If $f(z)$ is entire, why is $g(z)=\frac{f(z)-f(0)}{z}$ entire?

Question

I saw on another post that if $f(z)$ is entire, and $\lim\frac{f(z)}{z}=0$, then $g(z)=\frac{f(z)-f(0)}{z}$ is entire. Why is this true? I understand that by the way it's defined, $g$ is continuous (if you add on the point $f'(0)$ for when $z=0$).

Answer 1

Because if$$f(z)=a_0+a_1z+a_2z^2+\cdots,\tag1$$then$$\frac{f(z)-f(0)}z=a_1+a_2z+a_3z^2+\cdots\tag2$$and, since the series $(1)$ converges everywhere, then so does the series $(2)$. So, its sum is an entire function.