I have since seen more concise proofs, but am curious if this works.
If f is differentiable at an interior point $z_0$, then:
$\forall\epsilon>0,\exists \delta_1>0$ such that
$|z-z_0|<\delta_1\implies|\frac{f(z)-f(z_0)}{z-z_0}-f'(z_0)|<\sqrt{\epsilon/2}$.
My main concern is this part; is it valid to have $\sqrt{\epsilon/2}$ rather than $\epsilon$. I've seen $c\epsilon$ used with $c>0$. I thought doing this was valid because $\epsilon$ and $c\epsilon$ have the same range for $\epsilon>0$. If this is the case $\sqrt{\epsilon/2}$ should be valid too right?
I also use $\lim_{z\to z_0}(z-z_0)=0$:
$\forall\epsilon>0,\exists \delta_2>0$ such that
$|z-z_0|<\delta_2\implies|z-z_0|<\sqrt{\epsilon/2}$
and $\lim_{z\to z_0}f'(z_0)(z-z_0)=0$:
$\forall\epsilon>0,\exists \delta_3>0$ such that
$|z-z_0|<\delta_3\implies|z-z_0||f'(z_0)|<\epsilon/2$
Let $\delta=min\{\delta_1,\delta_2,\delta_3\}$.
$|f(z)-f(z_0)|=|z-z_0||\frac{f(z)-f(z_0)}{z-z_0}+f'(z_0)-f'(z_0)|$
$\leq |z-z_0||\frac{f(z)-f(z_0)}{z-z_0}-f'(z_0)|+|z-z_0||f'(z_0)|$
$<\sqrt{\epsilon/2}\sqrt{\epsilon/2}+\epsilon/2=\epsilon$
Thanks
EDIT:
Forgot the final line:
Therefore $\forall \epsilon>0$ $\exists \delta=min\{\delta_1,\delta_2,\delta_3\}$, such that:
$0<|z-z_0|<\delta\implies|f(z)-f(z_0)|<\epsilon$
Therefore differentiability at z_0 implies continuity at z_0
Best Answer
You proof seems fine.
You might want to be careful with $\color{blue}{0<}|z-z_0|<\delta$.
The reason why you can choose $\sqrt{\frac{\epsilon}2}$ is because it is positive.