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Determine the type of singularity at $z = 0$ of the following function and why?

$$f(z) = \sin \left(\frac{1}{\cos (\frac{1}{z})}\right).$$

I have no idea.

Thank you for your answer.

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The singularity at $z=0$ is not isolated (since $\cos(1/z) = 0$ for a sequence of points tending to $z=0$). The usual classification into removable singularities, poles and essential singularities is only applicable to isolated singularities.

(Your textbook may have another convention.)

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  • $\begingroup$ cos(1/z)=0 for a sequence of points tending to z=0???????? $\endgroup$ – Shailesh Jan 9 '14 at 11:54
  • $\begingroup$ @Shailesh Yes, $$z=\frac{2}{(4n+1)\pi}$$ $n \in \mathbb{Z}$ for example. $\endgroup$ – mrf Jan 9 '14 at 11:56
  • $\begingroup$ @mrf Thank you for your help. I get your answer. Please teach me how should I approach for this type of problem. Here how do you understand that you have to consider a sequence $\{z_n\}$ s.t. $\cos(\frac{1}{z_n})$ converges to $0$, this is the main trick for this problem. $\endgroup$ – Supriyo Jan 9 '14 at 12:58
  • $\begingroup$ @HopelessFool Start piece by piece. There is clearly a problem at points where the argument of $\sin$ has a singularity which happens when the inner denominator $\cos(1/z) = 0$. Solve this equation to find all problematic points and you will see that they accumulate at $z=0$. $\endgroup$ – mrf Jan 9 '14 at 13:00
  • $\begingroup$ @mrf So we shall do the same for the function of the type $f(g(z))$. Nice. $\endgroup$ – Supriyo Jan 9 '14 at 13:05

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