Laurent series... Complex
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Laurent series... Complex

[From: ] [author: ] [Date: 12-12-04] [Hit: ]
Σ -4ⁿ/z^(n+1).Σ n4^(n-1)/z^(n+2),Your solutions contain two errors. First,......
Find two Laurent for z^-1(4-z)^-2 involving powers of z and state where they are valid.


Answer
Z^-1(4-z)^-2 = 1/16z + Sum(n=0 to infinity) of ((n+2)z^n)/(4^(n+3)) for |z|<4,
Z^-1(4-z)^-2 = Sum(n=0 to infinity) of (n(4)^(n-1))/(z^(n+2)) for |z|>4,

-
d/dz 1/(4 - z) = 1/(4 - z)², and

1/(4 - z) = ¼/(1 - z/4) =


Σ ¼(z/4)ⁿ for |z| < 1
n=0

Differentiate term by term, and multiply through by 1/z for 0 < |z| < 4 to get

1/[z(4 - z)²] = 1/z d/dz(1/4 + z/16 + z²/4^3 + z^3/4^4 + ....) =

= 1/(16z) + 2/4^3 + 3z/4^4 + 4z²/4^5 + ...) = 1/(16z) +


Σ (n+2)zⁿ/4^(n+3), for 0 < |z| < 4
n=0

For |z| > 4, 1/(4 - z) = -(1/z)/(1 - 4/z) =


Σ -4ⁿ/z^(n+1).
n=0

Differentiate term by term and multiply through by 1/z to get 1/(z(4 - z)²) =


Σ (n+1)4ⁿ/z^(n+3) = <---reindex n + 1 -> n
n=0


Σ n4^(n-1)/z^(n+2), for |z| > 4
n=1

Your solutions contain two errors. First, the top series is valid for 0 < |z| < 4. And second, the bottom series for the terms as you've written it should have the index starting at 1 not at 0.
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keywords: Complex,series,Laurent,Laurent series... Complex
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