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The path rises from $(-1,1)$ to its maximum height $\sqrt{2}$ at $x=0$, then falls symmetrically to $(1,1)$. Its radius is $\sqrt{2}$, and its endpoint polar angles are $3\pi /4$ and $\pi /4$, giving angular span $\pi /2$. Hence its ordinary arc length isEquivalently, $\sqrt{1+y^{\prime 2}}=\sqrt{2}/\sqrt{2-x^2}$ and integrating over $[-1,1]$ gives the same value. This Euclidean length differs from the weighted destruction integral $D_{\min }\simeq 1.76275$: the slightly longer route saves undergrowth by travelling at greater height.

Interpret the density $1/y$ on the physical region $y>0$. It is singular at zero and would not describe a nonnegative density for $y<0$. The first integral gives $y\sqrt{1+y^{\prime 2}}=R$ for a positive constant $R$. To handle the horizontal tangent without dividing by $y^{\prime }$, use the full Euler-Lagrange equation:Two integrations give $y^2=-(x-x_0{)}^2+R^2$. The equal endpoint heights force $x_0=0$, and either endpoint then gives $R^2=2$. ThusThe path is the upper circular arc. A constant horizontal path solves the differentiated first integral trivially but fails $y{y}^{\prime \prime }+y^{\prime 2}+1=0$, so is not another extremum.

One can prove global minimality, rather than only stationarity. Define $\Phi (x,y)=\mathrm{arsinh}(x/y)$ on the upper half-plane. ThenThe Cauchy-Schwarz inequality gives, along any admissible path from $A$ to $B$,Our circle has tangent $(1,-x/y)$ for increasing $x$, so equality holds pointwise. This calibration for semicircular hyperbolic geodesics provesThe geometry is that of a hyperbolic geodesic in the Poincaré half-plane model. The proof even allows competing upper-half-plane paths beyond the class of monotonic graphs.

The monotonic horizontal coordinate allows the path to be written as a graph. Its arc length element is $ds=\sqrt{1+y^{\prime 2}}dx$, so the variational density is $f(y,y^{\prime })=g(y)\sqrt{1+y^{\prime 2}}$. It has no explicit horizontal-coordinate dependence. The Beltrami identity becomesConsequently every smooth minimizing path satisfiesThis is a first integral of the weighted arc-length problem; fixed endpoint values remove the endpoint variation terms.

Let $a_n=(1+p/n{)}^{n^2}$. For every fixed real $p$, the base is positive once $n$ is sufficiently large; any initial negative or zero bases affect only finitely many coefficients. For that tail,The Cauchy-Hadamard theorem therefore yieldsFor example $p=0$ recovers the geometric-series radius one. The stronger expansion $a_n=e^{pn-p^2/2+O(1/n)}$ also shows that, on $|z|=e^{-p}$, the magnitudes of the individual terms tend to $e^{-p^2/2}>0$. Thus this series diverges at every point on its convergence circle; the radius itself does not require that extra boundary analysis.

For coefficients $a_n=1/\ln n$, the Cauchy-Hadamard theorem gives $R^{-1}=\mathrm{limsup}|a_n{|}^{1/n}$. HereThereforeOmitting the first two powers has no effect on the radius of convergence.

Choose the common orthonormal eigenbasis established in part (ii), and let $U$ be the unitary matrix of its columns. ThenConsequentlyThus $N$ is unitarily diagonalizable, even though it need not be Hermitian. As a consistency check, commutation also gives $N{N}^{†}=N^{†}N=A^2+B^2$, so $N$ is a normal matrix, as also follows from unitary diagonalization of a normal matrix.

Suppose first that $AB=BA$, and take an eigenvector $v_j$ of $A$, with $A{v}_j=a_j{v}_j$. Commutation givesSince the eigenvalue $a_j$ is simple, its eigenspace is one-dimensional, so $B{v}_j=b_j{v}_j$ for some scalar $b_j$. Thus every vector in an eigenbasis of $A$ is also an eigenvector of $B$. The simple spectrum of $B$ ensures these account for its eigenvector directions as well. Eigenvectors are understood up to nonzero scaling and permutation, rather than as literally identical numerical vectors.

