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I need to find the series expansion at $$z=0$$ and the Laurent series at $$ \sqrt{ 2/3} <|z| < \infty $$ for the function $$ f(z)= \frac{1}{(2-3z^{2})^2} $$ I guess I first need to first find the partial fraction decomposition

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    $\begingroup$ It seems like you already know what you need to do. Maybe try computing the partial fraction decomposition and edit your question if you are still unsure of how to proceed. $\endgroup$
    – DMcMor
    Commented Jan 12, 2021 at 16:43

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Hint: $\;f(z)=\dfrac{\left(\dfrac12\right)^2}{\left(1-\dfrac32z^2\right)^2}=\dfrac {\left(\dfrac1{3z^2}\right)^2}{\left(1-\dfrac2{3z^2}\right)^2}\;$ and $\;\dfrac 1{(1-x)^2}=1+2x+3x^2+\cdots$

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