1
$\begingroup$

Find laurent series for $f(z)=\dfrac{1}{z^2-1}+\dfrac{1}{z(z-1)};z_0=0$ that converges in $0<|z|<1$.

I tried to find the solution for the first fraction like this.

\begin{align} \frac{1}{z^2-1}=\frac{1}{(z-1)(z+1)}&=-\frac{1}{2(z+1)}+\frac{1}{2(z-1)} \\ \end{align} The second term is $-\dfrac 12\dfrac{1}{1-z}$ and $|z|<1$ on $0<|z|<1$. So its $-\dfrac 12 \sum\limits_{k=0}^{\infty}z^k$.

Second fraction $$\dfrac{1}{z(z-1)}=\dfrac{1}{z-1}-\dfrac{1}{z}=-\dfrac 1z-\sum\limits_{k=0}^{\infty}z^k$$

all together $$-\dfrac 1z-\underbrace{\sum\limits_{k=0}^{\infty}z^k-\dfrac 12 \sum\limits_{k=0}^{\infty}z^k}_{-\frac 32 \sum\limits_{k=0}^{\infty}z^k } -\dfrac 12\dfrac{1}{1+z}$$

Am I on the right track? What tricks are needed for term $-\dfrac 12\dfrac{1}{1+z}$?

EDIT: I had an idea... $$-\dfrac 12\dfrac{1}{1+z}= -\dfrac 12\dfrac{1}{1-(-z)}=-\frac 12 \sum \limits_{k=0}^{\infty}(-1)^nz^k$$ so the series is...(maybe) $$-\dfrac 1z-\dfrac 32 \sum\limits_{k=0}^{\infty}z^k -\frac 12 \sum \limits_{k=0}^{\infty}(-1)^nz^k$$

$\endgroup$
1
$\begingroup$

Powers of $z$, right? One of them being a constant times $z^{-1}$, right? Just use geometric series: $(z-1)^{-1}=-1/(1-z)=-\sum_{n\ge0}z^n$, so that the troublesome part is $-\sum_{n\ge-1}z^n$, and the easy part is $-1/(1-z^2)=-\sum_{n\ge0}z^{2n}$. Add them up and I think I’ve gotten the same thing you did.

$\endgroup$

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.