Math, asked by kumarmayan2017, 8 months ago

Find the Laurent expansion of
(7z^2 +9z-18)/z^3-9z
in the region (i) 0<|z|<3
(ii)|z|>3

Answers

Answered by danu7962

Step-by-step explanation:

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

We approach using partial fraction decomposition:

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: 18B = 7(-3)2 + 9(-3) – 18 =

63 – 27 – 18 = 18 so that B = 1

For z = 3: 18 C = 7(3)2 + 9(3) – 18 = 63 + 27 – 18 = 72

So that C = 4

Then:

7 z 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 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 + …)

= 2/z + 1/z - 3/ z 2 + 32 / 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

Answered by prerita09