Past exam of the mathematics course of the University of Cambridge / 2009 / ib / Paper 2

The monomials $x^i{y}^j$ with $i,j\ge 0$ and $i+j\le n$ form a basis of the given vector space. There are $k+1$ of total degree $k$, soOn that basis the linear operator is diagonal:Both summands of the eigenvalue are nonnegative integers. Their sum is zero precisely when $i,j\in \{0,1\}$. Thus the kernel is the span of those members of $1,x,y,xy$ whose total degree is at most $n$. In particular it is $\mathrm{span}\{1,x,y\}$ for $n=1$ and $\mathrm{span}\{1,x,y,xy\}$ for $n\ge 2$; for $n=0$ it is the constants. The image of a linear map is the span of all the other permitted monomials, since each has a nonzero diagonal eigenvalue and hence is the image of its reciprocal multiple. No excluded monomial can occur in an image because its diagonal coefficient is zero.

The trace is the sum of these eigenvalues. At degrees zero and one all contributions vanish. At degree two the contributions are $2,0,2$, summing to four; at degree three they are $6,2,2,6$, summing to sixteen; at degree four they are $12,6,4,6,12$, summing to forty. ThereforeThis is the diagonal monomial action of a second-order Euler operator.

For a finite group of order $p^{a}m$ with $p$ prime and $p\nmid m$, the Sylow theorems assert that a subgroup of order $p^a$ exists; every subgroup whose order is a power of $p$ is contained in one such Sylow subgroup; all Sylow subgroups for that prime are conjugate; and their number $n_p$ divides $m$ and is congruent to one modulo $p$. A unique Sylow subgroup is a normal subgroup, since conjugation cannot move it to another one.

Here $56=7\cdot 8$, so $n_7$ divides eight and is congruent to one modulo seven. Hence $n_7=1$ or $8$. In the first case the Sylow subgroup of order seven is normal. In the second case, any two distinct order-seven subgroups intersect only at the identity, because a nontrivial intersection would generate both prime-order subgroups. Their nonidentity elements therefore account for $8\cdot 6=48$ elements.

Exactly eight elements of the group remain. Every Sylow subgroup of order eight consists entirely of these remaining elements: a nonidentity element in it has order a power of two and cannot be one of the order-seven elements already counted. Each such Sylow subgroup must consequently equal the remaining eight-element set. It is unique and hence normal. Thus a group of order 56 has a normal subgroup of order seven or of order eight.

The contraction mapping theorem states that a contraction mapping $T:X\to X$ on a nonempty complete metric space has a unique fixed point. Here contraction means that some $0\le q<1$ satisfies $d(Tx,Ty)\le qd(x,y)$ for all $x,y$.

For a proof, choose $x_0$ and iterate $x_{r+1}=T{x}_r$. Induction gives $d(x_{r+1},x_r)\le q^{r}d(x_1,x_0)$. For $m>n$, the triangle inequality and a geometric series giveThus the iterates form a Cauchy sequence, and completeness supplies a limit $a\in X$. A contraction mapping is continuous, so $Ta=\lim T{x}_n=\lim x_{n+1}=a$. If $b$ is another fixed point, $d(a,b)=d(Ta,Tb)\le qd(a,b)$, forcing $d(a,b)=0$. This proves existence and uniqueness.

For the exponential problem, use $h=f\circ f$ on the complete metric space $\mathbb{R}$. Its derivative isPutting $t=e^{-x}>0$, the function $t{e}^{-t}$ has maximum $1/e$ at $t=1$, as its derivative is $(1-t)e^{-t}$. By the mean value theorem, $|h(x)-h(y)|\le e^{-1}|x-y|$, so $h$ is a contraction mapping on all of $\mathbb{R}$. Let $a$ be its unique fixed point. The compositions commute, so $h(f(a))=f(h(a))=f(a)$; uniqueness of the fixed point of $h$ forces $f(a)=a$. Conversely every fixed point of $f$ is fixed by $h$. Hence there is exactly one real solution of $x=e^{-x}$. It lies in $(0,1)$ because $a=e^{-a}>0$ and then $e^{-a}<1$. This uses the iterated contraction argument without claiming that $f$ itself contracts on all of $\mathbb{R}$.

