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Determining Laurent Series expansion and residues of $f(z)=\frac{z}{(z+1)(z+2)}$ around $z = -2$. What is the validity of the expanded region? What is $res(f, -2)$??

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Well, what have you tried? – Silynn Jul 4 '14 at 1:44
up vote 3 down vote accepted

This is simple pole (because $z$ and $z+1$ do not have zeros at $z=-2$ and $z+2$ have zero of order 1), so Laurent series principal part converge everywhere.

Hence residue is $\lim\limits_{z\rightarrow -2}(z+2)\frac{z}{(z+1)(z+2)}=2$. So principal part is $\frac{2}{z+2}$.

Take that part off give you $\frac{z}{(z+1)(z+2)}-\frac{2}{z+2}=-\frac{1}{z+1}$.

A simple way to find Taylor series of this is to "shift" the whole thing so that instead of $z=-2$ we get $z=0$. In other word, we find Taylor series expansion of $-\frac{1}{z-1}=\frac{1}{1-z}$ around $0$. This, as it turn out, is a geometric series, so it's just $\sum\limits_{n=0}^{\infty}z^{n}$ with convergence in $|z|<1$. Now shift it back to get $-\frac{1}{z+1}=\sum\limits_{n=0}^{\infty}(z+2)^{n}$.

So the whole series is $\frac{2}{z+2}+\sum\limits_{n=0}^{\infty}(z+2)^{n}$ and converge at $0<|z+2|<1$.

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Hint: $$f(z)=\frac{(z+2)-2}{(-1)(1-(z+2))(z+2)}=\frac{2}{(z+2)(1-(z+2))}-\frac{1}{(1-(z+2))}$$

Then just apply the expansion for a geometric series.

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A related technique. Here is how you advance.

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

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

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

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

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

Now you need to expand $(1-1/(z+2))^{-1}$ using the geometric series. For residues see here.

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You're actually making it way harder than it needs to be. – Silynn Jul 4 '14 at 1:55

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