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Near an isolated essential singularity, a holomorphic function has dense image in the complex plane. If it avoided a disk about $w$, then $1/(f-w)$ would be bounded and extend across the singularity by the Riemann removable singularity theorem. If its extended value is nonzero, $f$ extends holomorphically; if zero, its finite-order zero makes $f$ a pole. Either contradicts an essential singularity.

For $z\ne 0$, the denominator zeroes are $z_n=1/[\pi (n+\frac{1}{2})]$. Its derivative is $\sin (1/z)/z^2$, nonzero at each such point. ThusThese poles accumulate at zero. Consequently zero is a non-isolated singularity, as described by reciprocal cosine with accumulating poles. It is not an isolated essential singularity: no punctured disc about zero is a domain of holomorphy for this reciprocal function. Although $\cos (1/z)$ itself has an essential singularity at zero, taking its reciprocal introduces this accumulation of poles.

After division by $z$, the coefficients are $P=2/z-1$ and $Q=-1/z$. At $0$, $zP$ and $z^{2}Q$ are analytic functions, so $0$ is a regular singular point. Its indicial equation is $r(r-1)+2r=r(r+1)=0$. There are no other finite singular points of a differential equation. At infinity the regular singular point criterion for a second-order equation would require $P=O(z^{-1})$, but $P\to -1$; hence infinity is an irregular singular point.

Substituting $y=f/z$ reduces the equation to $f^{\prime \prime }-f^{\prime }=0$, soA linearly independent pair is $1/z$ and $(e^z-1)/z$. Their Wronskian is $e^z/z^2\ne 0$. The second has a removable singularity at $0$ and value $1$, whereas the first has a simple pole, agreeing with the two characteristic exponents at a regular singular point, $0,-1$. The integer difference does not require a logarithm. At infinity $e^z/z$ has an essential singularity, consistent with the irregular singular point; the existence of the simpler solution $1/z$ does not make infinity regular.

One possible open contour begins at $t=0$ along a ray with $\mathrm{Re}(1/t)>0$, avoids the essential singularity at the origin, and ends at infinity along a ray with $\mathrm{Re}(zt)<0$. For example, when $z>0$ is real, it may leave the origin along the positive real axis, bend through the upper half-plane, and approach the negative real axis as $|t|\to \infty$. At its initial end, $e^{-1/t}\to 0$; at its final end, $e^{zt}\to 0$. More generally, the terminal ray may be chosen withso that $zt$ is negative real. This contour satisfies the endpoint condition from part (a) and therefore produces a solution of the differential equation.

Forboth $e^z-1$ and $\log (1+z)$ have a simple zero at zero. Hence $g$ has a simple pole, and its residue is

For $h(z)=\sin z\sin (1/z)$, multiplication of the two Laurent series shows that all powers are even:There are infinitely many negative powers, so zero is an essential singularity. There is no $z^{-1}$ term, and therefore

Termwise integration of the exponential series givesThere are infinitely many negative Laurent coefficients, so zero is an essential singularity.