$F:G\backslash \{d \}\rightarrow \mathbb{C}$ holomorphic , F' holomorphically extendable over d. It will now be shown that F is then also holomorphically extendable over d.

Assumption: $F:G\backslash\{d\}\rightarrow \mathbb{C}$ is holomorphic, but does not have a removable singularity at d, so it is not holomorphically extendable over d.

Let $h(z)= \begin{cases}(z-d)^{2}F'(z), & z\ne d,\\ z=0, & z=d. \end{cases}$

Then :$$h'(d) = \lim _{z\rightarrow d} \frac{(z-d)^2 F'(z)}{(z-d)} = 0$$

Is not true, but then there can't be a taylor series of $h(z)$ about d and so $F'(z)$ could not be holomorphically extendable over d.

So from this it follows that if F'(z) has a removable singularity at d, so does F(z).

Can anybody tell me if this is correct?


1 Answer 1


You assumed, to the contrary, that $F$ is not extendable. This implies various things about $F$, but you didn't use any of them. It follows that your argument is not correct.

A hint: Put $g(z):=F'(z)$ $(z\ne d)$ and $g(d):=\lim_{z\to d}F'(z)$. Then $g$ is analytic in a neighborhood of $d$. Now compare the function $G(z):=\int_d^z g(z)\ dz$ with $F$.

  • $\begingroup$ $G(z):=\int _{d}^{z} g(z) dz = F(z) - \lim_{z\rightarrow d} F(z) = F(z)$ So $\lim_{z\rightarrow d} F(z)$ has to be 0 and thus F(z) also extendable. ? ? ? $\endgroup$
    – VVV
    Commented Nov 29, 2011 at 15:02
  • $\begingroup$ @VVV: $G$ is analytic in a neighborhood $U$ of $d$, and $G'=g=F'$ in $\dot U$, which is connected. Therefore $F-G\equiv c$ for some $c\in{\mathbb C}$ in $\dot U$. It follows that $F$ can be extended to $d$ and has the value $G(d)+c=c$ there. $\endgroup$ Commented Nov 30, 2011 at 9:26

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