# Show Res(F'(z)/F(z), z0) = m if F(z) is analytic and has a zero of order m at z0.

1. Nov 2, 2008

### StumpedPupil

1. The problem statement, all variables and given/known data

Show Res(F'(z)/F(z), z0) = m if F(z) is analytic on the disc |z - z0| < R and has a zero of order m at z0.

2. Relevant equations

3. The attempt at a solution
We know that the kth derivation of F(z) is 0 for all k less than m, since F(z) has a zero of order m at z0. The kth derivative of F(z) is not 0 for all k greater than or equal to m.

We know F(z) can be written as a power series on the disc and that the coefficients are given by the kth derivative of F(z), evaluated at z0 and divided by k!. That means that the power series for F(z) = ∑(m to ∞) an(z-z0)^n since an = 0 for all n less than m.

I want to show F'(z)/F(z) has a residue at z0 which is m, but am not quite sure how I should begin. I tried solving by equating coefficients for some function G(z), where I showed F(z)G(z) = F'(z) but this seems wrong.

2. Nov 2, 2008

### HallsofIvy

Staff Emeritus
If F(z) is analytic and has a zero of order 3 at z0, then F(z)= (z- z0)3g(z) where g is analytic. F'(z)= 3(z-z0)2g+ (z-z0)3g'(z).
F'(z)/F(z)= 3/(z-z0)+ g'(z)/g(z). Does that help?

3. Nov 3, 2008

### StumpedPupil

It definitely helps, thank you. I do have a follow up question though. In the general case where F has a zero of order m and we represent F = (z-z0)^mg(z), like you did, then we can get F'/F = m/(z-z0) + g'(z)/g(z).
The only condition of g(z) is that it is analytic, so is it not possible that g'(z)/g(z) could include some other factor necessary to calculate the residue. If not, then why. And if not, then I see that Res( F'/F, z0) = m.