# What type of singularity is $z=\infty$ for $f(z)=\frac{1}{(sin(1/z))}$?

Consider the function $$f(z)=\frac{1}{(sin(1/z))}$$

At $$z=\infty$$ does $$f$$ have an isolated singularity or not? Or is $$z=\infty$$ a regular point?

$$f(1/t)=1/(sin(t))$$ has simple poles in $$t=k \pi$$, but those poles are for $$z=1/(k\pi)$$ which means there is an accumulation point of poles

So exactly what type of singularity is $$z=\infty$$ for $$f(z)$$?

Since $$0$$ is a pole of $$\dfrac1{\sin z}$$, your function has a pole at $$\infty$$.

The degree of a singularity of a function at $$z=\infty$$, can be found such that if $$\lim_{z\to\infty}{f(z)\over z^m}\ne 0$$for some $$m$$, then the degree is equal to $$m$$. In this case$$\lim_{z\to \infty}{1\over z^m\cdot \sin{1\over z}}=\lim_{u\to 0}{u^m\over \sin{u}}\ne 0$$only when $$m=1$$. Therefore $$\infty$$ is a degree-$$1$$ singularity.