Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 5 (Complex Analysis)

For a holomorphic function $h:\mathbb{D}\to \mathbb{D}$, the Schwarz-Pick lemma statesEquality in the first inequality for distinct points, or in the derivative inequality at one point, occurs exactly when $h$ is an automorphism of the unit disk.

Here is a proof, including the underlying Schwarz lemma. If $u:\mathbb{D}\to \mathbb{D}$ is holomorphic and $u(0)=0$, then $u(z)/z$ has a removable singularity at zero. On $|z|=r<1$ its modulus is at most $1/r$. The maximum modulus principle on this circle, followed by $r\uparrow 1$, gives $|u(z)|\le |z|$ and $|u^{\prime }(0)|\le 1$. Equality at a nonzero point, or $|u^{\prime }(0)|=1$, forces $u(z)/z$ to be a constant of modulus one by the same maximum modulus principle.

Set ${\varphi }_a(z)=(z-a)/(1-\overline{a}z)$, an automorphism of the unit disk. The composite ${\varphi }_{h(a)}\circ h\circ {\varphi }_a^{-1}$ fixes zero. The Schwarz lemma gives the first inequality, while its derivative at zero gives the second. The equality conclusion follows because a composite attaining equality is a rotation. Conversely, every automorphism of the unit disk attains equality. This also proves contraction of the hyperbolic metric.

The boundary lower bound first guarantees a zero inside the unit circle. If $f(0)=0$, there is nothing to prove about existence. Otherwise, if $f$ had no zeros on the closed unit disc, $1/f$ would be holomorphic on a neighborhood of it and satisfy $|1/f|\le 1/A$ on the boundary. The maximum modulus principle would give $|f(0)|\ge A$, contradicting the hypothesis. The boundary lower bound excludes boundary zeros, so there is a zero $a$ with $|a|<1$.

The upper boundary bound and the maximum modulus principle give $|f|\le B$ on the unit circle and its interior. The function is nonconstant under the stated hypotheses, so $f/B$ takes its interior values in the open unit disc. Apply the Schwarz-Pick lemma to the points $a$ and $0$:This is the zero location bound from Schwarz-Pick, and it applies to every zero inside the unit circle.

The bound is sharp. Given $0<a<1$ and $B>0$, chooseThis holomorphic function has no pole in ${\mathbb{D}}_R$, has constant modulus $B$ on the unit circle, and satisfies $|f(0)|=Ba<B$. With $A=B$, its zero at $a$ gives equality. The case $a=0$ is also attained by $f(z)=Bz$.

A normal family of holomorphic functions on a domain is one from which every sequence has a subsequence converging locally uniformly to a holomorphic function, or locally uniformly to infinity in the extended convention. For a bounded family, only finite limits occur.

For functions mapping a plane domain $D$ into the unit disc, the Cauchy integral formula supplies uniform derivative bounds on each compact subset. More explicitly, if a closed disc of radius $r$ around $z$ lies in $D$, the Cauchy estimate gives $|f^{\prime }(z)|\le 1/r$. A slightly larger compact neighborhood of any given compact set therefore gives uniform boundedness and equicontinuity there. Apply the Arzelà-Ascoli theorem on a compact exhaustion and take a diagonal subsequence. Its limit is locally uniform; the Cauchy integral formula shows that the limit is holomorphic. This proves normality rather than merely invoking it.

Now let $|g|\le M$ on the complex upper half-plane. Every dilation $g_t(w)=g(tw)$ has the same bound, so the argument just given, after scaling the bound if necessary, proves that $\{g_t:t\ge 1\}$ is a normal family. For any sequence $t_j\to \infty$, every locally uniform subsequential limit $G$ satisfiesThe identity theorem for holomorphic functions forces $G\equiv \ell$. It follows that $g_t\to \ell$ locally uniformly as $t\to \infty$: otherwise a sequence violating convergence on one compact set would have a convergent subsequence with this same constant limit, a contradiction.

For $0<\epsilon <\pi /2$, the arc $K_{\epsilon }=\{e^{i\theta }:\epsilon \le \theta \le \pi -\epsilon \}$ is a compact subset of the complex upper half-plane. Write $z=|z|w$ with $w\in K_{\epsilon }$. Local uniform convergence of the dilations givesFor $\epsilon \ge \pi /2$ the specified sector is empty. This is the same compactness mechanism as the normal-family proof of angular boundary convergence.

