I am teaching myself complex analysis and after learning about epsilon-delta limits, I saw an exercise that asked to prove that
$$\lim_{z\to z_0}z^2+c=z_0^2+c$$
I would like to know if this approach is correct, since there are no answers in the textbook, and this approach isn’t showcased in any supplementary examples:
Prove that
$$f(z)=z^2+c \Rightarrow |f(z)-(z_0^2+c)|=|z^2+c-(z_0^2+c)|=|z^2-z_0^2|=|(z-z_0)(z+z_0)|=|z-z_0||z+z_0|$$
So if for some $\epsilon>0$ and $0<|z-z_0|<\frac{\epsilon}{|z+z_0|}$, then $|f(z)-(z_0^2+c)|<\epsilon$. Thus, the limit definition is satisfied, taking $\delta$ to be $\frac{\epsilon}{|z+z_0|}$.
Any input would be appreciated.
Best Answer
You cannot make $\delta$ dependent on $z$. Use the fact that $|z-z_0||z+z_0| \leq |z-z_0|(|z-z_0|+2|z_0|) <|z-z_0|(|1+2|z_0|)<\epsilon$ if $|z-z_0| <1$ and $|z-z_0| <\frac {\epsilon} {1+2|z_0|} $. So we can take $\delta =\min \{1, \frac {\epsilon} {1+2|z_0|} \}$.