# Let $f$ be an entire function, with $f(0)=f'(0)=0$. Is $g(z)=\frac {f(z)}{z^2}$ for $z \neq 0$, $g(0)=0$ holomorphic?

Let $f$ be an entire function, with $f(0)=f'(0)=0$. Is $g(z)=\frac {f(z)}{z^2}$ for $z \neq 0$, $g(0)=0$ holomorphic?

So, I think not. It seems like $z=0$ is a pole of order 1, because if it seems like if it were $g(z)=\frac {f(z)}{z}$ for $z \neq 0$ and $g(0)=0$ then $g$ was holomorphic. How can I show it isn't holomorphic / or it is?

• since $0$ is a zero of $f$ of multiplicity at least $2$ (because of the vanishing of the function and its first derivative) you can write $f(z)=z^nh(z)$ with $h$ holomorphic and $n\ge2$ in a neighbourhood of $0$. Now conclude. Jun 15, 2015 at 17:50
• So $h(0) \neq 0$, and is actually entire? Or can I fix it and make it entire by setting $h(0)=0$? Jun 15, 2015 at 17:55
• I see. So $g$ is holomorphic at $z=0$. Hope I'm not missing anything Jun 15, 2015 at 18:02
Since $f$ is entire, for all $z\in\mathbb{C}$, $$f(z)=\sum_{n\ge0}\frac{f^{(n)}(0)}{n!}z^n= \sum_{n\ge2}\frac{f^{(n)}(0)}{n!}z^n= z^2\biggl(\sum_{n\ge0}\frac{f^{(n+2)}(0)}{(n+2)!}z^n\biggr)$$ so $$h(z)=\sum_{n\ge0}\frac{f^{(n+2)}(0)}{(n+2)!}z^n$$ is entire. Its value at $0$ is $$h(0)=\frac{f''(0)}{2}$$ So, unless $f''(0)=0$, the function $g$ is not holomorphic at $0$.
• Yeah, I deleted it right away when realizing my mistake. So if I set $g(0)=A$, which is some arbitrary value, I get a holomorphic function? Jun 15, 2015 at 18:18
• @Badstudent No, only $g(0)=f''(0)/2$ is good. Jun 15, 2015 at 19:14