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[SOLVED] Evaluating Integrals Using Cauchy's Formula/Theorem
Evaluate the given integral using Cauchy's Formula or Theorem.
\int_{|z+1|=2} \frac{z^2}{4 - z^2} \, dz
Cauchy's Theorem. Suppose that f is analytic on a domain D. Let \gamma be a piecewise smooth simple closed curve in D whose inside \Omega also lies in D. Then
\int_\gamma f(z) \, dz = 0
Cauchy's Formula. Suppose that f is analytic on a domain D and \gamma is a piecewise smooth, positively oriented simple closed curve D whose inside \Omega also lies in D. Then
f(z) = \frac{1}{2\pi i} \int_\gamma \frac{f(\zeta)}{\zeta - z} \, d\zeta
for all z \in \Omega.
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.
\int_{|z+1|=2} \frac{z^2}{4 - z^2} \, dz
Homework Equations
Cauchy's Theorem. Suppose that f is analytic on a domain D. Let \gamma be a piecewise smooth simple closed curve in D whose inside \Omega also lies in D. Then
\int_\gamma f(z) \, dz = 0
Cauchy's Formula. Suppose that f is analytic on a domain D and \gamma is a piecewise smooth, positively oriented simple closed curve D whose inside \Omega also lies in D. Then
f(z) = \frac{1}{2\pi i} \int_\gamma \frac{f(\zeta)}{\zeta - z} \, d\zeta
for all z \in \Omega.
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?