For ppt presentation

1. Power Series Representations
ECE 6382
Notes are from D. R. Wilton, Dept. of ECE
1
David R. Jackson
Fall 2023
Notes 7
0
z
a
z
b
z
a
b

 Analytic
z 0
z z


2. Geometric Series
Consider
2
0
1
N
N n
N
n
S z z z z

      
2
Consider the following sum:
Note that
2 1
N N
N
zS z z z z 
    
  1
1 1 N
N N N
S zS z S z 
    
Hence, we have
1
1
1
N
N
z
S
z




 
1
1 1 1
0 1
i N N
N N N
z r e r r z
  
  
  
  
iff
Note:
2
0
1
n
n
S z z z


    

Geometric series (GS):

3. Geometric Series (cont.)
3
x
y
1
z 
1
z 
1
1

2
0
1
1 1
1
n
n
z z z z
z


     

 ,
Hence, we have:
  2
1
=1 , 1
1
f z z z z
z
    

We can write this as:
(summation of geometric series)
(a power series expansion of the function f (z))
Complex plane for f (z)
The power series for f (z) converges inside the unit circle and diverges outside the
unit circle. It oscillates (does not converge) on the unit circle.

4. 4
x
y
1
z 
1
z 
1
1

For |z| > 1 we can use another series representation:
  2 3
1
2 3
1 1 1 1 1 1
1
1 1
1 1 1
z
z z z
z z z
z z z
  
     
 
   
 
   
 
 
GS
1
1 1
, z
z
 
i.e.,
This series converges iff
Geometric Series (cont.)
This is the geometric series
with z  1/z.

5. Geometric Series (cont.)
Consider
5
x
y
1
z 
1
z 
1
1

2
0
1
1 1
1
n
n
z z z z
z


     

 ,
Summary of Geometric Series Results
2 3
0
1 1 1 1 1 1
1 1
1
n
n
z z
z z z z
z z



 
       
 
  
 ,
The geometric series is important by itself, and it is also useful for later derivations.
(Geometric series)
This is an extension (or “continuation”) of the original geometric series.

6. Geometric Series (cont.)
Consider
2 3
1 1 1
1 1
1
p
p p p p p p
p
p
z z z z
z z
z
z z
z
z z z
z
z
z
 
     
  
       
     
     
  
       
 

 


 

if , i.e.
Generalize: (pole at zp):
6
x
y
p
z z

p
z z

R
p
z

2 3
1 1 1
1 1
1
p
p p p p
p
p
z z z z
z
z z z z z z z
z
z
z z
 
     
 
      
     
    
     
 

 



if , i.e.
Taylor series Laurent series

7. Consider
       
     
2 3
0 0 0
0 0 0
0 0 0
0
0
0
0
0 0
0
0
0 0
0
1 1 1 1
1
1
1
1 1 1
1
p p p p
p p
p
p
p
p
p
p p
z z z z z z
z z z z z z z z
z z z z z z
z z
z z
z z
z z
z z z z
z z
z z
z z z z z z
z z
 
     
  
  
       
     
     
 
   
 
   
      
 
 
 
 

 

   

 

   
 
Validif , i.e.
Similarly,
 
2 3
0 0 0
0 0 0 0
0
0
0 0
0
1
1
1
p p p
p
p
z z z z z z
z z z z z z z z
z z
z z
z z z z
z z
 
  
     
 
    
     
   
   
     
 
 

 

   

Validif , i.e.
Decide whether |zp-z0| or |z-z0| is larger (i.e., if z is
inside or outside the circle at right), and then factor
out the term with largest magnitude!
Geometric Series (cont.)
7
x
y
0
z z

0
p
z z

0
p
R z z
 
Radius of convergence :
p
z
0
z
z
R

The previous series were expanded about the origin. We can also expand about another point.
This is the case
illustrated here.
0 0
p
z z z z
  

8. Consider
 
2 3
0 0 0
0 0
0 0 0
0
1 1
1 p
p p p p
p
z z z z z z
z z z z
z z z z z z z z
z z
 
     
  