To transfer the result to the unit disc, use the Cayley transform between the half-plane and diskIt sends $r\omega$ to $iy$ with $y=(1+r)/(1-r)$, so $g(iy)\to \ell$ as $y\to \infty$. Choose the curvature-minus-one normalization of the hyperbolic metric, for which the disc line element is $2|dz|/(1-|z{|}^2)$ and the half-plane line element is $|dw|/\mathrm{Im}w$. The Cayley transform between the half-plane and disk preserves these metrics. For $w\in \mathbb{H}$ and $y>0$,Thus ${\rho }_{\mathbb{D}}(z,r\omega )<k$ implies $\mathrm{Im}w/|w|>\mathrm{sech}k$. This is the hyperbolic tube around a radial segment estimate. It places $w$ in a fixed sector bounded away from the real axis. Moreover $z\to \omega$ implies $|C_{\omega }(z)|\to \infty$. Applying the sector result givesA different constant normalization of the hyperbolic metric only rescales $k$ and gives the same conclusion.

For a polygonal domain with finite real prevertices $a_j$ and interior angles $\pi {\alpha }_j$, the Schwarz-Christoffel formula has the formwith branches chosen in the complex upper half-plane. A prevertex at infinity is interpreted by the corresponding limiting version of the formula. The exponent records the angle because locally $F-F(a_j)$ behaves as $(z-a_j{)}^{{\alpha }_j}$.

For a slit from $0$ to $\tau$ in the upper half-plane, put $\vartheta =\arg \tau \in (0,\pi )$. The two approaches to its base are separate boundary vertices with angles $\pi -\vartheta$ and $\vartheta$, and the tip has angle $2\pi$. With ordered prevertices $a<c<b$ mapping respectively to the base, tip and base, the Schwarz-Christoffel mapping therefore hasThe tip's factor is a simple zero of the boundary derivative, not an interior critical point.

For the specified power product, the principal branches givethroughout the complex upper half-plane, so $g$ maps into that half-plane. Logarithmic differentiation givesThere are no zeros of this derivative in the upper half-plane, so the map is locally conformal there. Its boundary values are negative real for $x<-1$, positive real for $x>1$, andThe positive scalar factor increases from zero to its maximum at $c$ and then decreases to zero, since its logarithmic derivative is $-k/(1-x)+(1-k)/(1+x)$. Therefore the tip isThe real boundary runs from the negative real axis to zero, out to $\tau$, back to zero, and then along the positive real axis.

For completeness, these boundary facts give global injectivity and surjectivity. At infinity,Apply the argument principle on a large upper half-disc, indenting the real prevertices by small semicircles. For a fixed $w$ in the upper half-plane off $[0,\tau ]$, the limiting image contour is the real boundary with the slit traversed once in each direction, followed by a large upper semicircle. Its winding number about $w$ is one. The two slit traversals cancel, so $g(z)-w$ has exactly one zero in the upper half-plane. There are no poles.

Nor can an interior point map onto the open slit. Each point of that slit has two real preimages, one on either side of $c$, with nonzero boundary derivatives. Local continuation across the real line maps the upper half-neighborhoods at these preimages onto the two different sides of the slit. An additional interior preimage would persist for nearby off-slit points and contradict the already proved one-preimage count. The tip cannot be an interior image either: the open mapping theorem would then place nearby open-slit points in the interior image. Finally, $g$ never vanishes inside the half-plane. Thus $g$ is a conformal bijection onto $S(\tau )$.

This is the power-product map to a slit half-plane. Its derivative is exactly the Schwarz-Christoffel formula with $a=-1$, $b=1$, $c=1-2k$, $\vartheta =\pi k$ and $C=1$. The additive constant is fixed by the boundary value $g(1)=0$. Thus the power-product expression is an explicit antiderivative of the corresponding Schwarz-Christoffel mapping, with this normalization.

Let $p:\mathbb{D}\to D$ be a universal covering map. On a small evenly covered neighborhood in $D$, choose a local inverse $s$ of $p$ and setAny two such lifts differ by a deck transformation of the disc. Every automorphism of the unit disk preserves the disc density: apply the derivative Schwarz-Pick lemma to the automorphism and its inverse to get equality. Hence the definition is independent of the local inverse. Uniqueness of the universal cover up to a covering isomorphism gives independence of the chosen cover as well. This is the Poincare metric on a Riemann surface, normalized to curvature $-1$.

