Complex Analysis Limit Proof

Question

Use the $\delta - \epsilon$ definition of a limit to prove the following $\displaystyle \lim_{z\rightarrow0 } \frac{\bar z^2}{z} = 0$.

I'm struggling with the proof. I wrote something up but not sure if it proves what I am asked to prove. My proof is as follows.

Answer 1

What you want to show should be

$\forall \epsilon >0, \exists \delta >0$, such that $\color{blue}{0< |z-0| < \delta}$ ,then $\left| \frac{\bar{z}^2}{z}-0\right|< \epsilon$

You are right that you can pick $\delta$ to be equal to $\epsilon$ for this question.

and indeed, we do have for $z \neq 0$, $\left| \frac{\bar{z}^2}{z}-0\right|=|z|=|z-0|$ but from the working, it's hard to judge whether you understand the definition.

Answer 2

Note that for all nonzero $z\in\mathbb{C}$, $$\left|{{\overline z}\over z}\right| \le 1.$$ Now it's obvious.