 
        
     
     
   
 
      
 
if
Geometric Series (cont.)
8
 
2 3
0 0 0
0 0
0 0 0 0
1 1
1
p p p
p
p
z z z z z z
z z z z
z z z z z z z z z z
 
  
     
 
       
     
    
 
     
 
if
Converges inside circle
Converges outside circle
Summary of Series
Taylor series
Laurent series
x
y
p
z
0
z

9. Uniform Convergence
Consider 3 2 1
3
3
10 0 10 0 10 0
10 0
1
1 00 0 001 0 000001 0 000000001
1 10
1 001001001001001
z i , i , i
z i
. . . .
.
  


   
 
    


.
:
Let's evaluate the geometric series for
Clearly, every additional term adds 3 more significant figure
2
2
1
1
10 0
1
1 00 0 01 0 0001 0 000001
1 10
1 0101010101
10 0
1
1 00 0 1 0 01
1 10
z i
. . . .
.
z i
. . .




 
    


 
  

:
:
s to the finalresult.
Here,however, each additional term adds only 2 more significant figures to the result.
0 001
1 11111
.
.
 

Andhere each additional term adds only1more significant figure to the result.
9
In general, for a given accuracy, the number of terms needed increases as |z| get larger
and approaches 1. The series is said to converge non-uniformly in the region |z| < 1.
2
1
1 1
1
z z , z
z
    


10.    
   
0
0
0
n
n
N
n
n
f z g z
N z
N N f z g z


 






  


, ,
A series is in a region if for any
there exists a number , dependent on but in the region
such that implies for a
R
uniformly convergent
independent of
z .
ll in R
10
Uniform Convergence (cont.)
Non-uniform convergence Uniform convergence
2 3
1
1
1
z z z
z
    

The series converges slower and
slower as |z| approaches 1.
x
y
1
z 
1
1
x
y
1
z R
 
R
1
Key Point:
Term-by-term integration of a series (switching the order of summation
and integration) is allowed over any region where
it is uniformly convergent. We use this property extensively later!

11. Uniform Convergence (cont.)
Consider
1
2
1
1 1
1
rel
1
1
1
1 1 1
1 1 1
N
N
N
N
N N
N N
N
N
z
S z z z
z
z
S S e z S z
z z z
S S
z
S



 


     

  
     
 
  
 

 
The partial sum error is
The relative error is
11
2 3
1
1 1
1
S z z z , z
z
      

The closer z gets to the boundary of the circle,
the more terms we need to get the same level
of accuracy (non-uniform convergence).
Note:
A relative error of 10-p
means p significant
figures of accuracy.
Example
Number of geometric series terms N vs. |z|
0
100
200
300
400
500
0 0.2 0.4 0.6 0.8 1
|z|
N
2 sig. digits
4 sig. digits
6 sig. digits
8 sig. digits
10 sig. digits
Number of terms N needed in series vs. |z|
|z|
N

12. Consider
12
Uniform Convergence (cont.)
1 1
rel
N N
N
S S
z R
S

 

  
The relative error is
1
z R
 
Assume:
For example:
  1
1
rel
0 95
0 95
N
N
R .
R .




 
Using N = 350 will give 8 significant figures everywhere inside the region.
x
y
1
z R
 
R
1
Example (cont.)
We now have uniform convergence for R = 0.95.
0
50
100
150
200
250
300
350
400
450
500
0 0.2 0.4 0.6 0.8 1
N
|z|
Number of geometric series terms N vs. |z|
2 sig. digits
4 sig. digits
6 sig. digits
8 sig. digits
10 sig. digits
0.95
N10
N8
N6
N4
N2
Number of terms N needed in series vs. |z|
|z|
N

13. Consider
13
Uniform Convergence (cont.)
Example
0
1
1
1
z
dz, z R
z
 


0 0
0 0 0
1
1
z z z
n n
n n
dz z dz z dz
z
 
 
 
 
   
   
    
 
  
Switch the order of integration and summation (integrate term-by-term)
1 2 3
0
0
1
1 1 2 3
z n
n
z z z
dz z
z n



