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 2^nd 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