilon=\frac12\delta$. However, on what basis can you state the last inequality? – RBS Mar 10 '14 at 11:15
• $x^2+2xy+y^2=(x+y)^2\ge 0$. – Martín-Blas Pérez Pinilla Mar 10 '14 at 11:24
substitute $x=r\cos \theta$ and $y=r\sin \theta$, the expression $f(z)$ becomes $r\cos \theta \sin \theta e^{i\theta}$, continuity follows from boundedness of $\cos \theta \sin \theta e^{i\theta}$ as $r \to 0$