Given $f(z)=\dfrac{4z-z^2}{(z^2-4)(z+1)}$ I need to find the Laurent series in the annulus: $A_{1,2}(0),\;A_{2,\infty}(0),\;A_{0,1}(-1)$

I found the following partial fractions: $f(z)=\dfrac{-3}{(z+2)}+\dfrac{1}{3(z-2)}+\dfrac{5}{3(z+3)}$,

the power series of these fractions are:

$\dfrac{-3}{(z+2)}=\displaystyle{\frac{-3}{2}\sum_{n=0}^\infty \left( \frac{-z}{2} \right)^n} $

$\dfrac{1}{3(z-2)}=\displaystyle{\frac{-1}{6}\sum_{n=0}^\infty \left( \frac{-z}{2} \right)^n} $

$\dfrac{5}{3(z+1)}=\displaystyle{\frac{5}{3}\sum_{n=0}^\infty \left( -z \right)^n} $

and the principle parts are:

$\dfrac{-3}{(z+2)}=\displaystyle{\frac{-3}{2}\sum_{n=0}^\infty \left( \frac{1}{-2z} \right)^n} $

$\dfrac{1}{3(z-2)}=\displaystyle{\frac{-1}{6}\sum_{n=0}^\infty \left( \frac{1}{-2z} \right)^n} $

$\dfrac{5}{3(z+1)}=\displaystyle{\frac{5}{3}\sum_{n=0}^\infty \left( \frac{1}{-z} \right)^n} $

In the first annuli I take the principle part only of $\dfrac{5}{3(z+1)}$, in the second annuli I take the principle part of all fraction. About the third one, I have $0<\vert z-1\vert<1$, I denoted $w=z-1$ and then I took the power series for all fractions and simply switched the $w$ back to $z-1$ at the end. Is it the right way of doing it?

I received $\displaystyle{\frac{-3}{2}\sum_{n=0}^\infty \left( \frac{1-z}{2} \right)^n - \frac{1}{6}\sum_{n=0}^\infty \left( \frac{1-z}{2} \right)^n + \frac{5}{3}\sum_{n=0}^\infty \left( 1-z \right)^n}$

  • $\begingroup$ For the 3rd, should you consider the power series consisting of the terms $(z+1)^n$ instead of $(z-1)^n$ because the annuli is centered at $-1$ not $+1$? FYI, lots of typos there. $\endgroup$ – xbh Apr 24 at 6:03

There's a typo in your partial fraction decomposition; it should be$$\frac{4z-z^2}{(z^2-4)(z+1)}=-\frac3{z+2}+\frac1{3(z-2)}+\frac5{3(z+1)}.$$If $z\in A_{1,2}(0)$, then:

  • $\displaystyle-\frac3{z+2}=-\frac32\frac1{1+\frac z2}=-\frac32\sum_{n=0}^\infty\frac{(-1)^n}{2^n}z^n$;
  • $\displaystyle\frac1{3(z-2)}=-\frac16\frac1{1-\frac z2}=-\frac16\sum_{n=0}^\infty\frac{z^n}{2^n}$;
  • $\displaystyle\frac5{3(z+1)}=\frac53\frac1{1+z}=-\frac53\sum_{n=-\infty}^{-1}(-1)^nz^n$.

In $A_{2,\infty}(0)$, the third expansion remains the same, but now we have:

  • $\displaystyle-\frac3{z+2}=-\frac32\frac1{1+\frac z2}=\frac32\sum_{n=-\infty}^{-1}\frac{(-1)^n}{2^n}z^n$;
  • $\displaystyle\frac1{3(z-2)}=-\frac16\frac1{1-\frac z2}=\frac16\sum_{n=-\infty}^{-1}\frac{z^n}{2^n}$.

In the case of $A_{0,1}(-1)$, you write\begin{align}\frac{4z-z^2}{z^2-4}&=-1+\frac1{z-2}+\frac3{z+2}\\&=1+\frac1{(z+1)-3}+\frac3{(z+1)+1}.\end{align}From this, you get the Taylor series of $\dfrac{4z-z^2}{z^2-4}$ in $D(1,1)$ and then, when you divide everything by $z+1$, you get the Laurent series that you're after.

  • $\begingroup$ How did you get that $\dfrac{4z-z^2}{(z^2-4)}=-1+\dfrac{1}{z-2}+\dfrac{3}{z+2}$? $\endgroup$ – Roni Ben Dom Apr 24 at 6:35
  • $\begingroup$ We have$$\frac{4z-z^2}{z^2-4}=\frac{4-z^2-4-4z}{z^2-4}=-1+\frac{-4-4z}{z^2-4}.$$Then I applied partial fraction decomposition to $\frac{-4-4z}{z^2-4}$. $\endgroup$ – José Carlos Santos Apr 24 at 6:51

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