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Messages - Amy Zhao

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Quiz-6 / Re: Q6 TUT 0201
« on: November 17, 2018, 06:32:49 PM »
$\frac{z^2}{z^2-1}$$=\frac{z^2-1+1}{z^2-1}$
$=1+\frac{1}{z^2-1}$
$=1+\frac{1}{(z-1)(z+1)}$
$=1+\frac{1}{2}\frac{1}{z-1}-\frac{1}{z+1}$, by partial fractions
$=1+\frac{1}{2}(\frac{1}{z-1}-\frac{1}{2+(z-1)})$
$=1+\frac{1}{2}(\frac{1}{z-1}-\frac{1}{2}\frac{1}{1+\frac{z-1}{2}})$
$=1+\frac{1}{2}((z-1)^{-1}-\frac{1}{2}\frac{1}{1-\frac{1-z}{2}})$
$=1+\frac{1}{2}((z-1)^{-1}-\frac{1}{2}\sum_{0}^{\infty} (\frac{1-z}{2})^n)$
Residue=$\frac{1}{2}$

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