A Hausdorff space is a topological space in which any two distinct points have disjoint open neighborhoods. Let $K$ be a compact subset of a Hausdorff space and $x\notin K$. For each $y\in K$, choose disjoint open neighborhoods $U_y$ of $y$ and $W_y$ of $x$. The $U_y$ cover $K$, so compactness gives a finite subcover $U_{y_1},\dots ,U_{y_m}$. Their paired intersection $W={\bigcap }_{r=1}^m{W}_{y_r}$ is an open neighborhood of $x$ disjoint from $K$. Thus every point of its complement has an open neighborhood in the complement, proving every compact subset of a Hausdorff space is closed. The empty set case is immediate.

In the cocountable topology on an uncountable set, any two nonempty open sets intersect: their combined complements are countable and cannot exhaust $X$. Thus the space is not Hausdorff.

Nevertheless every compact subset is closed here. In fact only finite subsets are compact. To prove this, suppose $K$ is infinite and choose distinct points $d_1,d_2,\dots \in K$. The open setsform an increasing open cover of $X$, and hence of $K$. Any finite selection is contained in $U_N$ for its largest index $N$, which misses $d_N\in K$. So $K$ is not compact. Finite sets are compact in any topological space, and in this cocountable topology they are closed because their complements are open. This proves the claimed behavior of compact subsets of a cocountable space.

The Fourier sine series coefficients of $x$ on $(0,\pi )$ areby integration by parts. The odd periodic extension is piecewise smooth, so the Fourier series converges to $x$ at every interior point. HenceTo justify integration from zero, the Fourier sine series partial sums converge to $x$ in L2 norm. By the Cauchy-Schwarz inequality, integration from zero to any $x\in [0,\pi ]$ has error at most $\sqrt{\pi }$ times that L2 norm error, uniformly in $x$. Integrating the partial sums and taking the limit therefore givesThe integrated series is absolutely and uniformly convergent, since its terms are bounded by $4/n^2$. Thus the resulting Fourier cosine series hasIndependently, the constant coefficient isComparison yields the requested exact sum

In SI units, Gauss's law relates the outward electric flux through a closed surface to its enclosed charge:The outward unit normal specifies the sign of the electric flux.

The uniform charge density is ${\rho }_e=3Q/(4\pi R^3)$. Spherical symmetry makes the electric field radial and constant in magnitude on each concentric sphere. A sphere of radius $r<R$ encloses charge $Q(r/R{)}^3$. Gauss's law therefore gives $4\pi r^2{E}_r=Q{r}^3/({\epsilon }_0{R}^3)$, soThis also defines the zero electric field at the centre by continuity.

Take the upper hemisphere to have $z>0$. Integrate the force $\mathbf{E}dq$ using the total interior electric field from part (a). This computes the net external force on the upper hemisphere, since the forces between pairs of its own charge elements cancel by antisymmetry of the Coulomb force. Horizontal components cancel by symmetry. The vertical component isA horizontal section at height $z$ has area $\pi (R^2-z^2)$, soThus the lower hemisphere repels the upper one vertically upward, withThe result is the same for either sign of $Q$, as expected for the force between hemispheres of a uniformly charged ball.

For a standard collinear Lorentz transformation, put $\beta =v/c$ and $\gamma =(1-{\beta }^2{)}^{-1/2}$, with $|v|<c$. Then $x^{\prime }=\gamma (x-vt)$ and $c{t}^{\prime }=\gamma (ct-\beta x)$. Define the rapidity $\varphi =\mathrm{artanh}\beta$. The identities ${\cosh }^2\varphi -{\sinh }^2\varphi =1$ and $\tanh \varphi =\beta$, with $\cosh \varphi >0$, give $\cosh \varphi =\gamma$ and $\sinh \varphi =\gamma \beta$. Substitution proves the desired hyperbolic form.