    
 


Hence, we have
Note: This is not valid for |z|  1.
(We do not have uniform convergence there.)
x
y
1
z R
 
R
1

14. The Taylor Series Expansion
Consider
This expansion assumes we have a function f(z) that is analytic in a disk.
   
0
0
n
n
n
f z a z z


 

14
 
  1
0
1
2
n n
C
f z
a dz
i z z
 





 
( )
0
!
n
n
f z
a
n

Here zs is the closest singularity to z0.
Rc = radius of convergence of the Taylor series (distance to closest singularity)
The Taylor series will converge within the radius of convergence and diverge outside.
“derivative formula”
Note: Both forms are useful.
The path C is any counterclockwise closed path
within the disk that encircles the point z0.
Summary of Taylor Series:
0 c
z z R
 
0
z
z
s
z
c
R
Analytic

0
z z

C

15. The Taylor Series Expansion (cont.)
15
Brook Taylor (1685-1731)
Taylor's theorem is named after the
mathematician Brook Taylor, who stated
a version of it in 1712. Yet an explicit
expression of the error was not
provided until much later on by Joseph-
Louis Lagrange. An earlier version of
the result was already mentioned in
1671 by James Gregory.
From Wikipedia

16. 16
The Laurent Series Expansion (cont.)
Review of Cauchy’s Integral Formula (from Notes 3):
 
 
0
0
1
2 C
f z
f z dz
i z z




 
 
1
2 C
f z
f z dz
i z z






Another way to write it:
 
 
 
( )
0 1
0
!
2
n
n
C
f z
n
f z dz
i z z
 





Take derivative n times.

17. Taylor Series Expansion of an Analytic Function (cont.)
 
 
 
   
 
 
 
 
0 0
0 0
0
0
0
0
0 0
0
1
2
1
2
1
2
1
1
2
C
C
C
n
n
C
f z
f z dz
i z z
f z
dz
i z z z z
f z
dz z z z z
i z z
z z
z z
f z z z
dz
i z z z z













  

 
   
 

 
 

 
  


  
 
 
 






uniform
conv
Write the Cauchy integral formula in the form
( )
 
 
 
 
 
 
 
 
     
 
0
0
( )
0 0 1
0 0
0 0
1
0
0
0
( )
1
2
!
!
2
1
2
n
n
n n
n
n C
n
n
n
n
n
C
n
z z
f z
n
z z f z dz
i z z
f z a z z f z z
f z
a
i z z
f z
dz
i z z
f z
n 











 


 
  








 
 






 


ergence
derivative
formulas
recall
Taylor series expansion of about
where
 
 
( )
0
1
0
!
n
n
C
f z
dz
n

 

17
0 0 0
s
z z z z z z

    
x
y
0
z z

0
z
z
z
0
z z

C
( )
s
z singularity
R
f is analytic inside R

Construct circle C so that

18. 0 0
0
s
s
z z z z
z z
  
;
Note that the result is valid for any
where is the singularitynearest hence f
will c
r o
e nv
h e
e r
i g
t s s e
e i
18
Taylor Series Expansion of an Analytic Function (cont.)
0 0
s
z z z z
  
Note: It can also be shown that the series will diverge for 0 0
s
z z z z
  
x
y
0
z z

0
z
z
0
z z
 C
s
z
R
z

Note that in the result for an, the integrand is analytic inside R away
from z0, and hence (from Cauchy’s theorem) the path C is now
arbitrary, as long as it encircles z0 and stays inside R.
The point z can even be outside the path C (see the note below).
 
  1
0
1
2
n n
C
f z
a dz
i z z
 





(Note: This integral does not have z in it.)