Conversely, if $v_1,\dots ,v_n$ are a common eigenbasis with $A{v}_j=a_j{v}_j$ and $B{v}_j=b_j{v}_j$, thenfor every basis vector. Hence $AB-BA=0$. ThereforeThis is simple spectrum gives simultaneous diagonalization. Without simple spectra, commuting Hermitian matrices still possess a common orthonormal eigenbasis, but not every arbitrary eigenvector of one matrix must be an eigenvector of the other.

Using $(AB{)}^{†}=B^{†}{A}^{†}$, the conjugate transpose of the proposed matrix isThusThe commutator of two Hermitian matrices is skew-Hermitian, and multiplication by $i$ changes it into a Hermitian matrix. No simplicity assumption on the spectra is needed for this part.

Use the branch from part (i) and close the real-axis path with a large upper semicircle, indenting clockwise around $-1$ and $1$. On the segment between them, take the upper-bank value of the branch cut. This is an instance of square-root endpoint integrals by an upper-half-plane contour.

The contributions can be controlled separately. Since $g(z)=O(z^{-2})$, the large arc contributes $O(R^{-1})\to 0$. Near either branch point, $g=O({\epsilon }^{-1/2})$, so an arc of length $O(\epsilon )$ contributes $O(\sqrt{\epsilon })\to 0$. On the upper bank of $(-1,1)$,which is odd in $x$, so its integral is zero. On the exterior real intervals $h(-x)=-h(x)$, while the rational numerator is odd. Hence $g(-x)=g(x)$ and the two exterior contributions sum to $2J$.

The enclosed residues use $h(i)=i\sqrt{2}$ and $h(2i)=i\sqrt{5}$:The residue theorem gives $2J=2\pi i[-i/(3\sqrt{2})-i/(6\sqrt{5})]$. ThereforeAs an independent check, $u=\sqrt{x^2-1}$ converts the integral towhich yields the same value. This check also confirms the branch signs and the factor of two in the contour argument.

The rational factors give four simple poles at $z=\pm i,\pm 2i$; none is canceled by the numerator. The square root has branch points at $z=\pm 1$, with local behaviour proportional to $(z\mp 1{)}^{-1/2}$ in the integrand.

For this part, choose $h(z)=\sqrt{z^2-1}$ analytic off $[-1,1]$ and satisfying $h(z)\sim z$ at infinity. Its values are $h(x)=\sqrt{x^2-1}$ for $x>1$, $h(x)=-\sqrt{x^2-1}$ for $x<-1$, andThus the chosen branch cut is $[-1,1]$, and the function is analytic elsewhere apart from the four poles. Infinity is not an intrinsic branch point: $g(z)=O(z^{-2})$ on this branch.

Another legitimate convention applies the positive-axis square root of part (a) to $w=z^2-1$; it places the cuts on $(-\infty ,-1]$ and $[1,\infty )$. Its upper-half-plane values agree with the branch used here, so the contour calculation below and the positive real square root in the required integral are unchanged. The branch choice must be stated because a square root alone does not specify a single-valued complex function.

For $x>0$, the upper-bank argument tends to zero, whereas the lower-bank argument tends to $2\pi$. ThusThe analytic function $f$ has identical boundary values across this part of the real axis. Hence the branch cut introduces a sign change:This compares integrand values. When two contour segments traverse opposite banks in opposite directions, their orientation supplies an additional minus sign.

Choose $0<\arg z<2\pi$, so $\sqrt{z}=\sqrt{R}{e}^{i\theta /2}$ on the arc. Its clockwise orientation givesBecause $f$ is an analytic function at zero, $f(R{e}^{i\theta })=f(0)+O(R)$ uniformly on the small arc. Integration then givesThereforeThe estimation lemma also gives $|I(R)|\le \pi \sqrt{R}{\max }_C|f|$. Hence a sufficient large-arc condition isFor example $f(z)=O(|z{|}^{-1/2-\eta })$ uniformly on the arc, with $\eta >0$, suffices. Under the stronger literal assumption that $f$ is analytic at infinity, its Laurent expansion is $f(z)=f(\infty )+O(1/z)$; the necessary and sufficient condition is then $f(\infty )=0$. A nonzero constant term would instead give the growing contribution $2\sqrt{2R}f(\infty )$.

