what type of singularity f has and what is the residue of f

**sorv1986** · December 6th 2011, 12:05 AM

Suppose f(z)=sin(1/cos(1/z))be a complex function.

Then what type of singularity f has at z=0 and what will be the residue of f at z=0?

thanks. regards

**Jose27** · December 6th 2011, 10:27 AM

Originally Posted by sorv1986

Suppose f(z)=sin(1/cos(1/z))be a complex function.

Then what type of singularity f has at z=0 and what will be the residue of f at z=0?

thanks. regards

Where did you get this question from? It seems like it isn't even "well-posed": $f$ has a bunch of essential singularities at the points $\left( \pm \frac{\pi}{2} \pm k\pi\right)^{-1}$ for all $k\in \mathbb{N}$ , and these singularities accumulate at $0$ , so for starters $0$ is not even an isolated singularity, and every closed curve that contains the origin in its interior will contain an infinite number of other singularities.

**chisigma** · December 7th 2011, 12:57 AM

The function has in z=0 an essential singularity. Setting $\frac{1}{z}=s$ the function becomes...

$f(s)=\sin \frac{1}{\cos s}$ (1)

... and the residue in z=0 is the term in s of its Taylor expansion, that is...

$\lim_{s \rightarrow 0} f'(s)$ (2)

The computation of the derivative is left to You and the result is that the limit (2) is zero...

Kind regards

$\chi$ $\sigma$

**sorv1986** · December 7th 2011, 09:05 PM

Originally Posted by chisigma

The function has in z=0 an essential singularity. Setting $\frac{1}{z}=s$ the function becomes...

$f(s)=\sin \frac{1}{\cos s}$ (1)

... and the residue in z=0 is the term in s of its Taylor expansion, that is...

$\lim_{s \rightarrow 0} f'(s)$ (2)

The computation of the derivative is left to You and the result is that the limit (2) is zero...

Kind regards

$\chi$ $\sigma$

i am a little bit confused. Why r u calculating the limit (2). Besides this, if s-> 0, then z->∞. You are calculating the residue of f(z) at z=∞ if i am right.How did you get z=0 is an essential singularity? sorry if i misunderstood. Thanks for kind reply.

**sorv1986** · December 7th 2011, 09:07 PM

Originally Posted by Jose27

Where did you get this question from? It seems like it isn't even "well-posed": $f$ has a bunch of essential singularities at the points $\left( \pm \frac{\pi}{2} \pm k\pi\right)^{-1}$ for all $k\in \mathbb{N}$ , and these singularities accumulate at $0$ , so for starters $0$ is not even an isolated singularity, and every closed curve that contains the origin in its interior will contain an infinite number of other singularities.

i got it from an exam paper. thanks for your kind reply.appreciated .

**chisigma** · December 8th 2011, 02:08 AM

Originally Posted by sorv1986...

... I am a little bit confused. Why r u calculating the limit (2). Besides this, if s-> 0, then z->∞. You are calculating the residue of f(z) at z=∞ if i am right.How did you get z=0 is an essential singularity?... sorry if i misunderstood...

I do hope that the following example will clarify the procedure to obtain an easy solution of Your question. Let's consider the function...

$e^{\frac{1}{z}}= 1 + \frac{1}{z} + \frac{1}{2\ z^{2}} + ... + \frac{1}{n!\ z^{n}} + ...$ (1)

The (1) is the Laurent expansion of the function $e^{\frac{1}{z}}$ around z=0 and, because the infinite number of term in $z^{-n}$ , it has in z=0 an essential singularity. By definition the residue in z=0 is the coefficient of the tem in $z^{-1}$ of the Laurent expansion and in this case in (1) You see that is r=1. Setting $z^{-1}=s$ the (1) becomes...

$e^{s}= 1 + s + \frac{s^{2}}{2} + ... + \frac{s^{n}}{n!}} + ...$ (2)

Now the (2) is the Taylor expansion of the analytic function $e^{s}$ around s=0 and, of course, the coefficient of the term in $z^{-1}$ in (1) and the coefficient of the term in s in (2) are the same. But the coefficient of the Taylor expansion of an analytic function f(s) around s=0 is f'(0) and that justifies the formula...

$r_{z=0} f(\frac{1}{z}) = \lim_{s \rightarrow 0} f^{'} (s)$ (3)

Kind regards

$\chi$ $\sigma$

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