Calculus of Residues Problem Solution:

  The Problem: Integrate:

  ò _C (z + 1)   dz / (2z +  i)

 Solution:

A simple (m=1) pole occurs at z = -i/2

Then:

Res (f(z)) =  (z – a)   f(z)  =

lim _{z ® -i/ 2}    (2z +  i)   [(z + 1)  / (2z +  i)]

  Res (f(z)) =   (z + 1)/ 2] _{z = -i/ 2}  =   (-i/2 + 1)/ 2 =   ½  -   i/4

If f(z) has the form p(z)/ q(z) and a first order pole exists at z = z _o, then Res f(z)  at z _o  =lim _{z ® z o}    p(z) / q'(z)  where q'(z) is the first derivative.   

Then: ò _C  (z + 1)   dz / (2z +  i)  =  2 pi   (sum of residues)

=  2 pi   (½  -   i/4)  =   pi -   (pi)²/ 2 =    pi   + p/ 2