# Classify the isolated singularities of $\frac{z^5}{z^3 + z}$ and $\frac{\cos z}{z^2 -1}$

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

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.
• 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? – NazimJ Jun 19 '19 at 12:33
• 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 $\frac z{\sin z}$. – José Carlos Santos Jun 19 '19 at 12:36
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}
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.