Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 7

The Cauchy-Riemann equations are $\boxed{u_x=v_y,u_y=-v_x}$. They give the gradient identityAt a regular intersection, where $f^{\prime }(z)\ne 0$, the two nonzero gradients are normal vectors to the two level sets, so their tangent directions are perpendicular. This is the orthogonality of regular real and imaginary level curves.

The regularity qualification is necessary: $|\nabla u|=|\nabla v|=|f^{\prime }|$, so at a critical point both normals vanish and the curves need not have unique tangents. For example $f=z^2$ gives diagonal lines for $u=0$ and coordinate axes for $v=0$, all meeting at the origin. Thus the conclusion is $\boxed{\text{right-angle intersection wherever }{f}^{\prime }(z)\ne 0}$, rather than an unrestricted claim at critical points.

Write $g=u+iv$ and let its constant modulus be $C$. If $C=0$, then $g=0$. For $C>0$, differentiating $u^2+v^2=C^2$ with respect to $x$ and $y$, then applying the Cauchy-Riemann equations, givesThusThe coefficient matrix has determinant $-(u^2+v^2)=-C^2\ne 0$, forcing $u_x=v_x=0$. The Cauchy-Riemann equations then also give $u_y=v_y=0$. Therefore $u,v$ are constant on a connected domain andThis proves the constant-modulus holomorphic function result. If a domain is disconnected, the conclusion applies separately to each connected component.

The zeros of $\sin z$ are $z=n\pi$ for $n\in \mathbb{Z}$, all simple, and $\cos (n\pi )=(-1{)}^n\ne 0$. Hence the finite singularities of $\cot z$ are exactlyLocally $\cot z\sim 1/(z-n\pi )$, so each pole has residue $1$.

For $z\ne 0$, singularities occur when $1/z=n\pi$, with $n\ne 0$. Each is a simple pole, since the substitution has nonzero derivative there:More explicitly, $1/z-n\pi \sim -(z-z_n)/z_n^2$, so $\cot (1/z)\sim -z_n^2/(z-z_n)$ and its residue is $-1/(n^2{\pi }^2)$.

These poles accumulate at zero. Consequently $\boxed{z=0\text{ is a non-isolated singularity}}$. The removable/pole/essential classification for isolated singularities cannot be applied at that accumulation point; it is not an isolated essential singularity.

At every $z=n\pi$, the inner function has a simple pole, with $\cot z=1/(z-n\pi )+h(z)$ for analytic $h$. ThusThe first factor is analytic and nonzero, while the second has infinitely many negative powers in its Laurent series. Therefore, by the exponential of a pole is an essential singularity result,At all other finite points the composition is analytic, so there are no additional finite singularities.

Set $w=z-1$. Partial fraction decomposition givesExpand both terms as geometric series:The two bracketed terms are complex conjugates. Using $1\pm i=\sqrt{2}{e}^{\pm i\pi /4}$ produces the required real coefficients:The initial terms are $1-\frac{1}{2}{w}^2+\frac{1}{2}{w}^3-\frac{1}{4}{w}^4+\cdots$. The nearest singularities are the simple poles $z=\pm i$, both at distance $\sqrt{2}$ from the center. Hence the Taylor series has