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I'm kind of stumped on a question here.

I've been asked to determine and classify the singularities of;

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

To me, it's pretty obvious that a singularity will occur when $z = -1$, however, I'm now having trouble being able to classify this, because I'm not sure how to go about determining my expansion.

I was given the hint in class to essentially "add zero", to make the function look more like a Laurent series, but I'm really not confident with what I'm meant to be looking for here.

Any help would be fantastic.

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  • $\begingroup$ Just by looking one can immediately tell that $-1$ is a pole of order $3$ because $\lim \limits_{z\to -1}\left((z+1)^3\dfrac{z^3}{(1+z)^3}\right)\in \mathbb C\setminus \{0\}$ and $\left|\lim \limits_{z\to -1}\left((z+1)^2\dfrac{z^3}{(1+z)^3}\right)\right|=\infty$. $\endgroup$ – Git Gud Jun 1 '14 at 10:46
  • $\begingroup$ If you want to find the laurent series around $-1$, just find the laurent series of $z\mapsto z^3$ around $-1$ and multiply by the obvious factor. $\endgroup$ – Git Gud Jun 1 '14 at 10:53
  • $\begingroup$ You can obtain a Laurent series expansion of $f(z)$ around $z=-1$ by re-writing it in the form $$ f(z) = \frac{(1+z-1)^3}{(1+z)^3} = \frac{(1+z)^3-3(1+z)^2+3(1+z)-1}{(1+z)^3} = 1 -\frac{3}{1+z} + \frac{3}{(1+z)^2} - \frac{1}{(1+z)^3},$$ but this is not necessary to determine that $z=-1$ is a triple pole of $f(z)$. You only need to observe that $f(z)$ is represented as a quotient of a holomorphic function, $z^3$, which does not vanish at $z=-1$, and another holomorphic function, $(1+z)^3$, which has a zero of multiplicity $3$ at $z=-1$ (This is nothing but what Git Gud has already suggested). $\endgroup$ – ivanpenev Jun 1 '14 at 11:02
  • $\begingroup$ Awesome!! Thank you both for the help!! In the classification of the residue, would it be correct to say that $z = -1$ is an isolated singularity, allowing me to calculate the residue as follows?? $$res_{z \rightarrow z_0} f(z) = \frac{ \phi^{(m-1)}(z_0)}{(m-1)!}$$ With $f(z) = \frac{\phi(z)}{(1+z)^3}$ and $\phi(z) = z^3$, which gives me an answer of $-3$ ?? (($m$ being the order of the pole)) $\endgroup$ – Jack Jun 1 '14 at 11:42
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    $\begingroup$ @dustin Reluctantly did so. In the end I concluded that I must take responsibility for (arguably) having killed the question in the comments. $\endgroup$ – Git Gud Jan 11 '15 at 19:14
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Note that for all complex numbers $z$ with $z\neq -1$, $\dfrac 1{(z+1)^3}=\dfrac 1{(z-(-1))^3}$ so if you find the laurent series (which will just be the taylor series) of $z\mapsto z^3$ around $-1$, multiplying the series by $\dfrac 1{(z-(-1))^3}$ will yield the laurent series of $f$ at $-1$.

This is what the hint is getting at, more specifically, add $0$ in the following way: $$z^3=((z+1)-1)^3=(z+1)^3-3(z+1)^2+3(z+1)-1.$$

One identifies the RHS as the taylor series of $z\mapsto z^3$ around $-1$ and so the laurent series of $f$ around $-1$ is $$-\dfrac 1{(z+1)^3}+\dfrac 3{(z+1)^2}-\dfrac 3{z+1}+1.$$

However this is all unnecessary to answer the question. One can immediately see that $$\lim \limits_{z\to -1}\left((z+1)^3\dfrac{z^3}{(1+z)^3}\right)\in \mathbb C\setminus \{0\}$$ and $$\left|\lim \limits_{z\to -1}\left((z+1)^2\dfrac{z^3}{(1+z)^3}\right)\right|=\infty.$$

Thus $-1$ is a pole of order $3$ of $f$. It is $f$'s only singularity.

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  • $\begingroup$ You can now put the link in Crusade for Answers!! since it is old and answered now. $\endgroup$ – dustin Jan 11 '15 at 19:15
  • $\begingroup$ @dustin I know about the existence of that group, but I don't know how it operates. What is the purpose of posting the link there? $\endgroup$ – Git Gud Jan 11 '15 at 19:16
  • $\begingroup$ So other "crusaders" are aware of another post being being taken out of unanswered purgatory. Generally, they will show their thanks by an upvote as well. $\endgroup$ – dustin Jan 11 '15 at 19:17

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