19. 19
Taylor Series Expansion of an Analytic Function (cont.)
0 0
c s
z z R z z
   
Converges for :
The radius of convergence of a Taylor series is
the distance out to the closest singularity.
Key point:
The point z0 about which
the expansion is made is arbitrary, but
It determines the region of convergence
of the Taylor series.
x
y
0
z
s
z
c
R


20. 20
Taylor Series Expansion of an Analytic Function (cont.)
Properties of Taylor Series
 A Taylor series will converge for |z-z0| < Rc (i.e., inside the radius of convergence).
 A Taylor series will diverge for |z-z0| > Rc (i.e., outside the radius of convergence).
 A Taylor series converges uniformly for |z-z0|  R < Rc.
 A Taylor series may be differentiated or integrated term-by-term within the
radius of convergence. This does not change the radius of convergence.
 A Taylor series converges absolutely inside the radius of convergence (i.e.,
the series of absolute values converges).
 Within the common region of convergence, we can add and multiply Taylor
series, collecting terms to find the resulting Taylor series.
 When a Taylor series converges, the resulting function is an analytic function.
Rc = radius of convergence = distance to closest singularity
J. W. Brown and R. V. Churchill, Complex Variables and Applications, 9th Ed., McGraw-Hill, 2013.

21. The Laurent Series Expansion
Consider
This generalizes the concept of a Taylor series to include cases
where the function is analytic in an annulus.
   
0
n
n
n
f z a z z


 

   
 
0
0 1 0
1
n
n n n
n n
f z a z z b
z z
 
 
  

 
or
n n
b a

where
21
Here za and zb are two singularities.
Note:
The point za may be the point z0.
The point zb may be at infinity.
 
  1
0
1
2
n n
C
f z
a dz
i z z
 




 (derived later)
(This is the same formula as for the Taylor
series, but with negative n allowed.)
Note: We no longer have the “derivative formula” as we do for a Taylor series.
The path C is any counterclockwise closed path that
stays inside the annulus and encircles the point z0.
Final Result:
0
z
a
z
b
z
a
b

 Analytic
z 0
z z

0
a z z b
  

22. The Laurent Series Expansion (cont.)
Consider
Summary of Laurent series:
   
0
n
n
n
f z a z z


 

22
0
a z z b
  
The Laurent series converges inside the region
The Laurent series diverges outside this
region if there are singularities at 0
z z a b.
  and
0
z
a
z
b
z
a
b

 Analytic
z 0
z z

 
  1
0
1
2
n n
C
f z
a dz
i z z
 






23. The Laurent Series Expansion (cont.)
23
Pierre Alphonse Laurent (1813 -1854)
The Laurent series was named after and first
published by Pierre Alphonse Laurent in
1843. Karl Weierstrass may have discovered
it first in a paper written in 1841, but it was
not published until after his death.
From Wikipedia

24. The Laurent Series Expansion (cont.)
Consider
Examples of functions with poles, and how we can
choose a Laurent series:
   
0
1
0: 0
f z z a , b
z
    
Choose
The Laurent series is particularly useful for functions that have poles.
24
x
y

: 0
z 
R
0
z
a
z
b
z
a
b

 Analytic
z 0
z z


25. The Laurent Series Expansion (cont.)
Consider
Examples of functions with poles, and how we
can choose a Laurent series:
   
0 1: 0
1
z
f z z a , b
z
    

Choose
25
   
0 0: 1
1
z
f z z a , b
z
    

Choose
x
y

1
x
y

1
: 1
z 
R
: 1 0
z  
R
0
z
a
z
b
z
a
b

 Analytic
z 0
z z


26. The Laurent Series Expansion (cont.)
Consider
Examples of functions with poles, and how we can
choose a Laurent series:
 
  
 
0 0: 1 2
1 2
z
f z z a , b
z z
   
 
Choose
26
x
y
1
 
2
x
y
1
 
2
 
  
 
0 0: 2
1 2
z
f z z a , b
z z
    
 
Choose
: 1 2
z
 
R
: 2
z 
R
0
z
a
z
b
z
a
b

 Analytic
z 0
z z


27. The Laurent Series Expansion (cont.)
Example: The singularity does not have to be a pole.
 
  
 
0
1 2
2
2: 3 4
2 1
/
z
f z z a , b
z z
    
 
Choose
27
0 2
z  
x
y
Branchcut
Pole
2
 1
 2
1

 
0 2: 4
z a , b
    
or choose
x
y
Branchcut
Pole
2
 1
 2
1

: 3 2 4
z
  
R : 2 4
z  
R
Choose “alternative” branch cut.

