Laurent series expansion around non-isolated singularities

Question

Is it possible to have a Laurent series expansion about non-isolated singularity? For example, take $$f(z) = \tan \dfrac{1}{z}$$ about $z = 0$.

Answer 1

Yes. A Laurent series around $z=a$ can converge in an annulus $r < |z-a| < R$. There is nothing to prevent the existence of infinitely many singularities near $a$. So your example will have a Laurent series in each annulus $\dfrac{2}{(2k+1)\pi} < |z| < \dfrac{2}{(2k-1)\pi}$ for positive integers $k$.

[EDIT to answer @aschepler's comment] $\tan(z)$ has simple poles at $\pm (k-1/2)\pi$ for each positive integer $k$, with residues $-1$. Thus $\tan(z)+\dfrac{1}{z-k+1/2}+\dfrac{1}{z+k-1/2}$ has removable singularities at $\pm (k-1/2)\pi$; removing the singularities you have a function analytic in $\dfrac{2}{(2k+1)\pi} < |z| < \dfrac{2}{(2k-3)\pi}$ (or $\infty$ if $k=1$). The difference between the coefficient of $z^n$ in the Laurent series of $\tan(z)$ in the two annuli $\dfrac{2}{(2k+1)\pi} < |z| < \dfrac{2}{(2k-1)\pi}$ and $\dfrac{2}{(2k-1)\pi} < |z| < \dfrac{2}{(2k-3)\pi}$ is the difference between these coefficients in the Laurent series of $-\dfrac{1}{z-k+1/2}-\dfrac{1}{z+k-1/2}$ in $|z| < \dfrac{2}{(2k-1)\pi}$ and $|z| > \dfrac{2}{(2k-1)\pi}$.