Answer to Question #146559 in Complex Analysis for Usman

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 :

"\\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)=\\frac{\\coth(z)}{z-i} = \\frac{\\cosh(z)}{(z-i)\\sinh(z)}" inside a "|z|=2" circle ?

We have only 2 residues inside this circle : "a_1 = i, a_2 =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=\\pi i" ). Let's calculate the residues at both of these points :

"res_{z=i} f(z) = \\coth(i) \\times res_{z=i}\\frac{1}{z-i} = -i\\ctg(1)\\times 1"

"res_{z=0} f(z) = \\frac{\\cosh(0)}{(0-i)}\\times res_{z=0}\\frac{1}{\\sinh(z)} = i"

Therefore we find :

"\\int_C \\frac{\\coth(z)}{z-i} dz = 2\\pi i (i-i\\ctg(1)) = 2\\pi(\\ctg(1) - 1)"


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