Question #146559
Q: Evaluate the following integral using residue theorem
∫ coth z / (z-i) dz ; C : |z| = 2
1
Expert's answer
2020-11-25T14:26:07-0500

Let's write the residue theorem :

Ccoth(z)zidz=2πii=1nresz=aif(z)\int_{C} \frac{\coth(z)}{z-i} dz = 2\pi i \sum_{i=1}^n res_{z=a_i} f(z)

What are the residues of a function f(z)=coth(z)zi=cosh(z)(zi)sinh(z)f(z)=\frac{\coth(z)}{z-i} = \frac{\cosh(z)}{(z-i)\sinh(z)} inside a z=2|z|=2 circle ?

We have only 2 residues inside this circle : a1=i,a2=0a_1 = i, a_2 =0 (as all singular points are of a form zi=0z-i=0 or sinh(z)=0\sinh(z)=0 and the latter gives the solutions of type z=πiz=\pi i ). Let's calculate the residues at both of these points :

resz=if(z)=coth(i)×resz=i1zi=ictg(1)×1res_{z=i} f(z) = \coth(i) \times res_{z=i}\frac{1}{z-i} = -i\ctg(1)\times 1

resz=0f(z)=cosh(0)(0i)×resz=01sinh(z)=ires_{z=0} f(z) = \frac{\cosh(0)}{(0-i)}\times res_{z=0}\frac{1}{\sinh(z)} = i

Therefore we find :

Ccoth(z)zidz=2πi(iictg(1))=2π(ctg(1)1)\int_C \frac{\coth(z)}{z-i} dz = 2\pi i (i-i\ctg(1)) = 2\pi(\ctg(1) - 1)


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