1. ## singularity proof

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)

2. 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] $\displaystyle \;\; \int_{S(0,1)}^{z^n}f(z)dz = 0$

or did you mean...

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

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...

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

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

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

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