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  ò _C  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  ò_C  exp( 1/z)  dz   and:  ò_C  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:

ò_C  f(z)  dz =       2 pi    åⁿ _{k = 1}    Res f (a _k) 

Alternately,  Res f (a _k)    =  1/2 pi   ò_C  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) ^m     :

(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 2^nd 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!)