- #1

- 63

- 0

**Something wrong with a "proof" relating to Laurent coefficients**

## Homework Statement

Let f(z) have an isolated singularity at z_0 and suppose that f(z) is bounded in some punctured neighborhood of z_0. Prove directly from the integral formula for the Laurent coefficients that a_(-j) = 0 for all j > 0. That is, f(z) must have a removable singularity at z_0

## The Attempt at a Solution

Right I know one correct solution but I can't find my fault in this one:

[tex]a_j = \frac{1}{2\pi i}\int \frac{f(z)}{(z-z_0)^{j+1}}[/tex]

As f(z) is bounded let M be the maximum value of f(z) in the punctured disk.

[tex]|\frac{1}{2\pi i}\int \frac{f(z)}{(z-z_0)^{j+1}}| \leq |\frac{M}{2\pi i} \int \frac{1}{(z-z_0)^{j+1}} |[/tex]

But I also know (except I suspect this is where problems arise) that

[tex]\int \frac{1}{(z-z_0)^n}dz [/tex]

Around any circle centered at z_0 will have the value 2∏i if n = 1, and 0 otherwise. That would imply the only Laurent coefficient that isn't 0 is a_0 but I'm pretty sure that's false.

I suspected that the last 'formula' is only valid simply connected domains, which I still believe to be the case (can be derived using Cauchy's integral formula for example).

Now here's my problem. I got to the chapter Residue theory and in the first page they, it seems to be, does exactly what I did above. I'll quote

Now we are applying the formula to a Laurent series expansion and as far as I understand the Laurent expansion is defined for an annulus around the singularity. Thus it's not a simply connected domain. But if I can use the formula on a domain that isn't simply connected how come my "proof" is false?Let us consider the problem of evaluating the integral

[tex]\oint_{\tau} f(z)dz[/tex]

where τ is a simple closed positively oriented contour and f(z) is analytic on and inside τ except for a single isolated singularity, z_0, lying interior to τ. As we know, the function f(z) has a Laurent series expansion

[tex]f(z) = \sum_{j = - \infty}^{\infty}a_j(z-z_0)^j[/tex]

converging in some punctured circular neighborhood of z-0. By the methods of an earlier section we know we can convert the integral to integration over a circle C where C is a circle centered at z_0 without affecting its value.

The last integral can be computed by termwise integration of the series along C. For all j =/= -1 the integral is zero, and for j = -1 we obtain the value 2∏ia_(-1). Consequently we have

[tex]\int_{τ} f(z)dz = 2\pi i a_{-1}[/tex]

Sorry, it was a bit long winded but explained the issue as thoroughly as I could. Any input would be appreciated, really at a loss here.

Edit: apparently \bigg{|} won't show at physicsforums?