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What are the singularities of $f(z) = e^{\frac{\sin z}{z}}$ ?

It clearly has a removable singularity at $0$. A textbook says that it has essential singularities at $kπ$ for $k \in \mathbb{Z}$. How and why ? If it was $f(z) = e^{\frac{z}{\sin z}}$ then I can see how but not in this case.

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    $\begingroup$ That textbook is wrong. $\endgroup$
    – Martin R
    May 15 at 13:32
  • $\begingroup$ @Martin Thanks a lot for confirmation. I just spend more than a hour utterly confused. $\endgroup$
    – Anonymous
    May 15 at 13:36
  • $\begingroup$ @Martin Sorry to bother you again, I have attached the image of the question. Is there any way I could have misinterpreted it? i.imgur.com/n7tte53.jpeg $\endgroup$
    – Anonymous
    May 15 at 14:05
  • $\begingroup$ @Martin Ex. 9.1.28 $\endgroup$
    – Anonymous
    May 15 at 14:06
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    $\begingroup$ That is definitely wrong in the book. I assume (as you did) that it is a typo and the author meant $f(z) = e^{z/\sin(z)}$. $\endgroup$
    – Martin R
    May 15 at 14:12

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