For a piecewise continuously differentiable path $\gamma$, define its length by $L_D(\gamma )=\int {\lambda }_D(\gamma )|{\gamma }^{\prime }|$, and the distance by the infimum of lengths of joining paths. Path reversal and concatenation prove symmetry and the triangle inequality. The distance between distinct points is positive: choose a small closed Euclidean disc around one point not containing the other, on which the positive continuous density is bounded below by $c>0$. Every joining path must leave that disc and has length at least $c$ times its radius before doing so. Joining paths of finite length exist because a plane domain is path connected and the density is locally bounded. Thus this construction gives an actual metric, not just a nonnegative path-length expression.

For the punctured unit disc, use $p(w)=e^{iw}$ from the complex upper half-plane. The half-plane line element is $|dw|/\mathrm{Im}w$, while $|dz|=|z||dw|$ and $\mathrm{Im}w=\log (1/|z|)$. ConsequentlyThe domain called an annulus in the paper has inner radius zero, so it is this punctured-disc metric, not the metric of an annulus with positive inner radius.

We now prove the Great Picard theorem. In its holomorphic form, an essential singularity takes every complex value, with at most one exception, infinitely often in every punctured neighborhood. It suffices to prove that a holomorphic function on a punctured disc omitting $0$ and $1$ extends meromorphically through the puncture: if two values occurred only finitely often, a smaller punctured disc would omit both, and an affine change of target would reduce them to $0,1$.

Use the universal covering by the modular lambda function $\lambda :\mathbb{H}\to Y=\mathbb{C}\setminus \{0,1\}$, whose construction is given in the following solution. Its deck transformation group is the projective level-two principal congruence subgroup $\Gamma (2)$. Rescale the source punctured disc to unit radius. Lift $f(e^{2\pi iw})$ to a holomorphic function $L:\mathbb{H}\to \mathbb{H}$. Uniqueness of lifts givesfor a fixed deck transformation $T\in \Gamma (2)$. The Schwarz-Pick lemma givesas $\mathrm{Im}w\to \infty$. A nonidentity deck transformation has no interior fixed point, so it is either a hyperbolic Möbius transformation or a parabolic Möbius transformation. The hyperbolic case is conjugate to a dilation $w\mapsto aw$ with $a>0$, $aeq1$. Since $|d\log \mathrm{Im}w|\le d{s}_{\mathbb{H}}$, its displacement is at least $|\log a|>0$, contradicting this inequality.

If $T$ is the identity, compose $L$ with the Cayley transform between the half-plane and disk. Its periodicity lets the resulting bounded function descend to the punctured disc and extend by the Riemann removable singularity theorem. A nonconstant extension takes its central value inside the disc by the maximum modulus principle. The covering map then gives an extension of $f$; a constant lift gives a constant $f$.

In the parabolic case, its fixed point is a rational modular cusp. Indeed, a parabolic integral matrix of trace $\pm 2$ has a rational eigenvector for its repeated eigenvalue. Conjugate by an element of the modular group sending this cusp to infinity. The level-two subgroup is normal, so the conjugated transformation is $\zeta \mapsto \zeta +2n$ for a nonzero integer $n$. The conjugated lift $P$ consequently satisfiesThe modular transformations permute the three target cusp values $0,1,\infty$, as seen from the lambda transformation formulas below. It is therefore enough to examine $\lambda (P(w))$.

To fix the sign of $n$, the function $\mathrm{Im}P$ is periodic in the real direction. Its mean $M(y)={\int }_0^1\mathrm{Im}P(x+iy)dx$ satisfies $M^{\prime }(y)=2n$ by the Cauchy-Riemann equations and the increment of $\mathrm{Re}P$. Positivity for all $y>0$ forces $n>0$. The functionis well-defined, nonvanishing and bounded by one on the punctured disc, so it extends holomorphically to zero. Its winding number along a circle around the puncture is $n$, because a unit increment in $w$ increases $\pi iP$ by $2\pi in$. The argument principle therefore shows that $Q$ has a zero of order $n$ at the origin.

