# singularity of a function.

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.

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