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I'm trying to compute the Laurent series of $f(z) = 1/z^2$ about the point $z_0=1$. Looking at my notes, it appears that I need to compute a series for $|z-1| < 1$ and one for $|z-1|>1$, due to the singularity at the point $1$.

Could someone show me how to compute each of these series? I'm a bit confused on where to begin.

EDIT: The problem says to "Write all Laurent series of the following functions on annuli centered at $z_0$," so I feel like there should be two series: one valid for $|z-1|<1$ and another valid for $|z-1|>1$.

user169067
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  • Note $f$ is analytic at $z_0=1$, so you just need to compute the taylor series; which will only converge in radius 1. – Clinton Bradford Aug 11 '14 at 02:40
  • I understand that's what I need to do for the series valid for $|z-1|<1$, but what about the other series, namely $|z-1|>1$? – user169067 Aug 11 '14 at 02:41
  • Of the following functionS, so I believe that your exercise was asking you to find the Laurent series of many functions, with respect to the point $x_0=1$. – Jack D'Aurizio Aug 11 '14 at 02:55
  • This is one part of a multipart question... – user169067 Aug 11 '14 at 02:56
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    I think you are a bit confused about the definition of a Laurent series. – Jack D'Aurizio Aug 11 '14 at 03:08
  • Here's my confusion...I've seen several examples where multiple Laurent series are computed for a function, based on the location of the singularities. Since $z=0$ is a singularity here, shouldn't we also compute a Laurent series on the annulus $|z-1|>1$ that vanishes at infinity? – user169067 Aug 11 '14 at 03:12
  • @JackD'Aurizio What I think OP means is that (for example) if $f(z)= 1/(z-1)$ at $z_0 = 0$ we have $\frac{1}{z-1} = -\sum_{n=0}^\infty z^n$ for $|z|<1$ and $\frac{1}{z-1} = \sum_{n=1}^\infty \frac{1}{z^n}$ for $|z|>1$, i.e. two series defined in two different annuli. – Winther Aug 11 '14 at 08:04

2 Answers2

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$z_0=1$ is not a singularity for $f(z)=\frac{1}{z^2}$, hence the Laurent series is just a Taylor series, and since: $$\frac{1}{(1-z)^2} = 1+2z+3z^2+\ldots = \sum_{j=0}^{+\infty}(j+1)z^j$$ around the origin, you have: $$\frac{1}{z^2}=\sum_{j=0}^{+\infty}(-1)^j(j+1)(z-1)^j$$ around one. You can check that the radius of convergence, one, is exactly the distance from the closest singularity, the origin.

Jack D'Aurizio
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The result you are after can be found using the same method as in the answer above.

For $|z| > 1$ we have

$$\frac{1}{(1-z)^2} = \frac{1}{z^2}\frac{1}{(1-1/z)^2} = \sum_{j=0}^\infty (j+1)\left(\frac{1}{z}\right)^{j+2}$$

so

$$\frac{1}{z^2} = \sum_{j=0}^\infty (-1)^j(j+1)\left(\frac{1}{z-1}\right)^{j+2}$$

for $|z-1| > 1$

Winther
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