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I have the following fraction:

$$ \frac{z}{(z-1)(z-2)} $$

When I try to decompose it to partial fractions, I get:

$$ -\frac{1}{z-1} + \frac{2}{z-2} $$

But the result in my book is:

$$ -\frac{z}{z-1} + \frac{z}{z-2} $$

Both results are correct, but, how am I supposed to get the second one?

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    $\begingroup$ The book's result is non-standard, because one tries to get the degree of the numerator less than the degree of the denominator. $\endgroup$ – almagest May 16 '16 at 16:57
  • $\begingroup$ I need it in the second form to perform the inverse Z-transform on it later. $\endgroup$ – Eenoku May 16 '16 at 16:59
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I agree with the comments that the result stated in the back of the book is somewhat non-standard. The way that you can see that the two results are the same is by cleverly adding and subtracting 1.

\begin{align*} \frac{-1}{z-1} + \frac{2}{z-2} &= \frac{-1}{z-1} -1 + \frac{2}{z-2} + 1 \\ &= \frac{-1}{z-1}-\frac{(z-1)}{z-1} + \frac{2}{z-2} + \frac{z-2}{z-2} \\ &= \frac{-z}{z-1} + \frac{z}{z-2} \end{align*}

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

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