At the infinite cusp, the modular lambda function has the convergent expansion $\lambda (\zeta )=16q+O(q^2)$ in $q=e^{\pi i\zeta }$. Thus $\lambda (P(w))=16Q(z)+O(Q(z{)}^2)$ extends and tends to zero. Undoing the target permutation gives a removable singularity or a pole for $f$. In every case the puncture is not essential, which proves the holomorphic Great Picard theorem. For a meromorphic function with an essential singularity, if three spherical values occurred only finitely often, a Möbius transformation would make the omitted values $0,1,\infty$. The same proof applies because omission of infinity makes the transformed function holomorphic. Hence the meromorphic version has at most two exceptional spherical values.

Take the Jacobi theta functions at zero, with $q=e^{\pi i\tau }$:Define the modular lambda function byThis is an explicit holomorphic expression on the complex upper half-plane. The product formsshow that all three are nonzero there. The theta identity ${\theta }_3^4={\theta }_2^4+{\theta }_4^4$ gives $1-\lambda ={\theta }_4^4/{\theta }_3^4$, so $\lambda$ omits $0,1$ and infinity on the half-plane.

The theta-constant inversion and translation laws, obtained from Poisson summation and changing indices in the defining sums, giveThese formulas imply invariance under the projective groupgenerated by $\tau \mapsto \tau +2$ and $\tau \mapsto \tau /(2\tau +1)$. This is the effective level-two principal congruence subgroup.

Its action is properly discontinuous because it is a discrete subgroup of $PS{L}_2(\mathbb{R})$. It is also free. An elliptic integral matrix has trace $0$ or $\pm 1$. The odd traces are excluded because $a+d$ is even. Trace zero would give $d=-a$ and $bc=-a^2-1\equiv 2(\mathrm{mod}4)$, impossible when $b,c$ are even. Thus the only elements acting trivially are the already quotiented central matrices $\pm I$.

A fundamental domain is the ideal quadrilateralwith vertices $-1,0,1,\infty$. The vertical sides are paired by translation by two, and the semicircular sides by $\tau \mapsto \tau /(2\tau +1)$. Reduction using these pairings gives the domain; equivalently it is a union of six translates of the standard modular fundamental domain, consistent with the six elements of $PS{L}_2({\mathbb{F}}_2)$. On adjoining the cusp points, there are three vertex classes, two paired edges and one face. The resulting orientable compact surface has Euler characteristic $3-2+1=2$, hence is a sphere. This is the compactification of the level-two modular curve.

All three cusp widths are two, so their local parameters are obtained from $q=e^{\pi i\tau }$ after a modular change of coordinate. The defining products giveat infinity: there is a simple zero. At the cusp zero, inversion gives $1-\lambda =16q_0+O(q_0^2)$, so the value is $1$. At the cusp one, combine inversion and translation; the value is infinity and the pole is simple. The three cusp classes are precisely the parity classes of primitive rational pairs $(p,q)$: odd-even, even-odd and odd-odd. A Bézout identity, with parity adjusted by adding multiples of $(p,q)$, places any such cusp in the orbit of $\infty$, $0$ or $1$, respectively.

The descended meromorphic function on the compactified quotient thus has exactly one simple pole. Its degree is one, since the degree equals its total pole multiplicity; it is therefore a biholomorphic coordinate on that sphere. Removing the three cusps identifies the quotient with $\mathbb{C}\setminus \{0,1\}$. Consequentlyis a covering map with deck group $\overline{\Gamma (2)}$. The source half-plane is simply connected, so it is the required universal covering map. This proves the universal covering by the modular lambda function, including the absence of interior branching. Three arbitrary distinct punctures are handled by a target Möbius transformation.

Treat $F$ as a vector-valued solution; matrix solutions are handled columnwise. It is important to distinguish the coefficient convention from the indicial exponent. In a fixed Fuchsian frame, a finite singularity at $a$ meanswhere $B_a$ is holomorphic near $a$. The residue is $R_a$, equivalently the residue of the matrix-valued one-form $A(z)dz$. Such a Fuchsian linear differential system has a regular singular point: on sectors, its solutions have at most power and logarithmic growth. Intrinsically a regular singular system can be made Fuchsian by a meromorphic change of frame; a higher coefficient pole in some other frame does not itself prove irregularity. Residues here refer to the displayed fixed coefficient one-form.

