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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? $\endgroup$ – Greg Martin Jan 18 '16 at 6:30
  • $\begingroup$ @gregmartin:can you find the expansion at thegiven point?? $\endgroup$ – Abhishek Shrivastava Jan 18 '16 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 '16 at 8:49
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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 $\endgroup$ – Abhishek Shrivastava Jan 18 '16 at 6:51
  • $\begingroup$ $a_j$ is the coefficient in Laurent series expansion where $j=-k, -k+1, ...$ $\endgroup$ – Meow Jan 18 '16 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 $\endgroup$ – Abhishek Shrivastava Jan 18 '16 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 $\endgroup$ – Abhishek Shrivastava Jan 18 '16 at 7:01
  • $\begingroup$ yes it is true. in some books this is used as a definition of pole. $\endgroup$ – Meow Jan 18 '16 at 7:01

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