Show that the integral of the analytic function is independent of radius.
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Let f be analytic on │z│> 1. Show that if r > 1, then the integral of f over C(0,r) is independent of r.
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Solution Summary
It is shown that the integral of the analytic function is independent of radius.
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Proof:
Let Int(f, C(0,r)) represents the integral of f along C(0,r) with counterclockwise direction.
To show that the integral of f over C(0,r) is independent of r, we only need to prove that
Int(f, C(0,r1)) = ...
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