28. Consider
Theorem:
The Laurent series expansion in the annulus region is unique.
(So it doesn’t matter how we get it; once we obtain it
by any series of valid steps, it is correct!)
 
 
 
0
cos
0 0
z
f z z , a , b
z
    
 
0
2 4 6
1
1
2! 4! 6!
z
z
z z z
f z
z



 
 
 
    
 
 
 
analytic
valid for
for
 
3 5
1
0
2! 4! 6!
z z z
f z , z
z
      
Hence
Example:
28
The Laurent Series Expansion (cont.)
This is justified by our Laurent series
expansion formula, derived later.
x
y
0 0
z  a
b

: 0
z 
R

29. Consider
A Taylor series is a special case of a Laurent series.
   
0
n
n
n
f z a z z


 

 
  1
0
1
2
n n
C
f z
a dz
i z z
 





If f (z) is analytic within C, the integrand is analytic for negative values of n.
Hence, all coefficients an for negative n become zero (by Cauchy’s theorem).
29
The Laurent Series Expansion (cont.)
Here f is assumed to be analytic within C.
0 1 2 3
n
a , n , ,
    
0
z
C
If f (z) is analytic:
Laurent series:

30. Pond: Domain of analyticity
Island: Region containing singularities
Bridge: Region connecting island and boundary of pond
30
The Laurent Series Expansion (cont.)
Derivation of Laurent Series
We use the “bridge” principle again
Pond, island, & bridge

31. Consider
 
 
2 1
1
2 C c
f z
f z dz
i z z
 




By Cauchy's IntegralFormula,
2
c

   
   
 
 
 
 
 
1 2
1
2 1
2 0 0
0 0
1
0 0 0 0 0
0 0 0
0
0
1 0 0
1 1
2 2
1 1 1 1
1
C
C C
n n
n n
n n
C ,C
f z f z
dz dz
i z z i z z
C z z z z
z z z z
z z z z z z z z
z z z z z z
z z
z z
C z z z z
 

 

 
 
 
 
 
 
  
 
   
  
    
 
  
 
 
 

 
  


 
 
(note the convergence regions for ove
, ,
,
On
On
   
 
 
 
 
 
1
0 0
1 1
0 0 0 1
0 0 0
0
0
1 1 1
1
n n ,
n n
n n
n n
n n
z z z z
z z z z z z z z z z z z
z z
z z

  
 

 
 
 

     
 
     
 
 
 
 




 
rlap!)
Contributions from the paths c1
and c2 cancel!
31
The Laurent Series Expansion (cont.)
x
y
2
C
1
C
1
c
2
c
z
0
z
2
s
z
1
s
z
z
z
f is analytic in R


2 1 1 2
C C c C c
   
The point
z is
between
C1 and C2.
Annulus (between C1 and C2)
R

32. Consider
 
 
2 1
1
2 C c
f z
f z dz
i z z
 




Hence,
2
c

 
 
 
 
 
 
1
2
1
0 1
0 0
0 1
1 0
1
2
1
2
C
n
n
n C
n
n
n C
f z
z z dz
i z z
f z
z z dz
i z z











 

 

 

  
 

 

 
 
uniform
convergence
32
The Laurent Series Expansion (cont.)
x
y
2
C
1
C
z
0
z
2
s
z
1
s
z
z


Note: The integrands are analytic in the blue region.
Because the integrands for the coefficient are analytic with R (the blue region), the
paths are arbitrary, as long as they stay within R. We can use the same path C,
which is arbitrary as long as it stays within R.
R

33. Consider
   
 
 
 
 
 
0 1
0 0
0 1
1 0
1
2
1
2
n
n
n C
n
n
n C
f z
f z z z dz
i z z
f z
z z dz
i z z










 



 

 
 
33
The Laurent Series Expansion (cont.)
   
0
n
n
n
f z a z z


 

We thus have
 
  1
0
1
2
n n
C
f z
a dz
i z z
 





where
The path C is now arbitrary, as long as it
stays in the analytic (blue) region R.
x
y
z
0
z
2
s
z
1
s
z
z
C


R

34. Examples of Taylor and Laurent Series Expansions
Consider  
 
1
1
f z
z z


Obtain expansions of about the origin.
Example 1: all
34
Use the integral formula for the an coefficients.
 
