Saturday, January 4, 2014

From Laurent Series to Residues


Recall we’d previously seen the function  f(z) expressed as the Laurent series

f(z) = å¥ n = -¥   c n (z – a) n  +    å¥ n =-¥   c -n /  (z – a) n 

We also noted that no positive powers of z appear, only negative, i.e. 1/z is  z - 1 .   So, in effect we could say that the coefficients of the positive powers are zero.  It was also important to note here that the coefficient of :

1/ 1! z     = 1/ z

is unity, so according to Laurent’s theorem we designate that coefficient:

c n  =  1/2 pi  òC  exp( 1/z)  dz

where C is any positively oriented simple closed contour around the origin. Further, since  n=  - 1 then, say for a function f(z) = exp(1/z):

 c - 1  =  1/2 pi  òC  exp( 1/z)  dz   and:  òC  exp( 1/z)  dz    =  2 pi 

And in this way, one can actually compute what is called “the residue” from a term of the Laurent series.  The “residue theorem” (due to Cauchy) is as follows:

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. Then:

òC  f(z)  dz =       2 pi    ån k = 1    Res f (a k) 


Alternately,  Res f (a k)    =  1/2 pi   òC  f(z)  dz

Thus, if we have some function f(z) centered at a we know we have a Laurent expansion:


f(z) =  = å¥ n = -¥   c n (z – a) n  +    å¥ n =-¥   c - n /  (z – a) n 

Then:  Res f(a) =  c - 1 

(There is no need to compute the actual integral)

Res f (a)  in the case of a singular point is independent of the choice of C and is called the residue of the function at the point z = a.


If f(z) is analytic at z = a the point z = a is then called a regular point and Res f(a) = 0. If, however, z = a  is an isolated singularity, then the residue may or may not = 0.


Things can be made even more straightforward by deriving a basic formula for computing the residue.

Assume f(z) has a pole of order m then the Laurent series of
 f(z) is:


f(z) =  a - m  / (z – a) m      +    a – m+ 1  / (z – a) m - 1      +   


a - 1  / (z – a)  +   a 0  + a 1 (z – a) +   + a 2 (z – a)2   +  ……



Now, multiply through by  (z – a)     :



(z – a) m    f(z) =     a - m     +    a – m+ 1    (z – a)   +  

……. a - 1  / (z – a) m - 1       +  ….



Which represents a Taylor series about z= a of the analytic function, i.e. on the left. Now, differentiate both sides (m – 1) times with respect to z and you get:

d m - 1    / dz m - 1    {(z – a) m    f(z) } =   (m – 1)! a - 1  +  m (m – 1) ……2 a 2 (z – a) + …

Then, let z ® a:

lim z ® a      d m - 1   / dz m - 1    {(z – a) m    f(z) } = (m – 1)! a - 1 

Where bear in mind that   a - 1   =   c - 1   seen earlier.

More generally, we may write:

c - 1     = 1/  (m – 1)!  lim z ® a      d m - 1   / dz m - 1    {(z – a k) m    f(z) }

Examples:

1)Find the residue for  f(z) =    4 / (1 - z)    

We have a simple pole (m=1)  at z = 1.

Then write:

Res (f(z)) =  (z – a)  f(z)  = lim z ® 1  (z – 1)    [4 / (1 - z)]    


 = 4/ -1 = -4

(2)    Find the residue for f(z) =  1/ 1 – z

Again, we have a simple pole at z = 1, so:

Res (f(z)) =  (z – a)     f(z)  = lim z ® 1      (z – 1)    [1 / (1 - z)]      = -1

3)
We have a fourth order pole (m = 4) at z = 0, and we know from the residue formula:
Res f(z) =    c - 1    

= 1/  (m – 1)!  lim z ® a     d m - 1  / dz m - 1    {(z – a k) m    f(z) }
Then:
Res f(z) = 1/ 3!  lim z ® 0      d 4 - 1   / dz 4 - 1    {(z) 4     ×  exp (z) / z 4  }
Res f(z) = 1/ 3!  lim z ® 0      d 3   / dz 3    { exp (z)}
= 1/ 3!    lim z ® 0   8  exp(2z) =   8/ 3! exp (0) =  8 / 6 =  4/3
(4) Find the residue of f(z) =   z 2 exp (z)/  (z – 3) 2    
At z = 3
We have a 2nd order pole, i.e. m = 2 so that;
Res f(z) =  lim z ® 3   d  / dz   {(z – 3) 2  [z 2 exp (z)/  (z – 3) 2 ]  }
Res f(z) = lim z ® 3     d    / dz      {z 2 exp (z)}
= lim z ® 3     exp (z) [ z2 + 2z ] =    9 exp (3) +  6 exp (3)


=   15 exp (3)

Problems for Math Mavens:

1) Find the residue for f(z) =  3 exp (z)/  z 4

Be sure to specify the order of the poles before proceeding!
2)Find Res f(z) for cos z/ z 5
3) Find all the residues at those singular points inside the circle  ÷ z  ÷     =    2  
For:  f(z) =   z 2    / ( z 4  - 1)
(Hint: remember your complex numbers!)

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