I'm a bit confused about how to calculate all the laurent series about a given point in the complex plane.

I have the complex function $$ f(z)=\frac{1}{z^2(z-3)}$$ I need to find all the Laurent series about $z=0$.

I understand $f(z)$ has singularities at $z=3$ and $z=0$, but I do not understand if the singularity at $z=0$ is special because of the $z^2$?

I also don't know which regions I should be looking at to determine the Laurent series, and which regions are Taylor series... and does finding all the Laurent series also mean finding the Taylor series...?

I'm pretty lost overall!

Please help :/


2 Answers 2

  • Yes, the singulaty at $z = 0$ is due ot $z^2$

  • Try to draw the regions $0 < |z| < 3$ and $|z| > 3$. And also notice that $\frac{3}{|z|} < 1$ so we may use Taylor series around the point zero.

The series are as follow:

For $0 < |z| <3 $ we have that

$$f(z)=\frac{1}{z^2(z-3)} = \frac{1}{z^2}\frac{1}{(z-3)} = -\frac{1}{3z^2}\frac{1}{(1-\frac{z}{3})} = -\frac{1}{3z^2} \sum_{n=0}^{\infty} \frac{z^n}{3^n} =- \sum_{n=0}^{\infty} \frac{z^{n-2}}{3^{n+1}}$$

For $|z| > 3$ we have that

$$f(z)=\frac{1}{z^2(z-3)} = \frac{1}{z^2}\frac{1}{(z-3)} = \frac{1}{z^3}\frac{1}{(1-\frac{3}{z})} = \frac{1}{z^3} \sum_{n=0}^{\infty} \frac{3^n}{z^n} = \sum_{n=0}^{\infty} \frac{3^n}{z^{n+3}}$$

Any questions leave in the comments.

  • $\begingroup$ So, the first series between 0 and 3 is a Laurent series or a Taylor series, it's a mix of both right? As far as I understand the Laurent series is just the part where the exponents on the z are negative (n=0 and n=1), so the Laurent series is just two terms and the rest is the Taylor series? $\endgroup$
    – Bitmaximus
    Commented Nov 16, 2014 at 19:53
  • 2
    $\begingroup$ Thanks for the edit Pauly, I was wondering what happened to the z^2. $\endgroup$
    – Bitmaximus
    Commented Nov 16, 2014 at 20:09
  • $\begingroup$ @Bitmaximus What you didn't understand? $\endgroup$ Commented Nov 16, 2014 at 20:25
  • $\begingroup$ @Bitmaximus This is how you calculate the Laurent series, over the points of singularity. Have you tried to draw the regions? $\endgroup$ Commented Nov 16, 2014 at 20:38

The $z^2$ simply indicates that the laurent series will have the lowest order term of $z^{-2}$. Note that the Taylor series of a function will be convergent in a disc to the closest singularity, so the function you have does not have a taylor series about $0$. As for finding the laurent series, consider that $$\frac{1}{z^2(z-3)}=-\frac1{3z^2}-\frac1{9z}+\frac{1}{9(z-3)}$$ Now the third term does have a Taylor series about $z=0$. We can find it by $$\frac{1}{9(z-3)}=-\frac{1}{27\left(1-\frac z3\right)}=-\frac 1 {27} \sum_{k=0}^\infty \frac{z^n}{3^n}$$ and by the properties of geometric series, the radius of convergence i s $\left|\frac z3\right|<1\implies|z|<3$. So we have $$\frac{1}{z^2(z-3)}=-\frac1{3z^2}-\frac1{9z}-\frac 1 {27} \sum_{k=0}^\infty \frac{z^n}{3^n}=-\frac1{27}\sum_{k=-2}^\infty\frac{z^k}{3^k}=-\sum_{k=0}^\infty\frac{z^{k-2}}{3^{k+1}}$$ for $|z|<3$.

As for the other region it can converge on ($|z|>3)$, consider that $$\frac{1}{9(z-3)}=\frac{1}{9z}\cdot\frac1{1-\frac3z}=\frac1{9z}\sum_{k=0}^\infty\frac{3^k}{z^k}=\sum_{k=0}^\infty\frac{3^{k-2}}{z^{k+1}}$$ which has a radius of convergence $\left|\frac3z\right|<1\implies |z|>3$. So we have $$\frac{1}{z^2(z-3)}=-\frac1{3z^2}-\frac1{9z}+\sum_{k=0}^\infty\frac{3^{k-2}}{z^{k+1}}=\sum_{k=2}^\infty\frac{3^{k-2}}{z^{k+1}}=\sum_{k=0}^\infty\frac{3^{k}}{z^{k+3}}$$ for $|z|>3$. Since the two annuli above cover the whole complex plane, we have found all the different Laurent series of the function.

  • $\begingroup$ So, the first step was to break the expression up with partial fractions. But then, I don't understand what happens afterwards. I see the Taylor series for the third term but how do the first two terms combine with this to give a Laurent series? It feels like the only term we cared about was the third one, why is this? $\endgroup$
    – Bitmaximus
    Commented Nov 16, 2014 at 20:22
  • $\begingroup$ The entire series is a Laurent series, not just the part with a negative exponent. The combining just happens to work in this case to give a single series; it is very common to leave it as the second expression in the final line for example. As for the final question, it's not that the other terms don't matter per se, but they are already powers of $z$, while the third term is not. So we are not doing anything to the first two terms because they are already in a valid form for the Laurent series! So we are fixing the one that is not $\endgroup$
    – Pauly B
    Commented Nov 16, 2014 at 20:34

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