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I would like to determine the Laurent series for the function

$$f(z) = \frac{1}{z(z-2)^3}$$

for two different domains

$$\vert{z}\vert<2$$

and

$$\vert{z}\vert>2$$

But, I am unsure when I am computing the Laurent series when I am to take into consideration the different domains. I use standard tricks such as taking out the $\frac{1}{z}$ term and rearranging the remaining term into the sum formula of the geometric series to compute the general Laurent series.

Thanks for all help!

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  • $\begingroup$ Hi, cool question ! Involves some extra stuff rather than the standard tricks ! I left you a complete answer down below, make sure to work yourself around and see how you derive the formulas explained. $\endgroup$ – Rebellos Dec 6 '17 at 1:16
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For $|z|>2$ :

$$f(z) = \frac{1}{z(z-2)^3} = \frac{1}{z[z^3(1-\frac{2}{z})^3]}=\frac{1}{z^4(1-\frac{2}{z})^3}$$

A known geometric series is :

$$\frac{1}{1-w} = \sum_{n=0}^\infty w^n ,\quad |w| < 1$$

From that, we can derive :

$$\frac{1}{(1-w)^3} = \sum_{n=0}^\infty\frac{1}{2}(1+n)(2+n)w^n, \quad |w|<1$$

So, applying that for $w=\frac{2}{z}$, we get :

$$f(z) = \frac{1}{z^4} \sum_{n=0}^\infty\frac{1}{2}(1+n)(2+n)\bigg(\frac{2}{z}\bigg)^n=\sum_{n=0}^\infty2^{n-1}(1+n)(2+n)z^{-(n+4)}$$

for $|\frac{2}{z}| < 1 \Leftrightarrow |z| > 2 . $

For $|z|<2$, we :

$$f(z)=\frac{1}{z(z-2)^3}= - \frac{1}{z(2-z)^3}= - \frac{1}{z}\sum_{n=0}^\infty2^{-(n+4)}(1+n)(2+n)z^n$$

$$=$$

$$-\sum_{n=0}^\infty2^{-(n+4)}(1+n)(2+n)z^{n-1}$$

which hols for $|z|<2$.

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  • $\begingroup$ Thank you very much. I was ignoring when the geometric series converges. Clever variable substitution. $\endgroup$ – Kyle Dec 6 '17 at 9:09
  • $\begingroup$ @Kyle No problem ! The variable substitution is a common trick but finding the more complicated series from the usual geometric ones isn't that simple/common, so it was a nice problem ! $\endgroup$ – Rebellos Dec 6 '17 at 10:47

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