Question #231357

18) Evaluate ∮(z)=1 Z² Sin1/z dZ?


1
Expert's answer
2021-09-25T11:36:32-0400

On the unit circle,z=sin(iθ) and dz=ieiθdθz = sin(iθ)\space and \space dz = ie^{iθ} dθ . We then have


z=1z+1z2dz=Z02π(eiθ+e2iθ)ieiθdθ\oint|z|=1 z + 1 z2 dz = Z\intop^ {2π} _0 (e−iθ+e−2iθ)ie^{iθ} dθ

=iZ02π(1+eiθ)dθ=2πi= i Z\intop^ {2π} _0 (1+e−iθ) dθ = 2πi


The integrand (z+1)/z2(z + 1)/z^2 has a double pole at z = 0. The Laurent expansion in a deleted neighborhood of z = 0 is simply 1(z+1)z21( z + 1) z^2 , where the coefficient of 1/z is seen to be 1. We have 


Res(z+1z2,0)=1Res (\frac{z+1}{z^2},0)= 1


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