Adding and subtracting those transformed coordinates uses $\cosh \varphi \mp \sinh \varphi =e^{\mp \varphi }$ and yieldsA second collinear Lorentz transformation of rapidity ${\varphi }^{\prime }$ multiplies these two null coordinates by $e^{-{\varphi }^{\prime }}$ and $e^{{\varphi }^{\prime }}$. Their total factors are therefore $e^{-(\varphi +{\varphi }^{\prime })}$ and $e^{\varphi +{\varphi }^{\prime }}$, provingThe hyperbolic tangent addition formula now gives the velocity composition lawAll velocities here are signed along the same axis; the speed is the absolute value if either frame moves in the opposite direction.

Use the ideal steady incompressible-flow model, neglecting height change across the smooth nozzle. Flow conservation gives inlet and outlet speeds $u_0=Q/A_0$ and $u_1=Q/A_1$. Set atmospheric pressure to zero. The Bernoulli equation between inlet and outlet givesChoose a control volume containing the fluid in the nozzle, with the positive axis pointing toward the outlet. The integral momentum equation givesThus the fluid's force on the nozzle is $p_0{A}_0-\rho Q(u_1-u_0)$, in the outlet direction. Substitute the pressure and velocities:The fireman's holding force is equal and opposite. Therefore the required force points upstream and has magnitudeThis is the holding force on a contracting nozzle; retaining the inlet pressure force is essential to its sign and value.

Label the towns $L,M,R$ from left to right, and the treatment plants $A,B,C,D$ as upper-left, upper-right, lower-left and lower-right. Model each treatment capacity as a sink arc to the sea. Incoming untreated sewage can be forwarded, so the numbers inside the plant circles constrain treatment, not total throughput.

The sum of treatment capacities is $16+20+14+12=62$, so no feasible flow network can handle more than 62. Here is a feasible routing attaining that upper bound. Send town inflowsEvery town-to-plant flow respects its directed pipe capacity. Send an additional five units from $C$ to $A$ through their bidirectional pipe, whose capacity is twelve, and two from $D$ to $B$ through their bidirectional pipe, whose capacity is fourteen. Use neither horizontal plant-to-plant pipe.

Now $A$ receives $6+5+5=16$ and treats all of it. Plant $C$ receives $9+10=19$, forwards five and treats fourteen. Plant $B$ receives $7+11+2=20$ and treats all of it. Plant $D$ receives $3+11=14$, forwards two and treats twelve. Thus flow conservation holds at every plant and all treatment capacities are saturated. The routing proves the upper bound is achievable:This is one optimal assignment, not a uniqueness assertion. The total bound comes from the cut consisting of all treatment-to-sea arcs; the explicit routing supplies a matching maximum flow certificate.

The ascending chain $\ker T\subseteq \ker T^2\subseteq \cdots$ must stabilize because $V$ is finite-dimensional. Once $\ker T^l=\ker T^{l+1}$, it stays stable: if $T^{l+2}v=0$, then $Tv\in \ker T^{l+1}=\ker T^l$, so $v\in \ker T^{l+1}=\ker T^l$; induction repeats the argument. Choose a positive such $l$, giving $\ker T^{2l}=\ker T^l$.

If $v\in \ker T^l\cap \mathrm{im}{T}^l$, write $v=T^{l}w$. Then $T^{2l}w=0$, so $w\in \ker T^l$ and $v=0$. By the rank-nullity theorem, the dimensions of those two subspaces sum to $\dim V$. HenceThis proves the finite-dimensional form of the Fitting lemma.

If $\det T=0$, the kernel contains a nonzero vector $u$, and $\mathrm{im}T$ is a proper vector subspace. Extend a basis of that image of a linear map to a basis of $V$ and choose a nonzero linear functional $\lambda$ which vanishes on the former basis. Define $S(v)=\lambda (v)u$. It is a nonzero endomorphism, since $\lambda$ takes a nonzero value somewhere, andThus a singular endomorphism has a nonzero two-sided annihilator, as required.

