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 $|z-1|>1$.

  • $\begingroup$ Note $f$ is analytic at $z_0=1$, so you just need to compute the taylor series; which will only converge in radius 1. $\endgroup$ – Clinton Bradford Aug 11 '14 at 2:40
  • $\begingroup$ I understand that's what I need to do for the series valid for $|z-1|<1$, but what about the other series, namely $|z-1|>1$? $\endgroup$ – user169067 Aug 11 '14 at 2:41
  • $\begingroup$ 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 $x_0=1$. $\endgroup$ – Jack D'Aurizio Aug 11 '14 at 2:55
  • $\begingroup$ This is one part of a multipart question... $\endgroup$ – user169067 Aug 11 '14 at 2:56
  • 2
    $\begingroup$ I think you are a bit confused about the definition of a Laurent series. $\endgroup$ – Jack D'Aurizio Aug 11 '14 at 3:08

$z_0=1$ 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.

  • $\begingroup$ But shouldn't there be a second Laurent series, valid for $|z-1|>1$? The problem statement says to compute ALL of the Laurent series. $\endgroup$ – user169067 Aug 11 '14 at 2:45
  • $\begingroup$ There is just ONE series, and it is a Taylor series, since the function $f(z)=\frac{1}{z^2}$ is holomorphic in a neighbourhood of $z=1$. $\endgroup$ – Jack D'Aurizio Aug 11 '14 at 2:46
  • $\begingroup$ Why isn't there a series for values beyond the singularity at $0$? $\endgroup$ – user169067 Aug 11 '14 at 2:48
  • $\begingroup$ ...because the problem statement says "on all annuli centered at $z_0$" $\endgroup$ – user169067 Aug 11 '14 at 2:49
  • $\begingroup$ Please see my edit $\endgroup$ – user169067 Aug 11 '14 at 2:53

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

For $|z| > 1$ we have

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


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

for $|z-1| > 1$


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