Laurent series expansion around non-isolated singularities Is it possible to have a Laurent series expansion about non-isolated singularity? For example, take $$f(z) = \tan \dfrac{1}{z}$$ about $z = 0$.
 A: 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 $|z| >  \dfrac{2}{(2k-1)\pi}$.
