Thursday, January 9, 2014

The Residue Theorem and Complex Integrals

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    Recall that we saw the “residue theorem” (due to Cauchy) Let f(z) be analytic on and inside a closed contour C (see diagram) except for a finite number of isolated singularities at z = a1, a2…..etc., which are enclosed by C.

òC  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:


òC  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:


òC  f(z)  dz =       òC1  f(z)  dz   + òC2  f(z)  dz  + òC3  f(z)  dz   +    ..........


Where:

òC1  f(z)  dz   =   2 pi   a - 1       



òC2  f(z)  dz  =  2 pi   b - 1       


òC3  f(z)  dz   =   2 pi   c - 1       


So that:


òC  f(z)  dz =   2 pi   [a - 1        +  b - 1          +  c - 1        + ..] =


2 pi   (sum of residues)


Example 1:

Evaluate the integral:  òC  cot (z)  dz

f(z) = cot (z)

For which: òC  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


  òC  cot (z)  dz  =    2 pi   (1) = 2 pi   


Example 2:

Evaluate the integral:

òC  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 2nd 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:

òC  exp (z)   dz  /  (z – 1) (z + 3)2    =  2 pi   a - 1   =   2 pi   (e/ 16)

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

Problems for Math Mavens:
1) Evaluate the integral:   òC  (z + 1)   dz / (2z +  i)
2) Consider Example (2) and obtain the integral if we have ÷ z  ÷   =  10   instead of
 ÷ z  ÷    =   3/2 

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

in the upper half z-plane

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