Laurent series question!
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# Laurent series question!

[From: ] [author: ] [Date: 11-05-22] [Hit: ]
or 1/(z + 1) = 1/[z(1 + 1/z)] = (1/z)/(1 - (-1/z)). The first has series ∞Σ (-1)ⁿ zⁿ = 1/(1 + z), convergent for |z| n=0The latter has series ∞Σ (-1)ⁿ (1/z)^(n+1) = 1/(1 + z), convergent for |1/z| 1. n=0The last term can similarly be arranged in a couple of ways. 1/(z - 2) = -1/(2 - z) = (-1/2)/(1 - z/2),......
f(z) = (z^2-2)/(z(z+1)(z-2)) = 1/z - 1/(z+1) + 1/(z-2)

a) Compute the Laurent Series of f(z) on 0 < |z| < 1
b) Compute the Laurent Series of f(z) on 1 < |z| < 2

I'm very confused as to what the question wants. Please help!

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You can rearrange the three terms in a variety of ways. What you want to do here is to find arrangements that allow you to use a generalization of a geometric series.

For |z| > 0, 1/z is it's own Laurent series. I hope that is clear. It has precisely the form

Σ a_n zⁿ
n=-∞

where a_(-1) =1 and all other a_n = 0.

Note that you can rewrite the second term in a couple of similar (yet different) ways.

1/(z + 1) = 1/(1 - (-z)), or

1/(z + 1) = 1/[z(1 + 1/z)] = (1/z)/(1 - (-1/z)).

The first has series

Σ (-1)ⁿ zⁿ = 1/(1 + z), convergent for |z| < 1.
n=0

The latter has series

Σ (-1)ⁿ (1/z)^(n+1) = 1/(1 + z), convergent for |1/z| < 1 ----i.e. for |z| > 1.
n=0

The last term can similarly be arranged in a couple of ways.

1/(z - 2) = -1/(2 - z) = (-1/2)/(1 - z/2), or

1/(z - 2) = 1/[z(1 - 2/z)] = (1/z)/(1 - 2/z).

Each of these can be expressed by a ("geometric like") series

Σ - zⁿ/2^(n+1) = 1/(z - 2), convergent for |z/2| < 1 ---i.e. |z| < 2
n=0

or

Σ - 2ⁿ/z^(n+1) = 1/(z - 2), convergent for |2/z| < 1 ---i.e. |z| > 2.
n=0

Okay, just pick amongst these to get the desired interval of convergence. It is okay if part of the series converges on a larger domain, you just want to be sure that all parts converge at least on the indicated domain.

a) 0 < |z| < 1, Choose the first series for each of the last two terms.

Σ - [(-1)ⁿ + 1/2^(n+1)] zⁿ + 1/z.
n=0

b) 1 < |z| < 2, Choose the second series for the term 1/(z+ 1) and the first for the term 1/(z - 2)

∞ . . . . . . . . . . . . . . . . . . . ∞
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