On the unit circle,z=sin(iθ) and dz=ieiθdθ . We then have
∮∣z∣=1z+1z2dz=Z∫02π(e−iθ+e−2iθ)ieiθdθ
=iZ∫02π(1+e−iθ)dθ=2πi
The integrand (z+1)/z2 has a double pole at z = 0. The Laurent expansion in a deleted neighborhood of z = 0 is simply 1(z+1)z2 , where the coefficient of 1/z is seen to be 1. We have
Res(z2z+1,0)=1
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