Use epsilon-delta definition to determine $\lim_{z \rightarrow 3+i} z^2 = 8 + 6i $ $$
\lim_{z \rightarrow 3+i} z^2 = 8 + 6i
$$
Is this correct? I am not sure how to go further.
$$
0 < |z-(3+i)| <  \delta
$$
$$
|z^2-(8+6i)| <  \epsilon
$$
$$
\delta = \epsilon
$$
A possible way forward, but I am not sure.
$$
 |z-(3+i)(z+(3+i)| <  \epsilon
$$
$$
 |6+2i| <  \epsilon
$$
$$
 |3+i| <  \frac{\epsilon}{2}
$$
$$
\delta = \frac{\epsilon}{2}
$$
 A: For any given $\epsilon > 0$, choose $\delta = \min\left\{1,\frac{\epsilon}{1+2\sqrt{10}}\right\}$. Then
\begin{align}
0 < |z-(3+i)| < \delta \quad\Rightarrow\quad |z^2 - (8+6i)| 
&= |z-(3+i)||z+(3+i)| \\
&< \delta\big[|z-(3+i)| + 2|3+i|\big] \\
&< \delta\big(\delta + 2\sqrt{10}\big) \\
&< \delta\big(1+ 2\sqrt{10}\big) \\
&< \epsilon
\end{align}
A: We will show that for any $z_0\in \mathbb{C}$, we have
$$\lim_{z\to z_0}z^2=z_0^2$$
We will restrict $z$ such that $|z-z_0|\le 1$ so that $|z-z_0|^2\le |z-z_0|$.  Then, we have for any $\varepsilon>0$
$$\begin{align}
|z^2-z_0^2|&=|z-z_0|\,|z+z_0|\\\\
&=|z-z_0|\,|z-z_0+2z_0|\\\\
&\le |z-z_0|\left(|z-z_0|+2|z_0|\right)\\\\
&\le |z-z_0| \left(1+2|z_0|\right)\\\\
&<\varepsilon
\end{align}$$
whenever $|z-z_0|<\delta =\min\left(1,\varepsilon/\left(1+2|z_0|\right)\right)$.

For the case $z_0=3+i$, $|z_0|=\sqrt{10}$, we have for any $\varepsilon>0$
$$|z^2-(3+i)^2|<\varepsilon$$
whenever $|z-(3+i)|<\delta =\min\left(1, \varepsilon/(1+2\sqrt{10})\right)\le\min\left(1,\varepsilon/7\right)$.
A: Recall that $\lim\limits_{z \to a}f(z) = L$ by definition means:
$$ \forall \epsilon > 0 \; \exists \delta > 0: 0 < |z-a|<\delta \implies |f(x)-f(a)| < \epsilon.$$
Make sure you fully understand the definition before looking at the approach below.


*

*Observe that to make $|z^2 - (8+6i)| < \epsilon$ one may factor LHS to $|z-(3+i)||z+(3+i)| < \epsilon$.


*We can make $|z-(3+i)|$ arbitrarily close to $0$ and noted that intuitively $|z+(3+i)|$ is bigger than a constant as $|z-(3+i)|$ being arbitrarily small(the proof is below).
(a.) The idea is $|z-(3+i)||z+(3+i)| < \epsilon$ is equivalent to
$$ |z-(3+i)| < \frac{\epsilon}{|z+(3+i)|}$$
where $\frac{\epsilon}{|z+(3+i)|}$ should be greater than$\frac{\epsilon}{C}$, where $C$ is a constant for $z$ we care(why? See below).
(b.) Noted that $|z+(3+i)| = |z-(-3-i)| \geq |(3+i)-(-3-i)|-|z-(3+i)|$ (by triangle inequality. Draw the triangle on x-y plane if you gets confused here, this might help a lot). If we only consider $\delta$ we choose that is $< 1$, then $|z-(3+i)|<1$, then $|(3+i)-(-3-i)|-|z-(3+i)| > 4\sqrt20 - 1$.

*

*Combine (a) (b) we shall only consider
$$ |z-(3+i)| < \frac{\epsilon}{|z+(3+i)|} < \frac{\epsilon}{|4\sqrt20 - 1|}.$$
For any $\epsilon > 0$ take $\delta < \frac{\epsilon}{|4\sqrt20 - 1|}$, then we are done.


*Conclusively, take $\delta = \min\{1, \frac{\epsilon}{|4\sqrt20 - 1|}\}$ should end the proof.
