## Friday, February 21, 2014

### More Difficult Complex Integrals

We now look at more difficult complex integrals:

Example:  Evaluate for  ÷ z ÷ < 1

o 2p   dq / (1 +  ½ cos q)

Here we make us of:   cos q  =  [exp(iq) + exp (-iq)]/ 2

But z =  exp (iq)  so that:  cos q  =   z + z -1/ z

And dz =  iexp(i q )  dq =   iz dq so that:  dq  = dz/ iz

We can then rewrite the integral in terms of z:

o 2p   dq / (1 +  ½ cos q)  =  ò C   dz/ iz / 1  +  ½ (z + z -1/ z)  =

1/2i ò C   2 dz/ (2z +  ½ z2 + ½)

First, obtain the poles from the quadratic in the denominator:

½ z2 + 2z +   ½  =  0

Solve for z using the quadratic formula:

z = [-b + {b2 - 4ac}1/2]/ 2a

To obtain:

z= -2 + Ö3  or z1= -2 + Ö3     and  z2= -2 -   Ö3

We can eliminate the 2nd root (pole)  since we require:  ÷ z ÷ < 1

Then:  we use -2 + Ö3

Because this is a simple pole we can use:

Res [z] z ® z o    = p(z)/ q’(z)

Where q’(z) = z + 2  and p(z) = 4

Then:  Res [z] z ® z o    =   4/ (z + 2)

Insert z = -2 + Ö3  :

Res [z] z ® -2 + Ö3    =   4/ [(-2 + Ö3)  + 2]  =   4/Ö3

Then:

ò C  f(z) dz  =    (1/ 2i)   [2 pi  c - 1  ]  =     4p / Ö3

Problems for Math Mavens:

1) Evaluate:

o 2p   dq / (37  -  12 cos q)  for  ÷ z ÷ < 6

2)Evaluate:

o 2p   dq / (13 -   5 sin q)     for  ÷ z ÷ < i