complex analysis - 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?

1 Answer

  1. Patrick- Reply

    2019-11-14

    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}

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