Let's write the residue theorem :
∫Cz−icoth(z)dz=2πi∑i=1nresz=aif(z)
What are the residues of a function f(z)=z−icoth(z)=(z−i)sinh(z)cosh(z) inside a ∣z∣=2 circle ?
We have only 2 residues inside this circle : a1=i,a2=0 (as all singular points are of a form z−i=0 or sinh(z)=0 and the latter gives the solutions of type z=πi ). Let's calculate the residues at both of these points :
resz=if(z)=coth(i)×resz=iz−i1=−ictg(1)×1
resz=0f(z)=(0−i)cosh(0)×resz=0sinh(z)1=i
Therefore we find :
∫Cz−icoth(z)dz=2πi(i−ictg(1))=2π(ctg(1)−1)
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