  1
0
1
2
n n
C
f z
a dz
i z z
 





The path C can be inside the circle or outside of it (parts (a) and (b)).
0 0
z 
y
1
x
z


z

35. Examples of Taylor and Laurent Series Expansions
Consider
35
  2 3
1
1 0 1
f z z z z z
z
        
,
Hence
 
 
 
 
1 2 2
0
2 2
0 0
1 (
0 1
1 1 1 1 1
1
2 2 2
1
2 1 2 1
1 1 1 1 1
0 1 2 1
2 2 2
m
n n n n
m
C C C
n m n m
m m
C C
a , b
f z
a dz dz z dz , z
i i i
z z z z
i, m n n m
dz dz
, m n n m
i i i
z z
  

  

  

 
   
 
 
 
 
    
   
   

    

   
 
   
    
  


  
 
 
a) Laurent series th
( )
wi
1 0)
n m
  
for
1 1
n
a , n
   
The path C is inside the blue region.
From previous example in Notes 3
From uniform convergence
y
1
x
z

C

 
 
1
1
f z
z z


: 0 1
z
 
R
From geometric series

36. Consider
 
   
 
1 2 3
1
3
0
3 3
0 0
1
1 1 1 1 1
2 2 2
1 1
1 1 1
1
2
2 2 3 1
1 1 1 1 1
0
2 2 2
n n n n
C C C z
n m
m
C
n m n m
m m
C C
a , b
f z
a dz dz dz
i i i
z z z z
dz , z
i z z
i, m n n m
dz dz
, m
i i i
z z
  


  
  




 
   
 
  

  
  
   
 
 

 
     
 
   
 

  


 
 
b) Laurent series with
( )
 
 
1
2 3 1
2 0
n n m
n m



     


    
for
36
Examples of Taylor and Laurent Series Expansions (cont.)
  2 3 4
1 1 1
1
f z z
z z z
    
,
Hence
1 2
n
a , n
  
The path C is outside the blue region.
From previous example in Notes 3
From uniform convergence
y
1
x
z


C
 
 
1
1
f z
z z


: 1
z 
R
From geometric series

37. Consider
37
Examples of Taylor and Laurent Series Expansions (cont.)
  2 3 4
1 1 1
1
f z z
z z z
    
,
  2 3
1
1 0 1
f z z z z z
z
        
,
Summary of results for the example:
 
 
1
1
f z
z z


y
1
x
z


z

38. Consider
38
Examples of Taylor and Laurent Series Expansions (cont.)
Note:
Often it is easier to directly use the geometric series (GS) formula
together with partial fraction expansions and some algebra,
instead of the contour integral approach, to determine the
coefficients of the Laurent expansion.
This is illustrated next (using the same example as in Example 1).

39. Consider  
 
 
 
 
0 0
l
1
1
1
1 1
im lim
z z
f
f
z
A B
z
z z z z
z
z f z
z z
A
 

  
 
 


Expand about the origin(we use partial fractions and GS):
Example 1
z  
   
1 1
1
1
1
lim 1 lim
z z
z
z
B z f z
 
 


  
 
1
z z 
 
 
 
2
1
1 1 1 1 1
1 1 1
1
1
f z
z z z z z z
z z
z

 
    
  

   


39
Examples of Taylor and Laurent Series Expansions (cont.)
  2 3
1
1 0 1
f z z z z z
z
        
,
Hence
0 1
z
 

40. Consider
 
 
 
1 1 1 1 1 1
1 1 1 1
1
f z
z z z z z z z
f z
z
 
 
      
  





1
1
z
 2
1 1
z z
 
  
 
 
40
Examples of Taylor and Laurent Series Expansions (cont.)
  2 3 4
1 1 1
1
f z z
z z z
    
,
Hence
Alternative expansion (|z| > 1): 1
z 
These are the same results that we got in the previous example by using the
integral formula for an in the Laurent series recipe.

