# If $\int_{|z|=1} z^n f(z)dz=0 \ \forall n = 0, 1, 2, …$, then $f$ has a removable singularity at $z=0$

True or false: $$f$$ holomorphic in $$A=\{z\in \mathbb{C}: 0\lt |z|\lt 2\}$$ and $$\int_{|z|=1} z^n f(z)dz=0 \ \forall n = 0, 1, 2, ...$$ then $$f$$ has a removable singularity at $$z=0$$.

I' m not sure if this is true or false because I can only prove that $$f$$ is either analytic at zero or has a removable singularity.

$$f$$ holomorphic in $$A$$ means that we can express $$f$$ as a Laurent series of the form

$$f(z)= \sum_{n=1}^{\infty}\frac{b_n}{z^n} + \sum_{n=0}^{\infty}{a_n}{z^n}$$

with $$a_n = \frac{1}{2 \pi i}\int_{\gamma} \frac {f(z)}{z^{n+1}}dz$$

$$b_n = \frac{1}{2 \pi i}\int_{\gamma} {f(z)}{z^{n-1}}dz$$

$$\int_{|z|=1} z^n f(z)dz=0 \ \forall n = 0, 1, 2, ...$$ means that $$b_n = 2 \pi i \cdot 0 = 0 \ \forall n$$, so $$f$$ has as worst a removable singularity at $$z=0$$.

Now to prove that $$f$$ has indeed a removable singularity one must prove that $$f$$ is not analytic at $$z=0$$. But $$f(z)=\sum_{n=0}^{\infty}{a_n}{z^n}$$ at $$A$$, so for $$f$$ not to be analytic at $$z=0$$ we must have that $$f(0) \neq a_0= \frac{1}{2 \pi i}\int_{\gamma} \frac {f(z)}{z}dz$$, which I haven't been able to prove.

• I think any entire $f$ is a counterexample. – Nitin Uniyal Aug 22 '19 at 12:34
• @user658409 Changing to $f\equiv 1$, you might just as well ask why $\frac zz$ isn't a counterexample. But that's an archetypal example of a function with a removable singularity. The answer is that the domain of $f$ has a puncture in it, and that puncture is a singularity. But there is a value you could assign to $f$ at that puncture to make the function analytic, and that makes the singularity removable. – Arthur Aug 22 '19 at 13:08
• First, as already pointed out you're confused about the definition of "removable singularity". Less to the point, since $f$ is only defined for $0<|z|<2$ it's not analytic at the origin - it can't be, since $f(0)$ is undefined! It's an extension of $f$ that's analytic at $0$ (which of course is precisely what it means to say $f$ has a removable singularity...) – David C. Ullrich Aug 22 '19 at 15:25
• @NitinUniyal What's the definition of "removable singularity"? – David C. Ullrich Aug 22 '19 at 16:30
• @David C.Ulrich..A singularity at $z=a$ is removable if there is no negative powers of $(z-a)$ in Laurent's expansion of $f(z)$. – Nitin Uniyal Aug 22 '19 at 16:44

Your interpretation of removable singularity is not the accepted one. You are not required to prove that $$f$$ is not analytic at $$0$$. For example $$f(z)=0$$ for all $$z \neq 0$$ has a removable singularity at $$0$$.

To expand a bit on Kavi's answer: Say $$A=\{z:0 and assume $$f\in H(A)$$.

You're concerned that $$f$$ may be analytic at the origin. Of course Kavi is exactly right when he says "so what?". But the situation has a curious aspect.

First, speaking carefully,

(i) $$f$$ is analytic at the origin

is impossible, because strictly speaking $$f(0)$$ is undefined. The sensible, formally correct version of (i) is

(ii) It's possible to define $$f(0)$$ in such a way that $$f$$ becomes analytic at the origin.

And here's the reason I'm posting this: Not only is (ii) no problem regarding whether $$f$$ has a removable singularity, in fact (ii) is precisely the definition of "$$f$$ has a removable singularity"!

So when you say "I can only prove $$f$$ has a removable singularity or (i)" you're really saying "I can only prove $$f$$ has a removable singularity or (ii)", and by definition that means you're saying this:

I can only prove $$f$$ has a removable singularity or $$f$$ has a removable singularity.

Moral: You need to know the definitions.

• for some reason I assumed $f$ defined in $A \cup \{ 0 \}$ and analytic in $A$ and thought that I could somehow infer $f(0)$ from the integral equality... so yeah, I should have paid more attention to what the question actually said. – Yagger Aug 22 '19 at 16:42

You are right: we have that $$b_n=0$$ for all $$n \ge 1$$ and therefore

$$f(z)=\sum_{n=0}^{\infty}{a_n}{z^n}$$ on $$A$$. Now define $$g(z):= \sum_{n=0}^{\infty}{a_n}{z^n}$$ for $$|z|<2.$$ Then $$g$$ is holomorphic and $$g$$ is a holomorphic continuation of $$f$$ on the disc $$|z|<2.$$

Conclusion: $$f$$ has a removable singularity at $$z=0.$$

• @user658409 Huh? If you know what a removable singularity is then it's utterly obvious that the function $f=0$ has a removable singularity. If you don't know what a removable singularity is you should look it up. – David C. Ullrich Aug 22 '19 at 15:50
• Of course this exactly correct, but it seems sort of beside the point. The OP appears to have got this far on his own; what you say here says nothing about what he's confused about, that is "Now to prove that $f$ has indeed a removable singularity one must prove that $f$ is not analytic at $z=0$. " – David C. Ullrich Aug 22 '19 at 16:14