Prove that $f(z)=z^2$ is continuous for all complex and real values of $z$.
What I've got so far is:
Given $ \epsilon >0$ and $|z-z_0|<\delta$ after some calculations (which I've checked with the answer key)
$$ |f(z)-f(z_0)|<\delta(\delta+2|z_0|) $$
Beyond this things get difficult when trying to create $\epsilon$ as a function of $\delta$, the answer reads:
$$\delta(\delta+2|z_0|)\leq \frac{\epsilon}{3|z_0|}(|z_0|+2|z_0|)=\epsilon $$
and I have no clue how to get there.
Best Answer
Given $\varepsilon > 0$, set $\delta = \min\{1, \varepsilon/(1 + 2|z_0|)\}$. For all $z$, $|z - z_0| < \delta$ implies $|z - z_0| < 1$ (which implies $|z + z_0| < 1 + 2|z_0|$ by the triangle inequality) and $$|f(z) - f(z_0)| = |z + z_0||z - z_0| < (1 + 2|z_0|)\frac{\varepsilon}{1 + 2|z_0|} = \varepsilon.$$