41. Consider
 
  
0
3
1
0 1 1
2
2
1
1
1
1 2
f
z
z ,
z
z
-
z
-
z
z





 
  

Expand about valid followingin the annular regions:
(a) ,
(b) ,
(c) .
Expand in a Taylor /Laurent series.
Example 2
41
Examples of Taylor and Laurent Series Expansions (cont.)
y
1 2 3
x
1 1 2
z
  
z
1 2
z  
1 1
z  


y
1 2 3
x
1 1 2
z
  
z
1 2
z  
1 1
z  


y
1 2 3
x
1 1 2
z
  
z
1 2
z  
1 1
z  



42. Consider
 
  
       
   
   
2
2
2
0 1 1
1 1 1
2 3 3 2
1 1 1 1
1 2 1 1 1 1
2 1 1 2
1 1
1
1 1 1 1
2 2 2
z
f z
z z z z
z z z
z
z z
z z
  
  
   

   
     
 
 
 
 
   
 
          
 
 
 
a) For :
Usingpartial fraction expansion and GS,
42
Examples of Taylor and Laurent Series Expansions (cont.)
       
2 3
1 3 7 15
1 1 1 0 1 1
2 4 8 16
f z z z z , z
          
Hence
(Taylor series)
y
1 2 3
x
1 1 2
z
  
z
1 2
z  
1 1
z  


Part (a)
 
  
1
2 3
f z
z z

 

43. Consider
 
         
1 1 2
1 1 1 1
1 2 1 1 2 1 1 2 1 1 1 1
z
f z
z z z z z
  

   
        
   
   
b) For :
43
Examples of Taylor and Laurent Series Expansions (cont.)
 
   
     
2
2 2
1 1
1 1 1 1
1 1
2 2 1 1
2 1
z z
f z
z z z
   
 
        
   
  
   
   
(Laurent series)
so
 
  
1
2 3
f z
z z

 
y
1 2 3
x
1 1 2
z
  
z
1 2
z  
1 1
z  


Part (b)
Part (b)

44. Consider
 
           
 
1 2
1 1 1 1
1 2 1 1 1 1 2 1 1 1 1 1
1
1
1
z
f z
z z z z z z
z
 
   
         
   
   


c) For :
     
2
2
2 2 1
1
1 1
1
z z
z
 
   
 
 

 
     2
1 1
1 1
z z
 
  
 
 
 
 
44
Examples of Taylor and Laurent Series Expansions (cont.)
 
     
2 3 4
1 3 7
1 1 1
f z
z z z
   
  
(Laurent series)
so
 
  
1
2 3
f z
z z

 
y
1 2 3
x
1 1 2
z
  
z
1 2
z  
1 1
z  


Part (c)
Part (c)

45. Consider
45
Examples of Taylor and Laurent Series Expansions (cont.)
 
     
2 3 4
1 3 7
1 2
1 1 1
f z , z
z z z
     
  
 
  
1
2 3
f z
z z

 
Summary of results for example:
       
2 3
1 3 7 15
1 1 1 0 1 1
2 4 8 16
f z z z z , z
          
 
   
     
2
2 2
1 1
1 1 1 1
1 1 1 1 2
2 2 1 1
2 1
z z
f z z
z z z
,
   
 
   
           
 
   

   
y
1 2 3
x
1 1 2
z
  
z
1 2
z  
1 1
z  


z
z

46. Consider
  2
1 cos
0
0
0
1 2
:
0
z
, z
f z z
z
z
, z





 
 



Find the s
)
eries expan
(
sion about
is a "removable" singulari y
t
Example 3
46
Examples of Taylor and Laurent Series Expansions (cont.)
 
