Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / 7

A complex function is holomorphic at a point if it is complex differentiable throughout some open neighbourhood of that point. Writing $\zeta =\xi +i\eta$ and $\psi =P+iQ$, a convenient sufficient and necessary criterion under continuously differentiable real components is the Cauchy-Riemann equationsThey must hold throughout an open neighbourhood, rather than at an isolated point. Merely having partial derivatives that satisfy these equalities at one point is not enough without an appropriate differentiability hypothesis. For the composition $\psi =G\circ f$, holomorphy of $f$ and $G$ on the relevant open sets suffices; the additional condition $f^{\prime }\ne 0$ is needed when using a local inverse conformal map, not for holomorphy of the composition itself.

Choose the branch of $\sqrt{(\zeta +1)(\zeta -1)}$ holomorphic in the upper half-plane whose boundary value at zero is $+i$. Thenand integration gives $\boxed{z=\arcsin \zeta ,\zeta =\sin z}$ on this branch. The normalization and intended upper-half-plane image fix the otherwise ambiguous square-root sign.

To see the image, put $z=x+iy$. For $-\pi /2<x<\pi /2$ and $y>0$,The bottom edge maps monotonically onto $[-1,1]$. The right wall maps by $\sin (\pi /2+iy)=\cosh y$ onto $[1,\infty )$, and the left by $\sin (-\pi /2+iy)=-\cosh y$ onto $(-\infty ,-1]$. These boundaries describe the conformal map of a vertical half-strip by sine.

For completeness, the map is one-to-one in the interior: equality of two sine values forces arguments differing by $2\pi m$ or summing to $(2m+1)\pi$, and neither alternative yields two distinct points in this half-strip. It is onto the upper half-plane as well. Setting $w=e^{iz}$ gives $w^2-2i\zeta w-1=0$; the root product is $-1$, and no root lies on $|w|=1$ because that would make $\zeta$ real. Thus exactly one root has $|w|<1$. The identity $\mathrm{Im}\zeta =(\mathrm{Re}w)(|w{|}^{-2}-1)/2$ then gives $\mathrm{Re}w>0$, so $z=-i\log w$ belongs to the half-strip. The closure gives the stated boundary-inclusive region; holomorphy is understood in its open interior.

In this region the transformed function is $G(z)=\log (2+\sin z)$. Its only finite intrinsic branch point in the closure occurs where $2+\sin z=0$:It lies on the left boundary. There are no singularities in the open half-strip. The points $\zeta =\pm 1$, where $f^{\prime }$ diverges, map to the bottom corners; $G$ is holomorphic through those corners since $2+\sin z\ne 0$ there.

For the principal logarithm, the boundary segment $z=-\pi /2+iy$, $y\ge \mathrm{arcosh}2$, is the preimage of its cut $\zeta \le -2$. Points on that cut above $z_{\ast }$ admit local analytic continuation after changing branch; $z_{\ast }$ itself is the genuine logarithmic branch point.

Writing $\zeta =\xi +i\eta$,ThereforeThe vector field points radially away from $(-2,0)$, with magnitude inversely proportional to distance from that point. It diverges on approach to that boundary point and tends to zero far away. This is the radial potential flow described by the logarithmic complex potential of a two-dimensional potential flow.

The chain rule givesTo make the directions explicit, set $z=x+iy$ and $D=(2+\sin x\cosh y{)}^2+{\cos }^{2}x{\sinh }^{2}y$. Multiplying by the complex conjugate of the denominator gives the logarithmic flow in a sine-mapped half-strip:In the open strip $U>0$. The vertical component is negative near the lower left wall where $\cosh y+2\sin x<0$, and positive above that curve. On the bottom the flow is rightward. On the right wall it is upward, whereas on the left wall it points downward below $z_{\ast }$ and upward above $z_{\ast }$. The field is tangent to the boundaries except at the source and vanishes at the two bottom corners. At large $y$, $U-iV\to -i$, so it tends to a uniform upward field $(0,1)$.

Near $z_{\ast }$, $G^{\prime }(z)\sim 1/(z-z_{\ast })$, showing a radial boundary source. The streamlines are the curves $\mathrm{Im}G=\arg (2+\sin z)=\text{constant}$; the figure shows their orientation without reusing the exam's graphics.