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!)

More Laurent Series Examples

No, we’re not yet finished with these exotic mathematical denizens! We want to give yet more examples, and especially showing different ways to obtain the Laurent equivalent of different functions.

 

Ex. (1):   exp(2z)/ (z - 1)³  about  z ₀  =   1

 

Let u = (z – 1)  so that z = (u -1)

Then:

 exp ( 2z) / (z - 1)³     =  exp(2 + 2u) /  u ³  =   [( e) ²  /  u ³   ] /  e ^2u 

 

=   e ²  /  u ³      [ 1  + 2u   + (2u) ² / 2!   +   (2u) ³ / 3!   + 

 (2u) ⁴ / 4! +     …….   

=  e ²  /  (z - 1)³     +   2 e ²  /  (z - 1)²      +   2 e ²  /  (z – 1)  +  

4 e ²  / 3  +  2 e ²  / 3 (z – 1)  +   …….

 

Note that z is a pole of order 3 or a “triple pole” (Why?)    Note also the series converges for all values of z  ¹  1

Ex. (2):   f(z) =  z – sin z /  z ³

We’ve already seen the series for sin z:

 

sin(z) = z -  z³ / 3!  +  z⁵ / 5!  …..  -  (-1) ^{n- 1} z ^{2n -1}/  (2n -1)!  +

 

Then:

z – sin z /  z ³     =  1/  z ³    [z -  (z -  z³ / 3!  +  z⁵ / 5!  …..  ) ] 

 

 =  1/  z ³    [z³ / 3!   -  z⁵ / 5!    +   z ⁷ / 7!  -    ……….

 

 

=   1/3!  -  z ² / 5!     +   z ⁴ / 7!  -    z ⁶ / 9!   + ……..

 

Ex. (3)

 Find a Laurent series for f(z) =  sin (4z) / z ⁴

 

Using the sine series we can write:

 

sin (4z) / z ⁴     =  1/   z ⁴   [4z -    4z³ / 3!  +  4 z⁵ / 5!  +  ……..]

 

=     4 /   z ³    -    4³ / 3!  z  +   4⁵  z  / 5!       +   …..

 

= 4 z  -³    -  4³  z -¹ /  3!   +   4⁵  z  / 5!        +   …..

 

 

 

Which on inspection is found to have the form:

 

å^¥ _{n = 0}    (-1) ⁿ   4 ^{2n+ 1}  z ^{2n -3}  /   (2n +  1)!

Check term number 3 (n = 2):

 

4 ^{2n+ 1}     = 4 ^{2(2)+ 1}      =      4⁵

 

z ^{2n -3}  =  z ^{2(2) - 3}      = z ¹=  z

 

(2n +  1)!   =  (2(2) +  1)!  = 5!

 

Then:  we get  4⁵  z  / 5!        

 

A removable singularity occurs at z = 0 so we expect the region of convergence to be:

 

÷ z  ÷      >   0   

 

Problems for Math Mavens:

 

Let f(z) = exp (-1/z ²) /  z⁵

 

a) Show the Laurent series can be written: 

 

å^¥ _{n = 0}    (-1) ⁿ   /   n !  z ^{2n + 5}   

b)Specify where the singularity would occur and the precise region of convergence

 

 

Challenge Problem: For über Math Mavens only –

 

Let f(z) = 7 z ²  + 9 z  - 18 /  z ³ -  9z

 

Find Laurent series for the convergence regions:

 

a)  0  < ÷ z ÷   <    3  and

b)    ÷ z  ÷      >   3  

 

(Hint: Approach the function f(z) by resort to partial fraction decomposition)