Write $r=|\mathbf{r}|$, $s=|{\mathbf{r}}^{\prime }|$. For a spherical source, the angular integral needed for the Green function of the Laplacian isThis is the angular form of the spherical shell theorem. Substituting the source into the Green function integral therefore yieldsThe elementary integrals areThe terms proportional to $e^{-\beta r}/\beta$ cancel, leavingThe value at the origin is the continuous limit. At large radius $\varphi \sim -1/({\beta }^{2}r)$, consistent with the integrated source $\int \rho dV=-4\pi /{\beta }^2$ and the factor $1/(4\pi r)$ in the far-field potential.

For a direct check, set $u=r\varphi$. Then $u^{\prime \prime }=e^{-\beta r}$, so ${\nabla }^2\varphi =u^{\prime \prime }/r=e^{-\beta r}/r=-\rho$ for $r>0$. Boundedness of $\varphi$ at zero cancels the singular $1/r$ terms and excludes an extra point source. The original density is itself singular but locally integrable; the equation at the origin is understood in the sense of distributions. This is the Poisson potential of an exponential-over-radius source with amplitude $A=-1$.

Translation and rotation invariance allow a radial Green function $G(R)$, with $R=|\mathbf{r}-{\mathbf{r}}^{\prime }|$. Outside the source, $G^{\prime \prime }+2G^{\prime }/R=0$, so $G=A/R+B$. Decay at infinity gives $B=0$.

Integrate the defining equation over a small ball centred on the source. The divergence theorem fixes its flux:ConsequentlyThis is the Green function of the Laplacian with the stated negative-delta sign. A homogeneous $1/R$ addition would change the delta normalization, so the singular coefficient is fixed, not an arbitrary integration constant.

Define the candidate potential by convolving the decaying Green function with the source. Acting with the Laplacian in the observation variable gives, in the sense of distributions,The final equality is the sifting property of the Dirac delta. The kernel found in part (ii) is locally integrable, and its convolution with a sufficiently regular rapidly decaying source also vanishes at spatial infinity. Thus it satisfies the Poisson equation and the required condition at infinity.

The difference from any other regular decaying solution is a harmonic function on all of space vanishing at infinity, hence is zero by the maximum principle. ThereforeOne can also obtain this expression from Green's second identity on larger and larger balls, with the singularity of the kernel treated distributionally. The surface term at infinity vanishes for the usual $1/r$ potential decay and $1/r^2$ gradient decay.

Equal endpoint heights implyso $x_0=0$. The height condition then gives $y_0=b-k\cosh (a/k)$ and the hanging catenary isSince $\sqrt{1+y^{\prime 2}}=\cosh (x/k)$, its length fixes $k$ throughFor the requested graphical argument, put $u=a/k>0$ and $q=L/a>1$. Intersections of $Y=\sinh u$ with $Y=qu$ determine the solution. Near zero the line has larger slope, while the hyperbolic sine eventually grows faster. Its slope $\cosh u$ increases strictly, so after turning upwards relative to the line it can cross it only once. The intersection at $u=0$ is excluded: it corresponds to infinite $k$ and the limiting length $L=a$.

More explicitly, $h(u)=\sinh u/u$ has derivative $(u\cosh u-\sinh u)/u^2>0$, because the numerator vanishes at zero and has derivative $u\sinh u>0$. Since $h(0+)=1$ and $h(u)\to \infty$, every $q>1$ yields exactly one positive $u$, hence exactly one positive $k$.

Introduce a constant Lagrange multiplier $\lambda$ for the fixed arc length. Up to an irrelevant constant term, the augmented functional isIt has no explicit $x$ dependence, so the Beltrami identity givesFor the hanging minimum choose $k>0$. Writing $y_0=-\lambda$, this becomes $1+y^{\prime 2}=(y-y_0{)}^2/k^2$. Its nonconstant smooth solution isOne may verify it without dividing by $y^{\prime }$ at the lowest point: the full Euler-Lagrange equation gives $(y+\lambda )y^{\prime \prime }=1+y^{\prime 2}$, which the expression satisfies everywhere. The branch with negative $k$ is the inverted stationary chain rather than the hanging minimum. The catenary minimizes the integral of height along a fixed-length curve, so the positive branch is also the global minimum, not merely a stationary candidate.

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.

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.

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.