Past exam of the mathematics course of the University of Cambridge / 2013 / ib / Paper 1 / 2D (Complex Analysis or Complex Methods)

The zeroes of $\cosh z$ occur at $z=i\pi (n+\frac{1}{2})$, $n\in \mathbb{Z}$. They are simple because $\sinh z\ne 0$ there. Squaring the denominator therefore givesThere are no other finite singularities; this follows from the local zero-order criterion for a pole.

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.

On the stated branch of the principal complex logarithm, the only denominator zero is $z=1$, and $\log z=(z-1)+O((z-1{)}^2)$. ThereforeZero is a logarithmic branch point, not a removable isolated singularity, even though $1/\log z$ tends to zero as $z\to 0$ within the cut plane. Continuation around zero changes $\log z$ by $2\pi i$. The nonpositive real axis is the chosen branch cut; its nonzero points are not intrinsically branch points and admit local analytic continuation. There are no other finite singularities inside the specified branch domain.

At $z=1$, numerator and denominator both have simple zeroes. Their derivatives give the removable valueAt each positive integer $n\ge 2$, the numerator is nonzero and the denominator has a simple zero, so these are simple poles. At zero the expansion on the principal sheet iswhich contains a fractional power. Thus zero is a branch point with pole-like leading growth, not an ordinary isolated pole. The negative real axis is the chosen branch cut. On local continuations across its nonzero points, negative integers also give simple poles because the continued square root is $\pm i\sqrt{|n|}$, never one. They lie outside the open principal branch domain; noninteger negative points are merely cut points, not intrinsic singularities.