For an idempotent linear map $T$, the decomposition $v=(v-Tv)+Tv$ shows $V=\ker T\oplus \mathrm{im}T$: the first term is killed by $T$, the second is fixed, and a vector both fixed and killed is zero. An adapted basis makes its matrix $\mathrm{diag}(0,I_r)$, with $r=\mathrm{rank}T$. Consequently equal ranks give identical adapted matrices and therefore a change of basis relating $T_1,T_2$. Conversely, similar matrices have the same rank because pre- and postmultiplication by invertible matrices preserve rank. Hence $T_1$ and $T_2$ are similar exactly when their ranks agree, proving the similarity classification of idempotent linear maps.

The centre of a group is $Z(G)=\{z\in G:zg=gz\text{ for every }g\in G\}$. It is a normal subgroup. For $|G|=p^r$ with $r\ge 1$, let $G$ act on itself by conjugation. The orbit-stabilizer theorem makes the size of the conjugacy class of $g$ equal to $[G:C_G(g)]$, a power of $p$. A class has size one exactly when $g\in Z(G)$; every other class size is divisible by $p$. The class equation therefore givesso $p$ divides $|Z(G)|$. Since the identity belongs to the centre of a group, its size is a positive multiple of $p$. This proves the nontrivial center of a finite p-group.

If $G/Z(G)$ is cyclic, choose a representative $a$ of a generator. Every element is $a^{m}z$ for an integer $m$ and $z\in Z(G)$. Any two such elements commute, since their central factors commute with everything and their remaining factors are powers of the same element. Thus $G$ is an abelian group, $Z(G)=G$, and the cyclic quotient by the centre of a group is trivial.

For a non-abelian group of order $p^3$, the centre of a group has order $p$, $p^2$ or $p^3$ by the proved result and Lagrange's theorem. The last would make the group abelian. The middle would make $G/Z(G)$ have prime order and hence cyclic, also making $G$ abelian by the preceding argument. Therefore its centre has order $p$.

Finally write the given upper-unitriangular matrix as $M(a,b,c)$. Direct matrix multiplication givesIt commutes with every $M(a^{\prime },b^{\prime },c^{\prime })$ exactly when $a{c}^{\prime }=a^{\prime }c$ for all $a^{\prime },c^{\prime }\in F$. Taking $(a^{\prime },c^{\prime })=(0,1)$ forces $a=0$, and taking $(1,0)$ forces $c=0$; those conditions also suffice. Thus the centre of a group in this Heisenberg group over a prime field is

For stereographic projection, choose the north pole $N=(0,0,1)$ of the unit sphere and identify the plane $Z=0$ with the complex plane by $z=X+iY$. Project a point $P=(X,Y,Z)\ne N$ along the line joining it to $N$; adjoin $z=\infty$ for the north pole to identify the whole sphere with the Riemann sphere. A point of that line is $N+t(P-N)$. Its last coordinate vanishes when $t=1/(1-Z)$, givingUsing $X^2+Y^2+Z^2=1$, or solving this line's second intersection with the sphere, gives the inverseA great circle lies in a plane $h_{1}X+h_{2}Y+h_{3}Z=0$ through the origin. Its projected equation isIt is a straight line only if $h_3=0$, meaning the plane contains the north pole and the south pole. Thus if all three sides of a nondegenerate spherical triangle projected to straight sides of a Euclidean triangle, all three distinct side planes would contain both poles. Each pair of side planes could then meet on the sphere only at those same two poles, which cannot supply three distinct triangle vertices. A nondegenerate spherical triangle cannot project to a Euclidean triangle. If a side passes through the projection pole, its image is unbounded and also cannot be a finite triangle side.

