Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 3

The denominator factors as $(z+1-i)(z+1+i)$ and neither factor cancels with the numerator. Thus there are simple poles at $z=-1\pm i$. At infinity the quotient tends to one and has a regular expansion in $1/z$, so it has a removable singularity at infinity, with extended value one.

The numerator is nonzero at $1$ and $-2/3$, so the finite singularities are a pole of order three at $1$ and a pole of order two at $-2/3$. The numerator degree exceeds the denominator degree by three, with leading behaviour $z^3/9$. Hence infinity is a pole of order three.

There are logarithmic branch points at $z=-3i$ and infinity. At zero, every local branch has $\log (3i)=\ln 3+i(\pi /2+2\pi k)\ne 0$, so division by $z^2$ gives a pole of order two at zero. Decay at infinity does not remove its branching: with $w=1/z$, the function is $w^2[\log (1+3iw)-\log w]$, which changes branch when $w$ encircles zero.

A holomorphic function is complex differentiable throughout an open region; equivalently it is locally represented by a convergent complex power series. For $f=u+iv$, the Cauchy-Riemann equations are $u_x=v_y$, $u_y=-v_x$. With continuous first partial derivatives, these equations imply holomorphicity and $f^{\prime }=u_x+i{v}_x$.

For $e^z=e^x\cos y+i{e}^x\sin y$, the equations hold and $u_x+i{v}_x=e^x(\cos y+i\sin y)=e^z$. For $\sin z=\sin x\cosh y+i\cos x\sinh y$, they also hold, and $u_x+i{v}_x=\cos x\cosh y-i\sin x\sinh y=\cos z$. Therefore

The Cauchy-Riemann equations give $v_x=2x$, $v_y=2-2y$. Integrating and matching both derivatives gives the harmonic conjugateThe real part of this polynomial is exactly the required $2x(1-y)$.

Here $v_x=-3x^2+3y^2+4y$ and $v_y=6xy+4x$. The harmonic conjugate and resulting polynomial areIndeed $\mathrm{Re}(-i{z}^3)=3x^{2}y-y^3$ and $\mathrm{Re}(2z^2)=2x^2-2y^2$.