Let $f$ be a holomorphic function on $\mathbb{C}\smallsetminus \{0\}$. Suppose $\int_{|z|=1}z^nf(z)\,dz=0$ for any $n=0,1,2,\ldots$. Prove that $f$ has a removable singularity at $z=0$. How to prove?

If $\lim_{z\to 0}z^nf(z)=0$ then I use the Laurent seris and solve it. But how to deal with the integral?

  • $\begingroup$ Are you familiar with Laurent series? $\endgroup$ – Jonathan Y. Jan 28 '14 at 11:23
  • $\begingroup$ You mean "for every $n$", not "for any $n$". $\endgroup$ – mrf Jan 28 '14 at 14:55

If a function $f$ is analytic in $\mathbb C\smallsetminus\{0\}$ then it is expressed as $$ f(z)=\sum_{n=-\infty}^\infty a_nz^n, $$ and it is readily proved that $$ a_n=\frac{1}{2\pi i}\int_{|z|=1}\frac{f(z)\,dz}{z^{n+1}}. $$ Hence, in your example $$ a_n=0, \quad \text{for all $n<0$}, $$ and consequently, the singularity at $z=0$ is removable.



Suppose we develope a Laurent series for $\;f\;$ in $\;0<|z|<r\;,\;\;r\in\Bbb R^+\;$ :

$$f(z)=\ldots+\frac{a_{-n-1}}{z^{n+1}}+\frac{a_{-n}}{z^n}+\mathcal O(z^{-n+1})\implies z^{n-1}f(z)=\ldots\frac{a_{-n-1}}{z^2}+\frac{a_{-n}}z+\mathcal O(1)$$

Now just remember that

$$\oint\limits_{|z|=1|}\frac{dz}{z^n}=\begin{cases}2\pi i&,\;\;n=1\\{}\\0&,\;\;n\neq 1\end{cases}$$

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    $\begingroup$ You should add lower terms in the development ($f$ is not assumed to be meromorphic at $0$). $\endgroup$ – user10676 Jan 28 '14 at 11:33
  • $\begingroup$ @user10676 It may be that either I am confusing definitions or you are: if the function's holomorphic in $\;\Bbb C^*\;$ then $\;z=0\;$ can be at most a pole (of order $\;n\;$), right? If this is true then I don't need lower terms in the Laurent series of $\;f\;$ . $\endgroup$ – DonAntonio Jan 28 '14 at 11:45
  • $\begingroup$ Oh, I know that @JonathanY., thanks. My doubt about definitions is whether $\;e^{1/z}\; $ would be considered holomorphic on the punctured plane or merely analytic. $\endgroup$ – DonAntonio Jan 28 '14 at 12:06
  • $\begingroup$ Oh, AFAIK holomorphic in an open domain simply means differentiable at every point of the set (and at a point, it means differentiable at some neighborhood of the point). $\endgroup$ – Jonathan Y. Jan 28 '14 at 12:10
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    $\begingroup$ @JonathanY. Yes, perhaps you and user10676 are right and I got it messed up, though I'm almost sure there was a difference...I shall check this later. Thanks. $\endgroup$ – DonAntonio Jan 28 '14 at 12:12

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