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Let $f(z)=z^5\cos(\frac{1}{z+1})$ then which kind of singularity $f$ have at $\infty.$

My attempt is that, since in the Laurent expansion of $f$ we have infinitely many terms like $\frac{n}{z^k}$. Hence essential singularity. Please tell me whether I am right or not.

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You have to look at $g(z):=f(1/z)$ !

We have $g(z)= \frac{1}{z^5} \cos( \frac{z}{1+z}).$

Since $g$ has a pole $0$, $f$ has a pole at $ \infty$.

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  • $\begingroup$ Thanking you sir, I think similarly, $z^5cos(z+1)$ will have pole of order $5$ at infinity. $\endgroup$ – Priya Pandey Oct 17 '19 at 11:08

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