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

 

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