A nonidentity rotation in three dimensions has an axis through the origin: an orthogonal three-by-three matrix of determinant one has an eigenvalue one, since nonreal eigenvalues pair as conjugates, real eigenvalues are $\pm 1$, and their product is one. Its axis meets the unit sphere in an antipodal pair, which is fixed. In the identity case every point is fixed, so an antipodal pair can again be selected. Projecting $P$ and $-P$ giveswhenever both points are finite. Thus the associated Möbius transformation fixes an antipodal pair satisfying the claimed relation. The universally valid relation is $q=-1/\overline{p}$ on the Riemann sphere, with $0$ and $\infty$ paired. A rotation about the projection axis fixes exactly that pair, so the PDF's literal finite product needs this extended-coordinate convention. This is the antipodal stereographic coordinate relation.

TakeIts defining matrix has nonzero determinant eight, so it is a Möbius transformation. Direct substitution shows that it fixes $p=1$ and $q=-1$, which satisfy $p\overline{q}=-1$. However, $M^{\prime }(1)=8/(3-1{)}^2=2$. Differentiating the inverse projection formula gives the round spherical line element $ds=2|dz|/(1+|z{|}^2)$. At a fixed finite point, its local length multiplier under a Möbius transformation is exactly $|M^{\prime }(p)|$, since the metric factors in numerator and denominator agree. Here that multiplier is two, so this map is not an isometry and cannot be a rotation in three dimensions of the sphere. It demonstrates that antipodal fixed points do not characterize sphere rotations.

Reflection of the sphere in the plane $Y=0$ sends $(X,Y,Z)$ to $(X,-Y,Z)$. It preserves dot products, hence the spherical distance between every pair of points. Its stereographic expression isThis is an isometry of the sphere but is not a Möbius transformation: at any finite point, its complex difference quotient is one along real increments and minus one along imaginary increments, so it is not holomorphic. A Möbius transformation is holomorphic away from its pole. The geometric distinction is that this reflection reverses orientation, whereas a Möbius transformation preserves it.

A subset $U\subseteq {\mathbb{R}}^n$ is an open set when every $x\in U$ has some radius $r>0$ for which $B(x,r)\subseteq U$. If a continuous function takes only the values zero and one, continuity at $x$ with tolerance $1/2$ gives a neighborhood on which $|f(y)-f(x)|<1/2$. The two possible values differ by one, so $f(y)=f(x)$ throughout that neighborhood. Thus $f$ is locally constant, and the zero linear map is its derivative, since the difference $f(x+h)-f(x)$ is exactly zero for all sufficiently small $h$. Hence $D{f}_x=0$ everywhere.

For the bound on $g$, parametrize the segment by $\gamma (t)=b+t(a-b)$, $0\le t\le 1$. The chain rule givesby the definition of operator norm. The mean value theorem applied to the real function $g\circ \gamma$ provesIf the endpoints agree the estimate is immediate. No continuity of the derivative is required.

For the finite-line complement, each removed line ${\ell }_i$ is closed, so $V$ is open. Because $a,b\notin {\ell }_i$, there is a unique affine plane $P_i$ containing $a$ and ${\ell }_i$, and a unique affine plane $Q_i$ containing $b$ and ${\ell }_i$. Choose $c$ outside the union of these finitely many planes, using the permitted fact about finite plane covers. In particular $c\in V$. If $[a,c]$ met ${\ell }_i$ at a point $d$, then $d\ne a$, and the line through $a,d$ lies in $P_i$. Since $c$ is on that same line, it would belong to $P_i$, a contradiction. The same argument using $Q_i$ applies to $[c,b]$. Thus $a$ and $b$ can be joined in $V$ by these two segments, proving polygonal connectivity of a finite-line complement.

Finally a continuous $f:V\to \{0,1\}$ has zero derivative by the first argument. Apply the proved segment estimate with $M=0$ separately on $[a,c]$ and $[c,b]$. It gives $f(a)=f(c)=f(b)$. Since $a,b$ were arbitrary, $f$ is constant on $V$.

For the Joukowski map $F(z)=z+b^2/z$, the derivative is $F^{\prime }(z)=1-b^2/z^2$. Thus it is a conformal map at every finite $z$ except its pole at zero and its critical pointsThe pole maps to $\infty$; as a map of Riemann spheres it is a simple pole and is not a branch point. Neither is the point at infinity, where the map behaves like $z$.

