Calculus of Residues Revisited

Let f(z) be analytic on and inside a closed contour C as shown below, except for a finite number of isolated singularities: z= a1, a2, etc. which are enclosed by C.

Then:   ò _C  f(z)  dz =      2 pi    åⁿ _{k = 1}    Res f (a _k) 

We now want to elaborate this a bit more by reference to the diagram shown. In this case we consider the function f(z) is analytic inside and on the simple closed curve C except at a finite number of specified points: a, b, c, etc.  at which there exist residues:   a _{- 1}  ,        b _{- 1} ,  c _{- 1}      , etc.

In which case we can write:

ò _C  f(z)  dz =   2 pi   [a _{- 1}        +  b _{- 1}          +  c _{- 1}        + …………………….]

That is, 2 pi    times the sum of the residues at all the singularities enclosed by C. To ensure this, one would respectively construct circles C1, C2, C3 etc. as I have done with respective centers at a, b, c etc. If we take care to do this properly then we can write:

ò _C  f(z)  dz =       ò_C1  f(z)  dz   + ò_C2  f(z)  dz  + ò_C3  f(z)  dz   +    ..........

Where:

ò _C1  f(z)  dz   =   2 pi   a _{- 1}       

ò _C2  f(z)  dz  =  2 pi   b _{- 1}       

ò _C3  f(z)  dz   =   2 pi   c _{- 1}       

So that:

ò_C  f(z)  dz =   2 pi   [a _{- 1}  +  b _{- 1}  +  c _{- 1}   + ..] = 2 pi   (sum of residues)

Example 1:

Evaluate the integral:  ò_C   cot (z)  dz

f(z) = cot (z)

For which: ò _C  f(z)  dz   =   2 pi   c _{- 1}       

Re-write: f(z) = cot (z) = 1/ tan z

For which singularities occur at tan z = 0

Or: o, + p, + 2p,+  3p  etc.

Then Res f(z) =   1/ sec² z ÷ _{z = + n p}     =    1/ (1/ cos² z)

= cos² z÷ _{z = + n p}     =    cos² (np)  

And :  cos² (np)    = 1   at z =  (2n + 1) p)/ 2

Therefore:    c _{- 1}  =  1, and

  ò _C  cot (z)  dz  =    2 pi   (1) = 2 pi   

Example 2:

Evaluate the integral:  ò _C  exp (z)   dz  /  (z – 1) (z + 3)²

Where C is given by  ÷ z ÷    =   3/2 

Solution:
Take the residue at the simple pole (z = 1) such that:
lim _{z ® 1}   [ (z – 1)  exp (z)    / ( z  -  1) (z + 3)²  ] =

exp(1)/ 16 = e/ 16

The residue at the 2^nd order pole (z = -3) is:
lim _{z ® -3}  d/ dz  [(z + 3)²    exp (z)    / ( z  -  1) (z + 3)²  ] = 

lim _{z ® -3}   [ (z – 1)  exp (z)    - exp(z) / (z – 1 )²  ]

   = - 5 exp (-3) / 16

The integral is therefore:

ò _C  exp (z)   dz  /  (z – 1) (z + 3)²    

=  2 pi   a _{- 1}   =   2 pi   (e/ 16)

(We do not add the 2^nd residue because it lies beyond the circle ÷ z  ÷    =   3/2  )

Practice Problems:

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

2) Evaluate the integral:   ò _C    z   dz / (z²  - 2z + 2)²