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?

(Bow)thanks. regards

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- Dec 6th 2011, 12:05 AMsorv1986what type of singularity f has and what is the residue of f
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?

(Bow)thanks. regards - Dec 6th 2011, 10:27 AMJose27Re: what type of singularity f has and what is the residue of f
Where did you get this question from? It seems like it isn't even "well-posed": $\displaystyle f$ has a bunch of essential singularities at the points $\displaystyle \left( \pm \frac{\pi}{2} \pm k\pi\right)^{-1}$ for all $\displaystyle k\in \mathbb{N}$, and these singularities accumulate at $\displaystyle 0$, so for starters $\displaystyle 0$ 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.

- Dec 7th 2011, 12:57 AMchisigmaRe: what type of singularity f has and what is the residue of f
The function has in z=0 an essential singularity. Setting $\displaystyle \frac{1}{z}=s$ the function becomes...

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Dec 7th 2011, 09:05 PMsorv1986Re: what type of singularity f has and what is the residue of f
- Dec 7th 2011, 09:07 PMsorv1986Re: what type of singularity f has and what is the residue of f
- Dec 8th 2011, 02:08 AMchisigmaRe: what type of singularity f has and what is the residue of f
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...

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

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

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

Now the (2) is the Taylor expansion of the analytic function $\displaystyle e^{s}$ around s=0 and, of course, the coefficient of the term in $\displaystyle z^{-1}$ 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...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$