An appropriate domain is the upper exterior half-discFor $z=x+iy$, $\mathrm{Im}F(z)=y(1-b^2/|z{|}^2)>0$ on $D$. To prove surjectivity and injectivity, solve $F(z)=\zeta$ as $z^2-\zeta z+b^2=0$. Its two roots have product $b^2$. For $\mathrm{Im}\zeta >0$, neither root lies on the real axis or on $|z|=b$, since either boundary maps to the real axis. Exactly one root therefore has modulus greater than $b$. The imaginary-part identity forces that root to have positive imaginary part. It is the unique root in $D$, so $F$ maps $D$ conformally and bijectively onto the upper half-plane.

HereThis entire function vanishes exactly at $z=inb$, $n\in \mathbb{Z}$, and those points map to $\zeta =(-1{)}^n$. The second derivative there is $(\pi /b{)}^2(-1{)}^n\ne 0$, so each is a simple critical point of the conformal map:Take the half-stripWriting $\pi z/b=u+iv$, its imaginary part maps to $\sinh u\sin v>0$. More precisely, $w=e^{\pi z/b}$ maps $D$ bijectively onto $\{w:\mathrm{Im}w>0,|w|>1\}$, with inverse $z=(b/\pi )\log w$ using arguments in $(0,\pi )$. ThenPart (i) with $b=1$, followed by positive real scaling, proves that this second map is a conformal bijection onto the upper half-plane. Their composition proves the required result and gives the half-strip conformal map by hyperbolic cosine.

For small transverse displacement, the wave equation on a string iswhere $T$ is the tension and $\rho$ is mass per unit length. Using separation of variables, put $y=X(x)F(t)$, the ratios $X^{\prime \prime }/X=F^{\prime \prime }/(c^{2}F)$ equal a constant. The fixed-end conditions $X(0)=X(L)=0$ exclude nontrivial solutions for nonnegative separation constants; for $-{\kappa }^2$ they give $X=\sin (\kappa x)$ and $\kappa =n\pi /L$, $n\ge 1$. The associated angular frequencies are ${\omega }_n=n\pi c/L$. Thus the general superposition of separated normal modes isThe string is initially straight and at rest before the impulse, so $A_n=0$. Assume the hammer interval lies inside $[0,L]$. The imposed velocity $g(x)=y_t(x,0)$ has Fourier sine series coefficientTherefore $B_n=g_n/{\omega }_n$, giving the amplitude of every excited normal mode.

The string energy is $\frac{\rho }{2}{\int }_0^L(y_t^2+c^2{y}_x^2)dx$. At the impulse it is $E_{\mathrm{total}}=\rho a{v}^2/2$. Orthogonality of the sine modes gives $E_n=\rho L{g}_n^2/4$ initially; each mode subsequently exchanges kinetic and elastic energy while retaining that total. Thus the modal energy of a localized velocity impulse on a string has fractionsFor $l=L/3$ and $a=L/10$, a mode is absent exactly when $\sin (n\pi /3)=0$ or $\sin (n\pi /20)=0$. Hence the unexcited modes are precisely

For a one-dimensional wavefunction, the probability density and probability current areThe Time-dependent Schrödinger equation with a real potential is $i\hslash {\psi }_t=-{\hslash }^2{\psi }_{xx}/(2m)+V\psi$. Multiply it and its complex conjugate by ${\psi }^{\ast }$ and $\psi$, respectively. Subtracting cancels the real-potential terms and givesThis derives the continuity equation ${\rho }_t+j_x=0$.

For $k>0$, the specified free-region wavefunction combines a right-moving incoming wave of unit amplitude and a left-moving reflected wave with complex reflection amplitude $R$. Applying the free Schrödinger operator to either plane wave gives $E={\hslash }^2{k}^2/(2m)$. Its time factor has modulus one, soThe density includes the stationary interference pattern; $|R{|}^2$, rather than $R$, is the reflection probability.

