= 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. In
$$
\sum_{j,k\ge0}\frac{t^{j+k}}{j!k!}z^{j-k-2},
$$
only $j-k-2=-1$ contributes. Set $k=m$, $j=m+1$. Thus
$$
\boxed{b_m=\frac{2\pi i}{m!(m+1)!},\qquad m\ge0.}
$$
The resulting power series converges for every complex $t$ by the ratio test. The <coefficients> are independent of the positive radius $R$.
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