 Solution

An isolated singularity at $a$ means that the function is holomorphic throughout some punctured disc $0<|z-a|<r$. Its residue is the Laurent coefficient $c_{-1}$, equivalently $(2\pi i{)}^{-1}{\int }_{|z-a|=\rho }f(z)dz$. The singularity is removable if all negative Laurent coefficients vanish, a pole of order $k$ if only finitely many are nonzero and the lowest power is $-k$, and essential if infinitely many negative coefficients are nonzero.

For the final contour integral, expand $e^{tz}{e}^{t/z}$. On the circle these exponential series converge absolutely and uniformly for $t$ in every bounded set, so their product can be integrated term by term. Inonly $j-k-2=-1$ contributes. Set $k=m$, $j=m+1$. ThusThe resulting power series converges for every complex $t$ by the ratio test. The coefficients are independent of the positive radius $R$.