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I'm trying to work through the following example:

Find the Laurent series of: $$ f(z) = \frac{1}{z(z-2)^3}, $$ about the singularities $z = 0$ and $z = 2$ (separately). Hence verify that $z = 0$ is a pole of order $1$ and $z = 2$ is a pole of order $3$, and find the residue of $f(z)$ at each pole.

I'm attaching the solution below:

To obtain the Laurent series about $z = 0$, we make the factor in parentheses in the

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I'm unsure of what motivates the change of variable etc. (going to $\zeta$, for example) that yields the Laurent series about the correct point. Is there a systematic way to write the Laurent series about specific points/poles/singularities.

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  • $\begingroup$ The point of making the switch frmo $z$ to $\xi$ is because working out a power series in terms of $z-2$ clutters everything and makes it easier to make mistakes and harder to spot patterns. $\endgroup$ – Arthur Jun 26 '16 at 11:05
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I also don't understand why he uses that cumbersome notation with all those minus signs for the simple geometric series...Anyway, around $\;z=2\;$ I'd go:

$$\frac1{z(z-2)^3}=\frac1{(z-2)^3}\cdot\frac1{2+z-2}=\frac1{2(z-2)^3}\cdot\frac1{1+\frac{z-2}2}=$$

$$\frac1{2(z-3)^3}\left(1-\frac{z-2}2+\frac{(z-2)^2}4-\ldots\right)=\frac1{2(z-2)^3}\sum_{n=}^\infty(-1)^n\frac{(z-2)^n}{2^n}$$

and observe the residue is easily obtained:

$$a_{-1}=\frac18$$

The author of the solution given seems to think that building the Laurent series aroud zeor is easier, and he's probably right in many case, and that's why that seemingly odd substitution $\;\xi=z-2\;$, but in this case, and many others, I think that the direct path is the easiest and shortest one.

Observe finally that the step from the first to the second calculations line is justified whenever $\;\left|\frac{z-2}2\right|<1\implies |z-2|<2\;$

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  • $\begingroup$ I still don't get what motivates the particular manipulations that you and the author has made to the function in order to find the Laurent series that has terms with negative integer coefficients to the correct order, hence yielding the residue of the desired singularity. :/ $\endgroup$ – Junaid Aftab Jun 26 '16 at 10:42
  • $\begingroup$ The only manipulation I did is to get the function "ready" to develop it as a power series around $\;z=2\;$. The term $\;\frac1{(z-2)^3}\;$ is already in that form so no need to touch it, but for the other one we had to put $\;\frac1z=\frac1{2\left(1+\frac{z-2}z\right)}\;$ to get a power series around the correct point $\;z=2\;$ . That's all. $\endgroup$ – DonAntonio Jun 26 '16 at 10:54
  • $\begingroup$ What about the $z = 0$ case? $\endgroup$ – Junaid Aftab Jun 26 '16 at 10:57
  • $\begingroup$ @JunaidAftab It is exactly as shown there. $\endgroup$ – DonAntonio Jun 26 '16 at 11:59

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