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Show that when $0 < |z-1| < 2$, $$\frac{z}{(z-1)(z-3)} = -3 \sum_{n=0}^\infty \frac {(z-1)^n}{2^{n+2}} - \frac{1}{2(z-1)}$$

I thought to attack this using a partial fraction decomposition and then breaking the partial fractions into Maclaurin series. I got the $$-\frac{1}{2(z-1)}$$ from that, but the other part has me a bit stumped.

I have: $$ \frac{3}{2(z-3)}$$

for the other partial fraction. And I factor out the 3/2 and then have:

$$\frac{1}{z-3}$$

So I get: $$\frac{1}{1 - (-(z-2))}$$

but this does not give me what I need.

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1 Answer 1

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$$\dfrac{z}{(z-1)(z-3)} = \dfrac1{(z-1)} + \dfrac3{(z-1)(z-3)}$$ $$\dfrac1{z-3} = \dfrac1{(z-1) -2} = -\dfrac12 \left( \dfrac1{1-(z-1)/2}\right) = -\dfrac12 \left( \sum_{k=0}^{\infty} \left( \dfrac{z-1}2\right)^k\right)$$ Hence, $$\dfrac3{(z-1)(z-3)} = - \dfrac32 \left( \dfrac1{z-1} + \sum_{k=0}^{\infty} \left( \dfrac{(z-1)^k}{2^{k+1}}\right)\right)$$ Hence, \begin{align} \dfrac{z}{(z-1)(z-3)} & = \dfrac1{(z-1)} + \dfrac3{(z-1)(z-3)}\\ & = \dfrac1{z-1} - \dfrac32 \left( \dfrac1{z-1} + \sum_{k=0}^{\infty} \left( \dfrac{(z-1)^k}{2^{k+1}}\right)\right)\\ & = - \dfrac12 \dfrac1{z-1}- 3 \left( \sum_{k=0}^{\infty} \left( \dfrac{(z-1)^k}{2^{k+2}}\right)\right) \end{align}

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Thank you very much. –  Rand Jul 5 '12 at 1:30

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