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I'm having trouble in finding the Laurent Series of this function:

$f(z)=\frac{1-z}{(1-2z)^2}$

Near the point $z=\frac{1}{2}$

I know the answer from Wolfram Alpha, but I don't understand how to get there myself. Is it by partial fractions?

Thank you!

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Note that $z=\dfrac12$ is an isolated singularity of the given function. Therefore this has a Laurent's expansion near centered at $z=\dfrac12.$ $$f(z)=\dfrac{1-z}{(1-2z)^2}=\dfrac{1+(1-2z)}{2(1-2z)^2}=\dfrac{1}{8(z-\frac12)^2}-\dfrac{1}{4(z-\frac12)}$$ is the required expansion and it follows from very simple simplification. Good Luck.

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  • $\begingroup$ This is of course, the partial fraction decomposition of $f.$ $\endgroup$ – Bumblebee Apr 27 '16 at 7:07
  • $\begingroup$ Thank you very much ! I don't know why I missed that simplification $\endgroup$ – ValientProcess Apr 27 '16 at 7:22

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