1
$\begingroup$

Find the Laurent series expansion of $$f=\frac {2z}{(z-1)(z-3)}$$ at the $1<|z|<3$.

Sol.: My centre is around zero.Ill use the known geometric series around zero.

$$f(z)=\frac {2z}{(z-1)(z-3)}=\frac{-1}{1-z} +\frac{3}{z-3}=\frac{1}{1-z} -\frac{1}{1-\frac{z}{3}}$$ Using $$-\sum_{n=1}^{\infty}\frac{1}{z^n}=\frac{1}{1-z}$$ for $|z| >1$ and $$\sum_{n=0}^{\infty}\frac{z^n}{3^n}=\frac{1}{1-\frac{z}{3}}$$ for $|z|<3$. Is it correct ?.If the centre was something else i would make a substitution in order to use the series around zero again?.Also fractions of the form $\frac{1}{z-centre}$ i don't need to change them since they will be able to get inside the sums?.Any other thing that needs noticing when calculating laurent series this way??

$\endgroup$
2
  • 1
    $\begingroup$ Minor complaint: When you write out your partial fractions for the first time, you have $-1/(1-z)$ instead of $1/(1-z)$. (You do then write it correctly in the very next expression, though.) $\endgroup$
    – Semiclassical
    Commented Aug 24, 2016 at 15:34
  • $\begingroup$ As a check on your result, note that for $z=2$ your Laurent series can be resummed as $-\sum_{n=1}^\infty \frac{1}{2^n}-\sum_{n=0}^\infty \left(\frac{2}{n}\right)^n = -(1)-(3)=-4$ which matches the initial expression. $\endgroup$
    – Semiclassical
    Commented Aug 24, 2016 at 15:40

0

Reset to default

You must log in to answer this question.