Quiz review rectification exercice

I want to rectify something about the quiz review, it's in exercise3: $\lim_{z \to z_0} f(z) = f(z_0)$ instead of $\lim_{z \to z_0} f(z) = f(z)$

I think you're totally right. We could then show the limit exists by choosing $\delta=\epsilon/\sqrt{a\bar{a}}$.

Proof. Let $a,b\in\mathbb{C}$ and $f(z)=az+b$. Suppose that $|z-z_0|<\delta$ and that $\delta=\epsilon/\sqrt{a\bar{a}}$ for some $\delta,\epsilon\in\mathbb{R}$ such that $\delta>0$ and $\epsilon>0$. Then $$\begin{align}|z-z_0|<\epsilon/\sqrt{a\bar{a}}&\implies\sqrt{a\bar{a}}|z-z_0|&<\epsilon\\&\implies|a||z-z_0|&<\epsilon\\&\implies|a(z-z_0)|&<\epsilon\\&\implies|az-az_0|&<\epsilon\\&\implies|az+(b-b)-az_0|&<\epsilon\\&\implies|(az+b)+(-b-az_0))|&<\epsilon\\&\implies|(az+b)-(b+az_0)|&<\epsilon\\&\implies|(az+b)-(az_0+b)|&<\epsilon\\&\implies|f(z)-f(z_0)|&<\epsilon\end{align}$$ Therefore, $\lim\limits_{z\rightarrow z_0}f(z)=f(z_0)$ exists. QED.

@dgallimo Definitely good enough for a "like". But, how do you choose $\varepsilon$ if $a=0$?

if $a=0$ we have $ f(z)= b$ for every z in $\mathbb{C}$
we have then $f(z)=f(z_0)=b \rightarrow f(z)-f(z_0)=0$
let $\epsilon >0$ then $ |f(z)-f(z_0)|< \epsilon$
if we take $\delta=\frac{\epsilon}{2}$ we have then
$0=|f(z)-f(z_0)|<|z-z_0|<\delta<\epsilon$. QED!

@mark Is it the case that $\delta$ can be any positive real number? If $a=0$, then $f(z)=b$, so we need to show that $|z-z_0|<\delta\implies|b-b|=|0|=0<\epsilon$. However, $0$ is always less than $\epsilon$ since $\epsilon$ is greater than $0$.