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Hennie

  • 3 years ago

Compute the Laurent expansion of z/((z-1)(2-z)) for each of the following regions: abs(z) <1 and abs(z-2)<1.

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  1. Jemurray3
    • 3 years ago
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    Partial fraction decomposition would probably be a good start.

  2. Hennie
    • 3 years ago
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    In the expression (A/(z-1)) and B/(z-2) is the value for A=1 and B=-2 ?

  3. Hennie
    • 3 years ago
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    For the laurent's expansion should I then use (-1/(1-z)) and ((1/(1-z/2)) and expand each term seperately ?

  4. Jemurray3
    • 3 years ago
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    It should be \[ \frac{1}{z-1} - \frac{2}{z - 2} \] so yes you're right... and now yes, you want to expand these two terms around those two points. For example, for |z|<1, we can rewrite that as \[ \frac{-1}{1 - z} + \frac{1}{1 - z/2} = -\sum_{n=0}^\infty z^n+ \sum_{k=0}^\infty (\frac{z}{2})^k \] you can simplify that if you'd like, bringing them both under the same summation sign and all that stuff, but that's the idea. Since |z|<1, these series converge. For |z-2|<1, they don't necessarily converge, so you have to expand them about the point z = 2 for the second part, but the same deal applies.

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