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How do I

(a) find the Laurent series for the following: $$ f(z)=\frac {z^2+z+1}{(z-1)^2}$$ (b) Find its pole and its order.

I suppose finding the Laurent series would make it easy to find the latter, but I think there's a short cut to finding the pole? Anyhow, I'm interested in both part (a) and part (b) above.

The only way I know to solve for part (a) is to use partial fractions, but in such a case I would still have a $(z-1)^2$ in one of the denominators since the above would split up as

$$f(z)=(z^2+z+1)\left[\frac{A}{z-1} +\frac{B}{(z-1)^2}\right]$$

Not sure how to proceed.

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hint: expand the numerator in powers of (z-1) – Raymond Manzoni Jul 18 '12 at 6:33
up vote 3 down vote accepted

$$ f(z)=\frac {z^2+z+1}{(z-1)^2} =\frac {(z-1)^2 + 3z}{(z-1)^2} = 1+ \frac {3z}{(z-1)^2} = 1+ \frac {3}{z-1}+\frac {3}{(z-1)^2}$$

From this onwards, you can find your solutions what you wanted. I think it will be helpful to you.

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My problem is with the $(z-1)^2$term in the denominator. I know how to write $(z-1)$ as a power series, but as far as I know $(z-1)^2$ would be the square of this power series, and I'm not quite sure how to deal with such a situation. – Joebevo Jul 18 '12 at 7:14
@Joebevo: an expression like $a+\frac b{(z-z_0)} + \frac c{(z-z_0)^2}$ is a Laurent series. A Laurent series doesn't need to have $z_0=0$ (it's perhaps your 'block'). – Raymond Manzoni Jul 18 '12 at 8:09
Ah..I see that now. I've been dealing with a bunch of infinite series so it didn't occur to me that it could be finite. – Joebevo Jul 18 '12 at 8:13


If you look at the last term provided by @Prasad G in the following way:

$$ \sum_{n=-2}^0 a_n (z-1)^n $$

with $a_{-2}=3$, $a_{-1} = 3$ and $a_0 =1$.

Then i am sure you will be close to finding your desired Laurent series ;-).

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