Help grading papers: Residue Thm. w/Rational fcns

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SUMMARY

The discussion focuses on applying the Residue Theorem to evaluate the integral of a rational function, specifically the integral of P(z)/Q(z) over a closed contour C. The theorem states that if P(z) and Q(z) are polynomials with specific conditions on their degrees and zeros, then the integral evaluates to zero. The participants confirm that by transforming the function using the substitution z = 1/w, the residue at z=0 can be shown to be zero, thus proving the integral's value.

PREREQUISITES
  • Understanding of complex analysis, specifically the Residue Theorem.
  • Familiarity with rational functions and polynomial degrees.
  • Knowledge of analytic functions and singular points.
  • Experience with contour integration techniques.
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  • Study the application of the Residue Theorem in complex analysis.
  • Learn about polynomial long division and partial fraction decomposition.
  • Explore examples of contour integrals involving rational functions.
  • Investigate the implications of singularities in complex functions.
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Students and professionals in mathematics, particularly those studying complex analysis, as well as educators looking for examples of applying the Residue Theorem in practical scenarios.

benorin
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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

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

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

[tex]\int_{C}\frac{P(z)}{Q(z)}dz=0[/tex]

Apply the theorem to get

[tex]\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][/tex]
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...
 
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I got it...

[tex]\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)}[/tex] 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.
 

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