Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2008 / 5

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 $R$.

Both numerator and denominator have simple zeros at zero, and $1-e^{-z}=z+O(z^2)$. Their quotient tends to one and extends holomorphically with that value. Hence the singularity is removable, and $\boxed{{\mathrm{Res}}_{0}f=0}$ because its Laurent series has no negative powers.

Use $\sin z=z+O(z^3)$ andTheir product is $z^{-2}+(3/2)z^{-1}+O(1)$. Thus the singularity is a pole of order two, with $\boxed{{\mathrm{Res}}_{0}f=3/2}$.

The denominator vanishes at $z_k=1/(2\pi ik)$ for every nonzero integer $k$. These points approach zero, the denominator has a simple zero at each, and $\sin z_k\ne 0$ because a nonzero purely imaginary number is not a zero of sine. Hence there are poles arbitrarily close to zero. The origin is a nonisolated singularity, so it has no residue defined by a Laurent series on a punctured disc. In particular it must not be classified as an isolated essential singularity merely because $e^{1/z}$ has one.