Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 7

For $f=u+iv$ analytic in $z=x+iy$, the Cauchy-Riemann equations areThey follow by computing the same complex derivative along the real and imaginary coordinate directions.

The Cauchy-Riemann equations giveAt a regular point, each level curve has tangent perpendicular to its nonzero gradient. The two gradients, and hence the two tangents, are orthogonal. Therefore regular level curves meet at right angles.

The qualification is necessary: $|\nabla u|=|\nabla v|=|f^{\prime }(z)|$, so at a critical point both gradients vanish and this argument gives no tangent direction. For example, for $f(z)=z^2$, the zero level curves of the real and imaginary parts are diagonals and coordinate axes; they meet at $45^{\circ }$ at zero. This is the regular-point statement of orthogonal level curves of harmonic conjugates.

Recognize that $z^2=(x^2-y^2)+2ixy$ and $e^{-z}=e^{-x}(\cos y-i\sin y)$. Their product has exactly the specified real part. Its imaginary part isOne can also recover it directly from the Cauchy-Riemann equations. Differentiating the specified real part givesIntegration of $v_y=u_x$ gives $v=v_0+h(x)$; the other equation $v_x=-u_y$ then forces $h^{\prime }(x)=0$. Thus the most general answer on a connected domain isIndeed the difference of any two analytic functions with the same real part has zero real part; its two Cauchy-Riemann equations force its imaginary part to be constant. This proves that no further analytic freedom is missing.

The denominator factors as $z^2+iz+6=(z-2i)(z+3i)$, with no cancellation against the numerator. Thus there areTo inspect infinity, put $w=1/z$:Hence infinity is a removable singularity at infinity, with limiting value zero. After this natural extension to the Riemann sphere it is also a simple zero. There are no other zeros or singularities.

The complex exponential never vanishes. The zeros of $\sinh z$ occur at $z=n\pi i$, $n\in \mathbb{Z}$, and they are simple because $\cosh (n\pi i)=(-1{)}^n\ne 0$. The numerator is $e^{2n\pi i}=1$ there, so no cancellation occurs. ThereforeFor example their residues are $1/\cosh (n\pi i)=(-1{)}^n$.

Infinity requires care: the poles themselves tend to infinity. Equivalently $f(1/w)$ has poles at $w=1/(n\pi i)$ accumulating at zero. Thus $\boxed{\infty \text{ is a non-isolated singularity}}$, and it is not a zero. In the usual classification of isolated singularities, it cannot be called a removable singularity, pole or isolated essential singularity. The non-isolated singularity distinction is necessary even though the function is meromorphic at every finite point.

Put $\zeta =z-3$ and apply the geometric series:ThusThe radius is the distance from the expansion centre $3$ to the nearest singularity $2i$, not the distance of that pole from the origin. On the boundary circle the series terms all have magnitude $1/\sqrt{13}$, so they do not approach zero: this particular series diverges at every boundary point.