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
For
z = 3: 18 C = 7(3)2 + 9(3) – 18 = 63 + 27 – 18 = 72
Then:
=
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
2nd and 3rd terms and expand using 1 / (1 – z) and
substituting:
2/
z + 1/3 (1 – z/ 3 + z 2 / 32 + …)
- 4/3 (1 + z/ 3 + z 2 / 32 + …)
Combining
Terms:
2/z
– 1 – 5z/ 32 + 3 z 2 / 33 + …)
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
2nd, 3rd etc terms using 1/ 1 – z:
Þ 2/ z +
1/z (1 – 3/z + 32 / z 2 +
….) +
......+ 4/z (1 + 3/z + 32 / z 2 + …)
......+ 4/z (1 + 3/z + 32 / z 2 + …)
=
2/z + 1/z - 3/ z 2 + 32 / z 3 + +…….
+ 4/z + 12/ z 2 + 36 / 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
n (4 + (-1)
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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