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)

 

 

Introduction to 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…..

Note this series is unique for a given annulus, i.e. the shaded region shown in the image.

 The integral can be treated by expanding 1 / (w – z) as we did already in the original blog on complex series, i.e.

1 / (w – z)  = 1/ {w – a) – (z – a) = 1/ (w – a)[ 1/ 1 –(z –a)/(w – a)]

Which will be:   - å^¥ _{m = 0}    (z – a) ^m  / (z – a) ^m+1

Which is convergent by the ratio test.  Then:

1/2 pi  ò_C1  f(w) dw /(w – z)  =   - 1/2 pi  ò_C1  f(w) dw /(w – z) 

=  - å^¥ _{m = 0}    1 / (z – a) ^m+1  1/2 pi  ò_C1  f(w) (w – a) ^m  dw 

Now, replace the positive index m by –(n+1) and rewrite the preceding as:

1/2 pi  ò_C1  f(w) dw /(w – z)  =   

 å ^-¥ _{n = -1}    (z – a) ⁿ  1/2 pi  ò_C1  f(w) (w – a) ^{n +1}  dw

Where the integrals:

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

ò_C1  f(w) dw /(w – a) ⁿ⁺¹            n = 0,1, 2….

Can also be evaluated over a common circle C, concentric with C1 and C2 and lying just within the annulus: R1 <  R < R2.   To prove uniqueness,  assume an expansion:

å^¥ _{n = -¥}   c _n (z – a) ⁿ 

exists and is valid in the annulus  R1 <  ÷ z – a ÷  < R2

Now choose some arbitrary integer k and  multiply both sides of the expression by

:

(z – a) ^{–k +1}  and integrate around a circle C about z = a lying inside the annulus..  Then:

ò_C  f(z) dz /(z – a) ^k+1   å^¥ _{n = -¥}   c _n  ò_C  f(z) dz /(z – a) ^{k+1 - n}  

Now, all integrals on the right side will vanish except for one, for which n = k and whose value is 2 pi . Therefore:

ò_C  f(z) dz /(z – a) ^k+1   =    c _k 2 pi

Note that the part of the Laurent series consisting of positive powers of (z – a) is called the regular part,  This resembles the Taylor series that we already saw – but it needs to be clarified that the nth coefficient can’t be disassociated in general with any nth derivative  f ⁿ (a) since the latter may not exist. (In most applications f(z) is not analytic at z = a).

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.

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