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Find out the points of singularities of the following function and classify them:

$f(z)=tan(\frac{1}{z}).$

my idea here is the following:

the singular points are $z=\frac{2}{(2n+1)\pi}$ (isolated).

also,z$\to$0 as n$\to$$\infty$

so, 0 is a point of non isolated essential singularity.

now, for the isolated singularities z=$\frac{2}{(2n+1)\pi}$, we can always find the Laurent series and classify the singularities. but, I am completely stuck in finding the laurent seires of following function:

f(z)=tan($\frac{1}{z}$) at say,z=$\frac{2}{3\pi}$

my idea was to put z-$\frac{2}{3\pi}$=t and find out the expansion of tan($\frac{1}{t+\frac{2}{3\pi}}$) at t=0 and then i am completely stuck.is this method coreect ??if not,suggest alternatives.

any help would be appreciated....

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  • $\begingroup$ Your method is correct. How does one find Laurent series in general?—taking derivatives, perhaps? $\endgroup$ Jan 18, 2016 at 6:30
  • $\begingroup$ @gregmartin:can you find the expansion at thegiven point?? $\endgroup$ Jan 18, 2016 at 6:40
  • $\begingroup$ the first step when searching for a Laurent series is to find if the coefficients are non zero when $n \to \infty$ and/or when $n \to -\infty$, and to notice that it is nothing more than a Fourier series $\endgroup$
    – reuns
    Jan 18, 2016 at 8:49

1 Answer 1

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For classification purpose your method is correct but alternatively one can do the following,

If $P$ is isolated singular point and $\lim_{z \to P}|f(z)|=+\infty $ then $f$ has pole at $P$. Further here in your example look at $g(z)=1/f(z)$ in a nbd around $z=2/3 \pi$ say, then $g$ has simple zero at $z=2/3 \pi$ so $f$ will have simple pole at that point.

Further if you need to find Laurent series expansion, you can use the following formula to find the Laurent series coefficients $a_j$ of $f$ expanded about point $P$, for $j=-k, -k+1, ...$

$a_j=\dfrac{1}{(k+j)!}\bigg(\dfrac{\partial}{\partial z} \bigg)^{k+j} ((z-P)^k.f)\bigg |_{z=P}$ where $k$ is order of pole and

In your example $k=1$.

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  • $\begingroup$ :how do we get the $a_j$ you described $\endgroup$ Jan 18, 2016 at 6:51
  • $\begingroup$ $a_j$ is the coefficient in Laurent series expansion where $j=-k, -k+1, ...$ $\endgroup$
    – Meow
    Jan 18, 2016 at 6:53
  • $\begingroup$ :is it true that if |f(z)|$\to$$\infty$ as z$\to$$z_0$ (where $z_0$ is isolated singularity),then,z=$z_0$ is a pole??because i know that this condition is true if we are given that z=$z_0$ is a pole $\endgroup$ Jan 18, 2016 at 6:56
  • $\begingroup$ :i am trying to find laurent series to show that at the given point we obtain the pole.the formula that you are giving gives the laurent series after knowing the orer of pole $\endgroup$ Jan 18, 2016 at 7:01
  • $\begingroup$ yes it is true. in some books this is used as a definition of pole. $\endgroup$
    – Meow
    Jan 18, 2016 at 7:01

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