## Saturday, January 11, 2014

### Solutions to Complex Integral Problems

1) Evaluate the integral:   òC  (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 =

lim z ® z o    p(z) / q'(z)  where q'(z) is the first derivative)

Then:

òC  (z + 1)   dz / (2z +  i)  =  2 pi   (sum of residues)

=  2 pi     -   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 residuesi.e.  for the simple and 2nd 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:   òC    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 2nd 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:

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