Prove f=0: Holomorphic Polynomial

[cheeez]

f is a holomorphic polynomial and if

$$\oint_{\partial D(0,1)}f(z)\bar{z}^{j}dz= 0 for j = 0,1,2,3...$$

where $$\partial D(0,1)$$ is the boundary of a disc of radius 1 centered at 0

prove f $$\equiv$$0

 

[cheeez]

any ideas? if we use any polynomial of order > j like j+1 so f(z) = z^(j+1)

then f(z)(z_bar)^j becomes z(z z_bar)^j = z |z|^2j which is just z since modulus is 1 on boundary of disk of radius 1. so obviously integral of z is 0 since its holomorphic but f(z) is not trivial. am i misunderstanding this.

 

[cheeez]

no one?

 

[Hyperbolful]

Does it happen that the degree of f is j or greater than j?

What I would do is appeal to the fact that if the integral around a boundary of a simple jordan curve is 0 then the function must be analytic (holomorphic), and then since you have an analytic function times a conjugate function that is still analytic it must be zero

Maybe better phrased as follows:
since the integral around the circle is zero the function being integrated is analytic
so
the analytic function depends on the conjugate function (a non analytic function), so if you take the "d-bar" differential (the dereivative with respec to the congugate of z) you should get zero, which will imply that f must be zero

This might not be as clear as you want, but I'm not sure what you're allowed to assume.

 

[mathwonk]

have you tried green's theorem?