Let $a,b, z_0$ denote complex constants. Use the definition of a limit to show that
$$\lim_{z \to z_0} (az + b) = az_0 + b.$$
Here is what I have done:
\begin{align*}
|az + b – (az_0 + b)| &= |az – az_0 + b – b|\\
&= |a(z – z_0)|\\
&= |a||z – z_0|.
\end{align*}
So for a positive number $\epsilon$,
$$|az + b – (az_0 + b)| < \epsilon \text{ whenever } |a||z – z_0| < \epsilon$$
or in other words $|az + b – (az_0 + b)| < \epsilon$ whenever $|z – z_0| < \delta$ where $\delta = \epsilon/|a|$.
Have I proved the statement correctly?
Best Answer
As mentioned in the comments, your proof works for $a \neq 0$. You can deal with the case where $a = 0$ separately or choose $\epsilon$ in such a way that the proof works for all $a$ (see N.S.'s comment for example). Aside from this your proof is correct.