## Saturday, May 6, 2017

### Math Revisited: Residue Calculus

We now revisit residue calculus.

Let f(z) be analytic on and inside a closed contour C, 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:

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:

ò C  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: 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

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

Practice Problem:

Integrate:
-¥  ¥    x  dx / (x2   - 2x + 2)