At infinity use $w=1/z$ and $\tilde{F}(w)=F(1/w)$. The equation becomesA Fuchsian singularity at infinity means $\tilde{A}$ has at most a simple pole at $w=0$, and $R_{\infty }={\mathrm{Res}}_{w=0}\tilde{A}(w)$. The residue theorem applied entrywise on the Riemann sphere givesFor a direct calculation when all coefficient poles are simple, write $A(z)=P(z)+{\sum }_j{R}_j/(z-a_j)$ by rational partial fractions. Regularity at infinity in this frame requires $A(z)=O(1/z)$, so $P=0$. Transforming coordinates gives $R_{\infty }=-{\sum }_j{R}_j$. More generally the entrywise residue theorem remains valid even when a different meromorphic frame has higher-order poles.

The printed nonconstant-solution claim is false. A counterexample satisfying its spectral hypothesis isIts residue has eigenvalues $0,-1/3$, with zero of largest real part. The equations are $u^{\prime }=0$ and $v^{\prime }-v/(3z)=0$. If $v$ is analytic at zero, coefficient comparison gives $(n-1/3)v_n=0$ for every nonnegative integer $n$, so $v=0$. Hence every analytic solution is $(c,0{)}^T$, which is constant. This disproves nonconstancy, rather than silently replacing it with a weaker conclusion.

A useful correct conclusion is that a nonzero analytic solution exists under the stated hypotheses. Write $A=R/z+{\sum }_{j\ge 0}{B}_j{z}^j$, and let $m$ be the largest nonnegative integer with $-m\in \mathrm{spec}R$; such an $m$ exists because $0\in \mathrm{spec}R$. Choose a nonzero $h_0\in \ker (mI+R)$ and seekThe Frobenius method givesAll left-hand matrices are invertible by maximality of $m$. For large $n$ their inverse norm is $O(1/n)$ by a geometric-series inverse; enlarging the constant bounds every $n\ge 1$ by $C/n$. If $\Vert B_j\Vert \le M{r}^{-j}$ on a smaller disc, the scalar majorant with generating functionsatisfies $v^{\prime }=CM(1-t/r{)}^{-1}v$ and dominates the recurrence coefficientwise. It converges for $|t|<r$, proving convergence of $H$. Thus the analytic solution at a maximal integer Fuchsian exponent is nonzero, and is nonconstant if $m>0$. When $m=0$, nonconstancy requires additional information, as the counterexample shows.

There is also a sign problem in the printed exponent terminology. Substitution of $F=z^{\mu }H$ with $H(0)\ne 0$ into the actual equation givesThus the indicial exponents are eigenvalues of $-R$, not $R$. With the printed sign, an exponent of maximal real part admits a pure Frobenius solution, because the corresponding coefficient matrices are invertible at all subsequent positive integer steps. Lesser exponents can meet resonance.

The supplied example does not demonstrate such failure. Solving it directly givesEvery solution is analytic. The residue eigenvalue $-1$ corresponds to the true exponent $1$, and setting $c_1=0$ gives $F=z(-c_{2}z/2,c_2{)}^T$ with nonzero leading coefficient. Even with the literal printed factor $z^{-1}$, every displayed solution has the form $z^{-1}H$ with $H=zF$ analytic, though $H(0)=0$.

Nevertheless the last existence-of-a-counterexample request, read literally, can be fulfilled by the diagonal counterexample above. For its nonmaximal residue eigenvalue $\lambda =-1/3$, substitution of $F=z^{-1/3}H$ yieldsfor every integer $n\ge 0$. Neither diagonal entry, $n-1/3$ or $n-2/3$, vanishes, so $H=0$. Therefore there is no nonzero solution of that literal form, even without imposing $H(0)\ne 0$.

For the intended logarithmic resonance phenomenon with correctly signed exponents, take insteadIts equation gives $v=c_2/z$ and $u=c_1+c_2\mathrm{Log}z$ on a chosen branch. The exponent $-1$ would require $H(0)=(0,c_2{)}^T\ne 0$, but $z\mathrm{Log}z$ prevents $H=zF$ from being analytic when $c_2\ne 0$. The exponent zero has the constant solution $(1,0{)}^T$. This supplies an explicit resonance obstruction under the correct convention and completes the qualified resolution of the printed defects.