Revisiting Laurent Series

We introduce the Laurent Series as follows:

If f(z) is analytic inside and on the boundary of the ring-shaped region R (see diagram) bounded by two concentric circles C1 and C2 with center at a and respective radii r1 and r2 (r1 > r2) then for all z in R:

f(z) = å^¥ _{n = -¥}   c _n (z – a) ⁿ  +    å^¥ _{n =-¥}   c _{-n /}   (z – a) ⁿ

where:

c _n  =  1/2 pi  ò _C1  f(w) dw /(w – a) ⁿ⁺¹            n = 0,1, 2…..

   c _{- n}  =  1/2 pi  ò _C2  f(w) dw /(w – a) ^{–n +1}            n = -1,-2, -3…..

 

The other part of the series, consisting of negative powers, is called the principal part.  Either part or both may terminate or be identically zero. If the principal part is identically zero then f(z) is analytic at z = a since the derivative exists and the Laurent series is identical to the Taylor series.

 

More intricacies – and singularities:

 

Point z = a is called a zero or root of the function f(z) if f(a) = 0. If then f(z) is analytic at at z = a then the Taylor series:

 

f(z) = å^¥ _{n = 0}   c _n (z – a) ⁿ 

must have  c ₀  = 0.  If c ₁  ¹ 0, the point a is called a simple zero (or a zero of order one). It could happen that c ₁ and perhaps several other next coefficients vanish. Then let  c _m be the next vanishing coefficient (unless f(z) = 0) then the zero is said to be of order m. The order of a zero may be evaluated – without any knowledge of the Taylor series – by calculating:

 

lim _{z® a}      f(z)  / (z – a) ⁿ 

 

For n = 1, 2, 3. The lowest value of n for which this limit doesn’t vanish is equal to the order of the zero.

 

 If a function f(z) is analytic in the neighborhood of some point z = a with the exception of the point z = a itself then it is said to have an isolated singularity (or isolated singular point) at z = a. It’s customary to distinguish isolated singularities by the following types of behavior of f(z) as z  ® a for an arbitrary function.

Examples:

 

1)     f(z)  remains bounded, i.e. ÷ f(z)÷  <B for a fixed B

 

2)     f(z) is not bounded and ÷ f(z)÷    approaches infinity. Namely,

 ÷ f(z)÷   > M for ÷ z – a ÷  <  e

 

3)     Neither of the two cases above, in other words f(z) oscillates.

 

Examples:

1)     f(z) =  1/ z – 1  (isolated singularity at z = 1)

 

a) If we demand ÷ z ÷  < 1 we can obtain a Taylor expansion.

 

b) ) If we seek:  1 < ÷ z ÷   <   0    we obtain a Laurent expansion

2)     f(z) = 1/ z (1 – z) 

 

a) If  0 <  ÷ z ÷   <   1  then we obtain a Laurent series

b)  If   1   < ÷ z ÷   <   0    we also obtain a Laurent expansion.

 

Problem for the Math Maven

 

Consider the function: f(z) = 1/ (z+ 1) ((z + 3)

 

a) Find a  Laurent series  for:   1< ÷ z ÷   < 3
 

b) Find a Laurent series for    0 <  ÷ z +  1 ÷  < 2   

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!)