# Laurent series of $f(z)=\frac{4z-z^2}{(z^2-4)(z+1)}$ in different annulus

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

• 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. – 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.

• How did you get that $\dfrac{4z-z^2}{(z^2-4)}=-1+\dfrac{1}{z-2}+\dfrac{3}{z+2}$? – Roni Ben Dom Apr 24 at 6:35
• 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 $\frac{-4-4z}{z^2-4}$. – José Carlos Santos Apr 24 at 6:51