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I am learning complex analysis, and still have trouble finding the "tricks" to deal with tricky denominators.. The question is:

Find and classify the isolated singularities of $$ (a) \qquad \frac{z^5}{z^3 + z} $$ $$ (b) \qquad \frac{\cos z}{z^2 -1} $$

I know I have to identify the isolated singularities, and then find the Laurent series expansion at each of the singularities and classify it as removable, a pole, or essential. But my problem is with all questions, is to actually find the series expansion... When it's centered at 0 it's usually easy, but otherwise I always have trouble

For part (a), $0$ is a singularity and it is removeable by factoring the denominator. Then we are left with finding the laurent expansion of $\frac{z^4}{z^2 + 1} $ centered at $i $ and $-i $. But finding the Laurent series expansion is not clear to me

For part (b), $1$ and $-1$ are singularities. The following is my analysis for the singularity $1$ which I think is a pole of order 1, but I don't think it's a laurent series expansion because of the extra term outside:

$$ \frac{\cos z}{z^2 -1} = \frac{1}{(z+1)(z-1)} \left( 1 - \frac{\cos^2 (1)}{2!} (z-1)^2 + \frac{\cos^4 (1)}{4!} (z-1)^4 - ... \right) = \frac{1}{(z+1)} \left( (z-1)^{-1} - \frac{\cos^2 (1)}{2!} (z-1)^1 + \frac{\cos^4 (1)}{4!} (z-1)^3 - ... \right) $$

Any hints or answers are greatly appreciated

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You don't need to find the Lauren expansion, at least not in these cases. In the first example,$$\lim_{z\to i}(z-i)\frac{z^4}{z^2+1}=\lim_{z\to i}\frac{z^4}{(z-i)(z-i)}=\frac{i^4}{2i}=-\frac i2\neq0$$and, since this limit exists and it is not $0$, $\frac{z^5}{z^3+z}$ has a simple pole at $i$. By the same argument, it has a simple pole at $-i$. And, again by the same argument, the other function also has simple poles.

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  • $\begingroup$ This is very interesting, never seen this before. So for $z_0$ to be a simple pole of $f (z) $, it means that $(z-z_0) f (z)$ has a non-zero constant term, and all higher powers of the Laurent expansion centered at $z_0$ are 0. So one question is, can we just plug in the singularity? Or do we have to take the limit as $z $ approaches the singularity? $\endgroup$ – NazimJ Jun 19 '19 at 12:33
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    $\begingroup$ You can't always plug in the singularity. Take $\frac1{\sin z}$, for instance, which has a singularity at $0$ (and at other places). You have to compute the limit $\lim_{z\to 0}\frac z{\sin z}$; it makes no sense to talk about pluging $0$ in $\frac z{\sin z}$. $\endgroup$ – José Carlos Santos Jun 19 '19 at 12:36
  • $\begingroup$ Please, check the first display. $\endgroup$ – egreg Jun 19 '19 at 12:42
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If you find it easier to take a Laurent expansion around $0$, do a linear change of variables to make the centre of expansion $0$, then change back to the original variable once you have the expansion. Thus to expand $z^4/(z^2+1)$ around $z=i$, let $s = z - i$ and write $$ \eqalign{\frac{z^4}{z^2+1} &= \frac{(s+i)^4}{(s+i)^2+1} = \frac{s^4 + 4 i s^3 - 6 s^2 - 4 i s + 1}{s^2 + 2 s i}\cr & = \frac{s^4 + 4 i s^3 - 6 s^2 - 4 i s + 1}{s}\left( -\frac{i}{2} + \frac{s}{4} + \frac{i}{8} s^2 + \ldots\right) \cr &= - \frac{i}{2} s^{-1} - \frac{7}{4} + \frac{17i}{8} s + \ldots\cr &= - \frac{i}{2} (z-i)^{-1} - \frac{7}{4} + \frac{17i}{8} (z-i) + \ldots\cr}$$

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  • $\begingroup$ Yes! I was thinking of this but I didn't want to try doing it in case it wasn't legal for whatever reason $\endgroup$ – NazimJ Jun 19 '19 at 12:38
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For Laurent series expansion the idea is the same for both functions and here is a hint: for $a \neq b$ we have $\frac 1 {z-b} =\frac 1 {(z-a)+(a-b)}=\frac 1 {a-b} \frac 1 {1+\frac {z-a} {a-b}}=\frac 1 {a-b}(1+\frac {z-a} {a-b}+(\frac {z-a} {a-b})^{2}+...)$ valid if $|z-a| <|b-a|$. In both case you will have to multiply two series.

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