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I have a function: $\frac{1+2z}{z^3 + z^2}$ for $0 < |z| < 1$ (about $z=0$)

I need to find the Laurent expansion of this function.

However, I'm a bit confused how to find the partial fractions of this function to begin with.

Can anyone help please? (sorry about poor coding)

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  • $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – Américo Tavares May 5 '14 at 23:43
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If I'm interpreting your post correctly:

$$\frac{1+2z}{z^3+z^2}$$

is the expression you're working with. Hint:

$$\frac{1+2z}{z^3+z^2} = \frac{A}{z}+\frac{B}{z^2}+\frac{C}{z+1}$$

for some $A, B, C \in \mathbb{C}$.

I got $A=B=1,\ C=-1$.

Edit: You seem to be having a lot of trouble with partial fractions here. Here are the steps involved in this one:

$$\begin{align*}\frac{1+2z}{z^2(z+1)} &= \frac{A}{z}+\frac{B}{z^2}+\frac{C}{z+1}\\ \frac{1+2z}{z^2(z+1)} &= \frac{Az(z+1)}{z^2(z+1)}+\frac{B(z+1)}{z^2(z+1)}+\frac{Cz^2}{z^2(z+1)}\\ \Rightarrow 1+2z &= Az^2+Az+Bz+B+Cz^2\\ 1+2z &= B + (A+B)z + (A+C)z^2\end{align*}$$ Comparing coefficients on both sides leads to: $$B=1,\ A+B=2,\ A+C=0$$

You should be able to solve this system with very little trouble.

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  • $\begingroup$ Thanks for your reply. I have that A(z^2)(z+1) + B(z)(z+1) + C(z)(z^2) = 1 + 2z. So letting z=-1 gives -C = -1 and so C=1. Not sure how to get A and B though. $\endgroup$ – user138059 May 5 '14 at 23:57
  • $\begingroup$ I notice that C=1 is also different to what you got, so I'm confused if I'm doing this correctly now. $\endgroup$ – user138059 May 6 '14 at 0:04
  • $\begingroup$ You need only multiply $C$ by $z^2$, not $z(z^2)$. Similarly, you need only multiply $A$ by $z(z+1)$ and $B$ by $(z+1)$. $\endgroup$ – Roger Burt May 6 '14 at 12:34
  • $\begingroup$ I understand this now and have found your solutions, thanks very much for your help! $\endgroup$ – user138059 May 6 '14 at 16:40

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