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[SOLVED] Evaluating Integrals Using Cauchy's Formula/Theorem
Evaluate the given integral using Cauchy's Formula or Theorem.
[tex]\int_{|z+1|=2} \frac{z^2}{4 - z^2} \, dz[/tex]
Cauchy's Theorem. Suppose that f is analytic on a domain D. Let [itex]\gamma[/itex] be a piecewise smooth simple closed curve in D whose inside [itex]\Omega[/itex] also lies in D. Then
[tex]\int_\gamma f(z) \, dz = 0[/tex]
Cauchy's Formula. Suppose that f is analytic on a domain D and [itex]\gamma[/itex] is a piecewise smooth, positively oriented simple closed curve D whose inside [itex]\Omega[/itex] also lies in D. Then
[tex]f(z) = \frac{1}{2\pi i} \int_\gamma \frac{f(\zeta)}{\zeta - z} \, d\zeta[/tex]
for all [itex]z \in \Omega[/itex].
For z equal to 2 and -2, the integrand is not defined and hence is not analytic in any domain that includes these points. The problem is that the curve |z + 1| = 2 contains the point -2 inside it so I can't apply either Cauchy's Theorem or Formula. What can I do?
Homework Statement
Evaluate the given integral using Cauchy's Formula or Theorem.
[tex]\int_{|z+1|=2} \frac{z^2}{4 - z^2} \, dz[/tex]
Homework Equations
Cauchy's Theorem. Suppose that f is analytic on a domain D. Let [itex]\gamma[/itex] be a piecewise smooth simple closed curve in D whose inside [itex]\Omega[/itex] also lies in D. Then
[tex]\int_\gamma f(z) \, dz = 0[/tex]
Cauchy's Formula. Suppose that f is analytic on a domain D and [itex]\gamma[/itex] is a piecewise smooth, positively oriented simple closed curve D whose inside [itex]\Omega[/itex] also lies in D. Then
[tex]f(z) = \frac{1}{2\pi i} \int_\gamma \frac{f(\zeta)}{\zeta - z} \, d\zeta[/tex]
for all [itex]z \in \Omega[/itex].
The Attempt at a Solution
For z equal to 2 and -2, the integrand is not defined and hence is not analytic in any domain that includes these points. The problem is that the curve |z + 1| = 2 contains the point -2 inside it so I can't apply either Cauchy's Theorem or Formula. What can I do?