Help grading papers: Residue Thm. w/Rational fcns

1. Feb 8, 2006

benorin

Ok, the quick and dirty: The given Theorem: If f is analytic everywhere in the finite complex plane except for a finite number of singular points interior to a positively oriented simple closed contour C, then

$$\int_{C}f(z)dz=2\pi i \mbox{Res}\left[ \frac{1}{z^2}f\left( \frac{1}{z}\right) ;z=0\right]$$

The problem: If $$P(z)=a_0+a_1z+\cdots +a_nz^n$$ and $$Q(z)=b_0+b_1z+\cdots+b_mz^m$$ where $$m\geq n+2,a_0\neq 0, b_0\neq 0$$ and all the zeros of Q(z) lie interior to C, prove that

$$\int_{C}\frac{P(z)}{Q(z)}dz=0$$

Apply the theorem to get

$$\int_{C}\frac{P(z)}{Q(z)}dz=2\pi i \mbox{Res}\left[ \frac{1}{z^2}\frac{P\left( \frac{1}{z}\right) }{Q\left( \frac{1}{z}\right) } ;z=0\right]$$
now what??? Pratial fractions? That is somewhat brutal, isn't there some simple way to do this. Thanks. --Ben

Papers due in a few hours, please hurry...

2. Feb 8, 2006

benorin

I got it...

$$\frac{1}{z^2}\frac{P\left( \frac{1}{z}\right) }{Q\left( \frac{1}{z}\right) } ;z=0\right] = \frac{p(z)}{q(z)}$$ where deg(p) +2<= deg(q), and the constant term of q is b_m which is non-zero and hence the residue is 0.