EXAMPLE.
As an example of the formulas obtained let us consider the following fraction:
3z 4  4
z 2 ( z 2  1)
In this case :
P( z )  3 z 4  4
Q(z)  z 2 (z 2  1) 3
and
The roots of Q(z)=0 are :
Root
z=0
z=i
z=-i
Multiplicity
two
three
three
Then the fraction can be expanded as:
L1 L2 A1 z  B1 A2 z  B2 A1 z  B1
3z 4  4





z ( z 2  1) 3 ( z 2  1) 2 ( z 2  1)
z 2 ( z 2  1) 3 z 2
(I)
Moreover :
P' ( z )  12 z 3
(a)
P' ' (z)  36z 2
(b)
To calculate L1 we use (5) :
L1 
P(a)
T (a)
(5)
Where :
T ( z) 
Q( z )
( z  a) r
(4)
In our case a =0 and r=2, then :
T ( z )  ( z 2  1) 3
from this
Therefore we can obtain :
T' (z)  6z(z 2  1) 2
P(0)  4
P' (0)  0
T(0)  1
T' (0)  0
Thus, we have for L1 :
L1 
P(0) 4

T (0) 1
L1  4
To obtain L2 we make k=1 in (15b):
L2 
P' (0)  T ' (0) 0  (0)( 4)

1!T (0)
1
L2  0
To calculate B1 , B2 , B3 , C1 , C2 and C3 we use the following formulas :
U ( z) 
Q( z )
( z  bz  c)
(19)
P(a)
Q(a )
(21)
2
A1 a  B1 
m
F ( m) (a)  {C mh U (m-h) (a)
h 0
m -1
h!
[( z  a * ) k ( Ak 1 z  Bk 1 )] ( h  k )
k  0, h (h - k)!

z a
}
P ( m ) (a)  F ( m ) (a)
Am 1 a  Bm 1 
m!(a  a * ) m U (a )
m  1,2,3,....,  1.
(27)
In this case :
ai
U ( z) 
z 2 ( z 2  1) 3
( z 2  1) 3
U(z)  z 2
(c)
Then :
U ' ( z)  2z
U' ' ( z )  2
(d)
(e)
Whence along with P(z), (a) and (b) we have :
P(i )  7
P' (i )  -12i
(26)
P' ' (i )  -36
Now we use (21) with a  i :
U(i )  -1
U' (i )  2i
U' ' (i )  2
A1i  B1 
P(i )
U (i )
A1i  B1 
7
 -7
-1
Likening both sides we get :
A1=0
B1=-7
To calculate A2 and B2 we make m=1 in (26) :
1
F ' (a )   {C1h U (1-h) (a )
h 0
F ' (i )  C10U ' (i )
0
h!
[( z  (i )) k ( Ak 1 z  Bk 1 )] ( h  k )
k  0, h (h - k)!

0
0!
[( z  (i)) k ( Ak 1 z  Bk 1 )] ( 0 k )
k  0 , 0 (0 - k)!

z i
z i
}

0
C11U ' (i )
1!
[( z  (i )) k ( Ak 1 z  Bk 1 )] (1 k )
k  0 ,1 (1 - k)!

1
F ' (i )  C10U ' (i ) [( z  i) 0 ( A1 z  B1 )] ( 0 )
0!
z i
z i
1
 U (i ) [( z  i ) 0 ( A1 z  B1 )]'
1!
z i
Substituting values we obtain :
F ' (i )  2i (0  7)  A1
z i
F ' (i )  14i  0
F ' (i )  14i
Next we make m=1 in (27) :
A2i  B2 
P' (i )  F ' (i )
1!(i  i )U (i )
If we substitute values it becomes :
A2 i  B2 
 12i  (14i )
(2i )( 1)
A 2 i  B 2  1
A2  0
B 2  1
To obtain A3 and B3 we make m=2 in (26) :
2
F ' ' (a)  {C U
h 0
h
1
1
(2- h)
(a)
h!
[( z  (i)) k ( Ak 1 z  Bk 1 )]( hk )
k 0, h (h - k)!

z i
}
As similarly done for F’(z) we get :
F ' ' (i )  U ' ' (i )( A1 z  B1 )
z i
U (i ){[( z  i ) 0 ( A1 z  B1 )]' '
1
[( z  i )( A2 z  B2 )] ( 0) } 
0!
 2[( z  i )( A2 z  B2 )]' z i }
 2U ' (i ){( A1 z  B1 )'
z i
z i

Substituting values :
F ' ' (i )  2(0i  7)  2(2i ){0  (i  i )(0i  1)}  {0  2[( z  i ) A2  A2 z  B2 ] z i
F ' ' (i )  14  4(2i )  2[(i  i )(0)  0i  1]
F ' ' (i )  4
With m=2 (27) remains as :
A3i  B3 
A3  0
P' ' (i )  F ' ' (i )
2!(i  i ) 2 U (i )
B 3  -4
A3i  B3 
 36  (4)
2(4i 2 )( 1)
Therefore expansion (1) becomes :
3z 4  4
4
7
1
4
 2  2
 2
 2
2
2
3
3
2
z ( z  1)
z
( z  1)
( z  1)
( z  1)
A 3i  B 3  -4