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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?

Thanks in advance!

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  • $\begingroup$ 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. $\endgroup$
    – b00n heT
    Jun 15, 2015 at 17:50
  • $\begingroup$ So $h(0) \neq 0$, and is actually entire? Or can I fix it and make it entire by setting $h(0)=0$? $\endgroup$
    – Badstudent
    Jun 15, 2015 at 17:55
  • $\begingroup$ no. I'm not stating that. The statement is that locally your function looks like the above, not globally. $\endgroup$
    – b00n heT
    Jun 15, 2015 at 17:57
  • $\begingroup$ I see. So $g$ is holomorphic at $z=0$. Hope I'm not missing anything $\endgroup$
    – Badstudent
    Jun 15, 2015 at 18:02
  • $\begingroup$ yep. That should do. $\endgroup$
    – b00n heT
    Jun 15, 2015 at 18:06

1 Answer 1

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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 $0$.

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  • $\begingroup$ 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? $\endgroup$
    – Badstudent
    Jun 15, 2015 at 18:18
  • $\begingroup$ @Badstudent No, only $g(0)=f''(0)/2$ is good. $\endgroup$
    – egreg
    Jun 15, 2015 at 19:14

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