Solve ∫|z|=3tan⁡(πz)dz using argument principle∫|z|=3tan⁡(πz)dz=∫d(cos⁡πz)(−π)cos⁡πzdz=−2i(N−P)where N = num of zeroes inside C:|z| = 3 and P is num of poles inside C (Is this correct or should we also consider on C???)zeroes for cos⁡πz=−0.5,−1.5,−2.5,0.5,1.5,2.5⇒N=6No poles for cos⁡πzvalue of given integral = −2i(N−P)=−2i(6−0)=−12iIs this correct? pls correct me if i am doing wrong

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utloverej · Accepted Answer

It is correct, yes. And the zeros and polse are inside C. Actually, if there were zeros or poles on C, the expressin ∫Cf′(z)f(z)dz would make no sense.

Solve \int_{|z| = 3} \tan (\pi z) dz using argument principle \int_{|z| = 3} \tan (\pi

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