^{2}+ 9 z - 18 / z

^{3}- 9z

a) 0 <

**÷**z

**÷**

**< 3 and**

b)

**÷**z**÷****> 3**^{ }

7z

^{2}+ 9 z - 18 / z

^{3}- 9z = z (z + 9) + 18/ [z {z + 3) (z – 3) ]

=
A/ z + B/ (z + 3) + C / (z – 3)

7
z

^{2}+ 9 z - 18 = A(z + 3) (z – 3) + Bz (z – 3) + Cz (z + 3)
For
z = 0: - 9A = -18 so that A = 2

^{2}+ 9(-3) – 18 =

For
z = 3: 18 C = 7(3)

^{2}+ 9(3) – 18 = 63 + 27 – 18 = 72
Then:

^{2}+ 9 z - 18 / [z {z + 3) (z – 3) ]

=
2/ z + 1 / (z + 3) + 4/ (z -3)

Rewrite
as:

2/
z + 1/3 (1 / 1 + z/3) – 4/3 (1 / 1 –
z/3)

For
term 1:

**÷**z**÷****< 1**
For
term 2 :

**÷**z**÷****< 3**
For
term 3:

**÷**z**÷****< 3**
Expand
2

^{nd}and 3^{rd}terms and expand using 1 / (1 – z) and substituting:
2/
z + 1/3 (1 – z/ 3 + z

^{2}/ 3^{2}+ …) - 4/3 (1 + z/ 3 + z^{2}/ 3^{2}+ …)
Combining
Terms:

2/z
– 1 – 5z/ 3

^{2}+ 3 z^{2}/ 3^{3}+ …)
Which
series can be represented:

2/
z + [ å

^{¥}_{n = 0}(-1)^{n }– 4^{ }**/**3^{n + }^{1 }] z^{4 }^{ }
For:
0 <

**÷**z**÷****< 3**
Now,
rewrite the original partial-fraction f(z) in the form:

2/
z + 1/ z (1 / 1 + 3/z) + 4/z ( 1 / 1 – 3/z)

Expand
2

^{nd}, 3^{rd}etc terms using 1/ 1 – z:
Þ 2/ z +
1/z (1 – 3/z + 3

......+ 4/z (1 + 3/z + 3

^{2}/ z^{2}+ ….) +......+ 4/z (1 + 3/z + 3

^{2}/ z^{2}+ …)
=
2/z + 1/z - 3/ z

+ 4/z + 12/ z

^{2}+ 3^{2}/ z^{3}+ +…….+ 4/z + 12/ z

^{2}+ 36 / z^{3}+ …..
=
2/z + [5/z + 9/ z

^{2}+ 45 / z^{3}+ ………]
Which
can be represented in the form:

2/
z + å

^{¥}_{n = 0}3**(4 + (-1)**^{n}^{n }**) /**z^{n + }^{1 }
For:

**÷**z**÷****> 3**
## 2 comments:

2 years to late but the expansion si z^n not z^4 made a little typo in z<3

Thanks for the correction!

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