Identify the isolated singularities of the $f(z) = \frac{\cos(1/z)}{(z-1)^3}$. If the isolated singularity is a pole, find its order.

Question

I am trying to solve this problem.

Identify the isolated singularities of the $f(z) = \frac{\cos(1/z)}{(z-1)^3}$. If the isolated singularity is a pole, find its order.

First what I did is I solve the zero for $(z- 1)^3$ and found out that the zero is $z = 1$. I don't have to be bother with the $\cos \frac{1}{z}$ because it is in the numerator so whatever is, it does not affect the analyticity of the function $f$. Right?

Now what I did instead on the numerator is the $\frac{1}{z}$ of the $\cos \frac{1}{z}$ because such function won't be analytic when $z = 0$.

Does it mean that the points of singularities of $f$ are z = 0, 1? IF YES. Is that a pole? meaing the order of $z = 0$ is 1 and the order of $z = 1$ is 3?

Please help me. Thanks

Answer 1

If you note that $$ \cos \frac 1z = \sum_{n=0}^{\infty} \frac{(1/z)^{2n}}{(2n)!} $$

you'll see that the Laurent series has an infinite number of negative order terms. Therefore, $z=0$ is not a pole, it is an essential singularity.

Answer 2

The singularities of $f$ are located at $z=0$ and at $z=1$, yes. And $1$ is a pole of order $3$, since $\cos(1)\ne0$. On the other hand, $f$ has an essential singularity at $0$, not a pole. In fact, if $f$ had a pole or a removable singularity at $0$, then $(z-1)^3f(z)$ would also have a pole or a removable singularity there. But$$\cos\left(\frac1z\right)=1-\frac1{2!}z^{-2}+\frac1{4!}z^{-4}-\frac1{6!}z^{-6}+\cdots$$