For the finite square well, let $k=\sqrt{2mE}/\hslash$ outside and $q=\sqrt{2m(E+V_0)}/\hslash$ inside. Write the stationary parts as $e^{ikx}+R{e}^{-ikx}$ to the left, $A{e}^{iqx}+B{e}^{-iqx}$ inside, and $T{e}^{ikx}$ to the right. At both finite potential steps, the wavefunction and its first spatial derivative are continuous. Integrating the stationary Schrödinger equation across a shrinking step proves derivative continuity; a jump of the wavefunction would create an unmatched distributional derivative and is excluded.

Propagate the boundary values at $a$ back through the well. The values at zero areMatching them to $1+R$ and $ik(1-R)$, and adding the two equations after division of the second by $ik$, givesSince the exterior wave numbers agree, the transmission probability is the current ratio $|T{|}^2$, soWhen $V_0=3E$, $q=2k=\sqrt{8mE}/\hslash$, yieldingClassically an incoming particle of positive energy crosses the attractive well and emerges with its original energy, so transmission is one. The quantum result agrees exactly when $qa$ is an integer multiple of $\pi$:These are widths with resonant transmission through a square well; the degenerate width zero also has unit transmission.

The vacuum Maxwell equations are $\nabla \cdot E=\nabla \cdot B=0$, $\nabla \times E=-B_t$ and $\nabla \times B=E_t/c^2$. Taking the curl of Faraday's equation and using $\nabla \times (\nabla \times E)=\nabla (\nabla \cdot E)-{\nabla }^{2}E$ gives $-{\nabla }^{2}E=-E_{tt}/c^2$. The same calculation for Ampère's equation gives ${\nabla }^{2}B=B_{tt}/c^2$. Hence each Cartesian component satisfies the electromagnetic wave equation.

Model the walls as a perfect conductor, with no oscillating fields in its interior. The tangential electric field is zero at the surface, otherwise it would drive unbounded conductor currents. The normal magnetic field is zero, since the divergence-free field has a continuous normal component and the interior oscillating field is zero. Thus $n\times E=0$ and $n\cdot B=0$. Normal $E$ and tangential $B$ may be nonzero because surface charge and surface current support their jumps.

The PDF writes the full magnetic field as $B=B_0(x,y)\hat{z}{e}^{i(kz-\omega t)}$. Taken literally, this cannot be a nonzero travelling wave: $\nabla \cdot B=ik{B}_z$, so for $k\ne 0$ it forces $B_z=0$. The subsequent requests require transverse components. The consistent intended reading is $B_z=B_0(x,y)e^{i(kz-\omega t)}$, with transverse fields determined by the Maxwell equations. Use that reading below.

All fields have the same $z,t$ factor. Put $\Delta =(\omega /c{)}^2-k^2$. The harmonic Maxwell equations are $\nabla \times E=i\omega B$ and $\nabla \times B=-i\omega E/c^2$. Their $x$ and $y$ components givetogether with ${\partial }_y{E}_z-ik{E}_y=i\omega B_x$ and $ik{E}_x-{\partial }_x{E}_z=i\omega B_y$. Substitution and solving, for $\Delta \ne 0$, givesThe corresponding electric components areFor a transverse electric mode of a rectangular waveguide, $E_z=0$. The electromagnetic wave equation becomesAt $x=0,a$, the tangential component $E_y=-i\omega {\partial }_x{B}_z/\Delta$ vanishes, so ${\partial }_x{B}_0=0$; at $y=0,b$, the corresponding condition $E_x=0$ gives ${\partial }_y{B}_0=0$. These are Neumann boundary conditions, and the normal-$B$ conditions give the same conclusions through the displayed transverse magnetic components. Separation giveswhere $m,n\ge 0$ are integers and $(m,n)\ne (0,0)$. The constant candidate is not a physical oscillating mode: its transverse derivatives vanish, so it has $E=0$ and Faraday's equation forces $B=0$ for $\omega \ne 0$. More generally a hollow single-conductor guide has no transverse-electromagnetic zero-cutoff mode: its transverse electrostatic potential is harmonic with the same constant boundary value, hence constant, leaving zero transverse field.