Inquiry regarding the $\delta$-neighborhoods of $|z-z_0|$ and $|z+z_0|$ I am attempting to prove that:
$$\lim_{z \rightarrow z_0} (z^2 + c) = z^2_0 + c$$
From the limit definition, we want to show that $0 < |z-z_0| < \delta \implies |(z^2+c)-(z^2_0+c)| < \varepsilon$. We observe that:
\begin{align}
|(z^2+c)-(z^2_0+c)| &= |z^2-z^2_0| \\
&= |(z-z_0)(z+z_0)| \\
&= |z-z_0||z+z_0| \\
&< \varepsilon
\end{align}
I am unsure as to how to deal with the $|z+z_0|$ term. There are a few thoughts I would like to clarify:


*

*Is it fair to say that if $0 < |z-z_0| < \delta$ holds by premise, then we also have $0 < |z+z_0| < \delta$, since the only thing that changes is the center of the circle on the complex plane? My strategy to finish the proof would then be to write:


$$|z-z_0||z+z_0| < \delta^2$$
Then set $\delta = \sqrt\epsilon$. More generally, does $|z+z_0|$ have any relevance to the inequality $0 < |z-z_0| < \delta$?


*If not, then is it legal to fix some radius $R = |z+z_0|$, thus $|z-z_0| R < 
\varepsilon$ and finishing the proof by setting $\delta = \frac{\varepsilon}{R}$?

 A: You can show by the Triangle inequality that:
$$ |z + z_0| = |z - z_0 + 2z_0 | \leq |z - z_0| + 2|z_0| $$
and so
$$ |z-z_0||z + z_0| \leq |z - z_0|^2 + 2|z_0||z-z_0| $$
Therefore
\begin{align*}
|z-z_0| < \delta \implies |z-z_0||z + z_0| &\leq |z - z_0|^2 + 2|z_0||z-z_0| \\
&< \delta^2 + 2|z_0|\delta \\
&< \epsilon
\end{align*}
where
$$ \delta^2 + 2|z_0|\delta - \epsilon= 0 $$
You can prove that for any $\epsilon> 0$, there exists a $\delta > 0$ that solves that equation.
The determinant is: $\Delta = 4|z_0|^2+4\epsilon > 0$, so there is always a solution.
Furthermore, the solution
$$ \delta = \frac{-2|z_0| + \sqrt{4z_0^2+4\epsilon}}{2} $$
is positive since $4z_0^2+4\epsilon> 4z_0^2$, so $\sqrt{4z_0^2+4\epsilon} > 2|z_0|$ 
So for any choice of $\epsilon > 0$, we can choose an $\delta > 0$ by the formula above, such that the required inequality holds.
A: First, $|z-z_0|<\delta$ does not imply that $|z+z_0|<\delta$.  To use an example to show this, suppose that $z_0=1$ and $\delta=1$.  Then, we see that for $z=3/2$, $|z+z_0|=5/2>\delta$.
We can set, as you suggested, an initial bound. But we set that initial bound for $|z-z_0|$.  For example, we require that $|z-z_0|\le 1$.  Then, note that
$$|z+z_0|=|z-z_0+2z_0|\le |z-z_0|+2|z_0|<1+2|z_0|$$
Can you finish now?  
