Prove the limit below by using epsilon-delta definition Prove $$\lim_{z\rightarrow4+3i}\overline{z}^2 =(4-3i)^2$$ by using epsilon-delta definition.
My working:
Assume $|z-(4-3i)|<1 \implies |z|<6$
\begin{align}
|\overline{z}^2-(4-3i)^2| &=|[\overline{z}-(4-3i)][\overline{z}+(4-3i)]|\\
&=|\overline{z}-(4-3i)||\overline{z}+(4-3i)|\\
&=|\overline{z}-(4+3i)+6i||\overline{z}-(4+3i)+8|\\
&=|\overline{z}-(4+3i)|\left[\left|1+\frac{6i}{\overline{z}-(4+3i)}\right|\left|1+\frac{8}{\overline{z}-(4+3i)}\right|\right]\\
&\le \delta \left[\left(1+\frac{|6i|}{|\overline{z}-(4+3i)|}\right)\left(1+\frac{8}{|\overline{z}-(4+3i)|}\right)\right]\\
&< 63\delta
\end{align}
Is this correct?

Careful, you can assume $|z - (4+3i)| < \delta$, not $|\overline{z} - (4+3i)| < \delta$.
If we pick $0 < \delta < -5+\sqrt{25+\varepsilon}$ we have
\begin{align}
\left|\overline{z}^2 - (4-3i)^2\right| &= |\overline{z} - (4-3i)|\cdot |\overline{z} + (4-3i)|\\
&= \left|\overline{z - (4+3i)}\right|\cdot \left|\overline{z + (4+3i)}\right|\\
&= |z - (4+3i)|\cdot |z + (4+3i)|\\
&= |z - (4+3i)|\cdot |z - (4+3i) + (8+6i)|\\
&\le  |z - (4+3i)|\cdot \big(|z - (4+3i)| + |8+6i|\big)\\
&<  \delta(\delta + 10)\\
&< \varepsilon
\end{align}
