singularity proof

**foolsgold** · August 26th 2010, 09:42 AM

Unfortunately I'm not very good with singularities, so any help with this proof would be very much apprectiated. Kudos in advance!!!!!

S (a,r) denotes a positively orientated circular contour of radius r centered at a
D'(a,r) denotes a disk such that {z: 0< |z-a|< r}

Let f be an analytic function on the domain D'(0,2) and suppose that for all n=0,1,2...

S(0,1)∫ (z^n)f(z)dz = 0

Show that f has a removable singularity at z=0

(The s(0.1) is meant to be below the integral sign)

**mfetch22** · August 26th 2010, 08:04 PM

Originally Posted by foolsgold

Unfortunately I'm not very good with singularities, so any help with this proof would be very much apprectiated. Kudos in advance!!!!!

S (a,r) denotes a positively orientated circular contour of radius r centered at a
D'(a,r) denotes a disk such that {z: 0< |z-a|< r}

Let f be an analytic function on the domain D'(0,2) and suppose that for all n=0,1,2...

S(0,1)∫ (z^n)f(z)dz = 0

Show that f has a removable singularity at z=0

(The s(0.1) is meant to be below the integral sign)

I have no clue how to solve this, but I thought I'd rewrite that last equation for you with the pretty 'latex' program so that other posters that know how to solve it can more easily read it. So, I think you were saying something like this for that last equation:

[A] $\;\; \int_{S(0,1)}^{z^n}f(z)dz = 0$

or did you mean...

[B] $\;\; \int_{S(0,1)}(z^n)f(z)dz = 0$

Or did you mean something else entirely? I'm just trying to ensure "equational clarity" for the posters who can actually prove this problem for you. I hope it helps, just maybe...

**foolsgold** · August 29th 2010, 07:13 AM

hey thanks a lot. it's the second one!

**Jose27** · September 1st 2010, 06:33 PM

I'm gonna be ommiting the curve over which we are integrating. Now expand $f(z)$ in a Laurent series, notice the unifrom convergence and use that $z^m$ has a primitive in $D'(0,2)$ iff $m\neq -1$ to conclude that, for $n\geq 0$ :

$0= \int z^nf(z)dz = \sum_{-\infty }^{\infty } a_k\int z^{n+k}dz = a_{-1-n} 2\pi i$

(where $a_k$ are the coefficients of the Laurent series) and so $a_{-1-n}=0$ for all $n\geq 0$ which is saying $a_k=0$ for all $k<0$ . So the Laurent series is a Taylor one, and so $f$ is analytic in the disc $D(0,2)$ .

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