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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 $f(z)$?

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

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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.

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