I have to find all of the Laurent series for the function

$$ f(z)=\frac{z}{(z+1)(z-2)} $$ about $z=0$

I'm a little confused about the regions that I'm dealing with. I started with the partial fraction decomposition: $$ f(z)=\frac{1}{3(z+1)}+\frac{2}{3(z-2)} $$

Coming out from $z=0$ there's a Taylor series up until $z=-1$ that I guess works for both singularities? i.e. for $|z|<1$ the Laurent series is: $$\frac{1}{3}\sum_{n=0}^{\infty}((-1)^n-(\frac{1}{2})^n)z^n$$

Is the above right? If not how do you go about finding the rest of the Laurent series for $f(z)$? What I really need to understand is how you need to look at the regions because I'm finding it really confusing for this problem.

Thank you!

  • 3
    $\begingroup$ You have singularities at $-1$ and at $2$, so on the circles with radius $1$ resp. $2$ (and centre $0$). Thus the three Laurent series are for 1. $\lvert z\rvert < 1$, 2. $1 < \lvert z\rvert < 2$, 3. $\lvert z\rvert > 2$. $\endgroup$ Nov 29, 2014 at 19:47
  • $\begingroup$ Thank you, this is what I thought. Can you help me understand what the components are for each region? $\endgroup$
    – Bitmaximus
    Nov 29, 2014 at 20:08
  • $\begingroup$ You have the partial fraction decomposition. So look at each part separately. $\frac{1}{z - a}$ is holomorphic on the disk $\lvert z\rvert < a$, so you Taylor-expand it there. On $\lvert z\rvert > a$ you expand it into powers of $\frac{1}{z}$. Combine the expansions accordingly. You get three out of the for possible combinations. $\endgroup$ Nov 29, 2014 at 20:13

1 Answer 1


As @DanielFischer told you, the regions to consider are $(1)$ $\lvert z\rvert<1$, $(2)$ $1<\lvert z\rvert < 2$, and $(3)$ $\lvert z\rvert >2$. For $(1)$, your solution is correct.

For $(2)$, we want Geometric series with $\lvert r\rvert <1$ so $1<\lvert z\rvert\Rightarrow \frac{1}{\lvert z\rvert} < 1$ and $\frac{\lvert z\rvert}{2}<1$. Then \begin{align} \frac{1/3}{z+1}+\frac{2/3}{z-2} &=\frac{1}{z}\frac{1/3}{1+\frac{1}{z}}-\frac{1/3}{1-\frac{z}{2}}\\ &=\frac{1}{3z}\sum_{n=0}^{\infty}(-1)^n\Bigl(\frac{1}{z}\Bigr)^n-\frac{1}{3}\sum_{n=0}^{\infty}\Bigl(\frac{z}{2}\Bigr)^n\\ &=\frac{1}{3}\sum_{n=0}^{\infty}\biggl[(-1)^n\Bigl(\frac{1}{z}\Bigr)^{n+1}-\Bigl(\frac{z}{2}\Bigr)^n\biggr] \end{align}

For $(3)$, we have $\lvert z\rvert >2$ so $\frac{1}{\lvert z\rvert} < 1$ which we found in $(2)$ and $\frac{2}{\lvert z\rvert}<1$. We only need to determine the Laurent series for $\frac{2}{\lvert z\rvert}<1$. \begin{align} \frac{1}{3z}\sum_{n=0}^{\infty}(-1)^n\Bigl(\frac{1}{z}\Bigr)^n+\frac{2/3}{z-2}&=\frac{1}{3z}\sum_{n=0}^{\infty}(-1)^n\Bigl(\frac{1}{z}\Bigr)^n+\frac{1}{z}\frac{2/3}{1-\frac{2}{z}}\\ &=\frac{1}{3z}\sum_{n=0}^{\infty}(-1)^n\Bigl(\frac{1}{z}\Bigr)^n+\frac{2}{3z}\sum_{n=0}^{\infty}(-1)^n\Bigl(\frac{2}{z}\Bigr)^n\\ &=\frac{1}{3}\sum_{n=0}^{\infty}(-1)^n\biggl[\Bigl(\frac{1}{z}\Bigr)^{n+1}+\Bigl(\frac{2}{z}\Bigr)^{n+1}\biggr] \end{align}


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