With complex $z$, prove $\lim\limits_{z \to z_0} z^2 = z_0^2$

Question

With complex $z$, prove $\lim\limits_{z \to z_0} z^2 = z_0^2$

This is obviously true at an intuitive level but I'm having trouble with the actual proof. I suspect I'm missing a rather obvious step. This seems an absurdly simple problem.

My proof set up:

We must show that for any positive number $\mathcal{E}$ that there exists a positive number $\delta$ so that:

$\left|z - z_0\right| < \delta \implies \left|z^2 - z_0^2\right| < \mathcal{E}$

So that's just the basic setup for a limit proof. There is one step to do the actual proof and I can't see what it is. Thank you for any help.

Answer 1

The simplest solution is to use the fact that

\begin{align*} \lim\limits_{z \to z_0} f(z)g(z) &= \left( \lim\limits_{z \to z_0} f(z) \right) \left( \lim\limits_{z \to z_0} g(z) \right) \\ \lim\limits_{z \to z_0} z^2 &= \left( \lim\limits_{z \to z_0} z \right) \left( \lim\limits_{z \to z_0} z \right) \\ &= z_0^2 \end{align*}

The proof for the first statement is in my textbook book, and it's based on some other proofs that are best understood in separate steps. Once you have that general theorem proven, problems like the original question I posted are quite trivial.

Anyway, thanks for the help.

Answer 2

HINT:

$$|z^2-z_0^2|=|z-z_0|\,|z+z_0|$$

Now, choose a neighborhood around $z_0$ to bound $|z+z_0|$ from above. Note that $|z+z_0|=|z-z_0+2z_0|$ and apply the triangle inequality.