$$ \lim_{z \rightarrow 3+i} z^2 = 8 + 6i $$
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.
$$ |z-(3+i)(z+(3+i)| < \epsilon $$ $$ |6+2i| < \epsilon $$ $$ |3+i| < \frac{\epsilon}{2} $$ $$ \delta = \frac{\epsilon}{2} $$
On
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 $|z-(3+i)|<\delta =\min\left(1, \varepsilon/(1+2\sqrt{10})\right)\le\min\left(1,\varepsilon/7\right)$.
On
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.
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}