Answer to Question #146559 in Complex Analysis for Usman

Question

Q: Evaluate the following integral using residue theorem
∫ coth z / (z-i) dz ; C : |z| = 2

Answer 1

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 :

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