2 4
1
2! 4! 6!
z z
f z , z
     
 
2 4
sin
1
3! 5!
z z z
f z z
z
      
Similarly, we have
Hence
3 5
sin
3! 5!
z z
z z z
     
Note :
1 cos 1
z
  1

2 4 6 2 4 6
2! 4! 6! 2! 4! 6!
z z z z z z
 
       
 
 
Analytic everywhere!

47. Consider
   
     
 
 
 
sin sin
sin cos cos sin sin
s
n
in 0
cos 1
si
f z z z
z z z
f
f
f
z z .
 
   




 



   
 
 
      
 
  



Alternatively,directly use the derivative formula for Taylor seri
Find the seri
es :
es for about
Example 4
 
 
 
 
 
sin 0
cos 1
sin 0
cos 1
iv
v
f
f
f

 
 
 
 
    
 
  
47
Examples of Taylor and Laurent Series Expansions (cont.)
   
   
3 5
3! 5!
z z
f z z , z
 
 
 
        
   
   
3 5
3! 5!
z z
f z z , z
 
 
 
        
3 5
sin
3! 5!
z z
z z z
     
Note :
   
0
0
n
n
n
f z a z z


 

 
 
 
( )
0
1
0
1
2 !
n
n n
C
f z
f z
a dz
i n
z z
 


 



48. Consider
 
   
 
3
2
2
2 4
0
0 0
2
2 3 4
2
sin ln 1
l
1
1 1
1
n 1 1
2 3 4
ln 1 1
2 3 4
0
si
1
1
n
1
z z
z
z z , z
.
z
z z z
dz z d
z
z z
z
z
z
z z , z
z z
z z ,
z
z
z
z 

    

         
     

 


 



 
Find the first three terms of the r
se ies for b
Since then
Al o
a out
s
Example 5
 
3 5 3 5
4
2 6
4 2 3 4
2 2 6
4
3
3! 5! 3! 5!
2
3 45
2
sin ln 1
3 45 2 3 4
1 1
2 3 3
z z z z
z
z
z z
z z z z
z z z z z
z
z
   
    
   
   
   
  
         
  
  
 
     
 
 
Hence
5 6
1 1
1
4 6
z z , z
 
    
 
 
48
Examples of Taylor and Laurent Series Expansions (cont.)
The branch cut is chosen
away from the blue region.
y
1 x
z
1

49. Summary of Methods for Generating
Taylor and Laurent Series Expansions
Consider
   
 
 
   
 
 
0
0
0
0 1
0
0
!
1
2
n
n
n n
n
n
n n n
n C
f z
f z a z z a
n
f z
f z a z z a dz
i z z
z z ,









  


  



 
Taylor ( Laurent) series, , canbe generatedusing
Taylor Laurent series, , canbe generatedusing
To expand about first
not
and
   
0 0
f z f z z z

 
 
 
write in the form , rearrange
and expandusing geometric series or other methods.
Use partial fraction expansion and geometric series to generate series for rational functions
(ratios of polynomials,degree of numerator less than degree of denominator).
49
Summary of Methods

50. Summary of Methods for Generating
Taylor and Laurent Series Expansions (cont.)
Consider


Taylor / Laurentseries can be integrated or differentiated term-by - term within their region
of convergence.
Two Taylor series can be multiplied term -by - term :
within their common region of convergence
       
         
0 0
0 0
0 0 0
0 0 0 0
n m
n m
n m
p
n m p
n m p p n p n
n m p n
f z a z z g z b z z
f z g z a z z b z z c z z c a b
 
 
  

   
   
   
     
   
   
 


 
 
   
where
Two Laurent series can be - t
,
e
multiplied term-by rm within their common an
       
         
0 0
0 0 0
n m
n m
n m
n m p
n m p p n p n
n m p n
f z a z z g z b z z
f z g z a z z b z z c z z c a b
 
 
   

   
   
   
     
   
   
 

 
 
   
:
,
where
nulus of convergence
(if there is one)
50
Summary of Methods (cont.)