Solutions to Complex Integral Problems

1) Evaluate the integral:   ò_C  (z + 1)   dz / (2z +  i)

 

Solution:

 

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

 

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

(Remember 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

2) Consider Example (2) and obtain the integral if we have

÷ z ÷   =  10   instead of   ÷ z  ÷    =   3/2 

 

If we now have  ÷ z ÷   =  10    then both residues,  i.e.  for the simple and 2^nd order poles can be added since the singularities are now contained in the circle.

 

Then:
 

Sum of residues =  e/ 16  - 5 exp (-3) / 16

 

= e/16 -   5e ^-3/ 16 

So that the re-evaluated  integral is now:

2 pi   (sum of residues) = 2 pi   (e/16 -   5e ^-3/ 16 )

 

 

=  pi (e -   5e ^-3)/8

3)Evaluate the integral:   ò_C    z   dz / (z²  - 2z + 2)²

 

Solution,    first rewrite:

 

f(z)  =    z/  (z  - 1 + i)² (z  - 1 -   i)²

 

 

In the upper half plane we need to obtain Res(f(z) for z = 1+i

 

For which we have a 2^nd order pole. 

 

Then:

 

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

 

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

 

And from  the product rule for differentiating,  this yields:

 

2 + a – 2z/ (z + a)³    for  a =    (-1 + i)

 

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

 

=   (1 + i) + (-1 +i) – 2(1 + i)/ [(1 + i) + (-1 +i)] ³

 

=   -2/ -8i  = 1/ 4i  so that:

ò_C    z   dz / (z²  - 2z + 2)²   = 2 pi   (1/ 4i) =  p/  2