Laurent Series Watch

TheEd
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Report Thread starter 7 years ago
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Q If \displaystyle f(z) = \frac{z}{(z-1)(z-2)} find the Laurent Series of f around the simple pole z=2. What is the radius of convergence of the holomorphic part of this series?

Can anyone see if I've done this correct?

Splitting f into partial fractions: \displaystyle f(z) = -\frac{1}{z-1} + \frac{2}{z-2}.

The 1st fraction is holomorphic in the neigbourhood of z=2. And

\displaystyle - \frac{1}{z-1} = \frac{1}{1-z} = \sum_{n=0}^{\infty} z^n

i.e. the geometric series.

So the Laurent Series of f around the simple pole z=2 is:

\displaystyle \sum_{n=0}^{\infty} z^n + \frac{2}{z-2}

and the radius of convergence of the holomorphic part of this series (the geometric series) is R=1.
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nuodai
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Report 7 years ago
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The series \displaystyle \sum_0^{\infty} z^n doesn't converge around z=2, and anyway, your Laurent series should take the form \displaystyle \sum_{-1}^{\infty} a_n(z-2)^n, whereas yours has the n=-1 term expanded at z=2 and the others (the 1/(1-z) bit) expanded at z=0. You need to expand \dfrac{1}{1-z} about n=2 instead, to get something of the form \displaystyle \sum_0^{\infty} a_n(z-2)^n.
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