1
$\begingroup$

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 $res(f, -2)$??

$\endgroup$
  • $\begingroup$ Well, what have you tried? $\endgroup$ – Silynn Jul 4 '14 at 1:44
3
$\begingroup$

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 $0<|z+2|<1$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.