- #1
jamilmalik
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Find the Laurent series of the following function in a neighborhood of the singularly indicated, and use it to classify the singularity.
[itex]f(z) = \frac{1}{z^2-4} ; z_0=2 [/itex]
Laurent series
[itex]\sum_{-\infty}^{\infty} a_n (z-c)^n [/itex]
I started by doing a partial fraction decomposition of [itex] f(z) = \frac{1}{z^2-4} ; z_0=2 [/itex] which I got to be [itex] f(z) = \frac{-1}{4} \cdot \frac{1}{z+2} + \frac{1}{4} \cdot \frac{1}{z-2} [/itex] .
Then I used [itex] 0< \vert z+2 \vert < 2 \Rightarrow \vert \frac{z+2}{2} \vert < 1 [/itex] to apply the geometric series and get [itex] \frac{1}{4} \sum_{k=0}^{\infty} (\frac{2}{z})^k [/itex].
I am not sure if I made a mistake somewhere or if this method is not applicable. Can somebody please provide a hint or point out an error on my behalf?
Homework Statement
[itex]f(z) = \frac{1}{z^2-4} ; z_0=2 [/itex]
Homework Equations
Laurent series
[itex]\sum_{-\infty}^{\infty} a_n (z-c)^n [/itex]
The Attempt at a Solution
I started by doing a partial fraction decomposition of [itex] f(z) = \frac{1}{z^2-4} ; z_0=2 [/itex] which I got to be [itex] f(z) = \frac{-1}{4} \cdot \frac{1}{z+2} + \frac{1}{4} \cdot \frac{1}{z-2} [/itex] .
Then I used [itex] 0< \vert z+2 \vert < 2 \Rightarrow \vert \frac{z+2}{2} \vert < 1 [/itex] to apply the geometric series and get [itex] \frac{1}{4} \sum_{k=0}^{\infty} (\frac{2}{z})^k [/itex].
I am not sure if I made a mistake somewhere or if this method is not applicable. Can somebody please provide a hint or point out an error on my behalf?
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