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Using the complex Euclidean norm and the normal matrix identity,Both sides are nonnegative. Therefore $\Vert Cx\Vert =0$ if and only if $\Vert C^{†}x\Vert =0$. This proves the stronger reusable fact that normal matrices have equal adjoint norms.

Since $A$ is an invertible matrix, so is its Hermitian conjugate, andIf $A$ is a normal matrix, substitute $A^{†}A=A{A}^{†}$ into the second identity to obtain $B{B}^{†}=I$, so $B$ is a unitary matrix. Conversely, if $B$ is a unitary matrix, multiply $B{B}^{†}=I$ on the left by $A$ and on the right by $A^{†}$. This gives $A^{†}A=A{A}^{†}$, the defining normal matrix identity. Thus $B$ is a unitary matrix exactly when $A$ is a normal matrix.

A branch of the complex logarithm must be specified to obtain a unique locus. First use the principal complex logarithm. Away from the negative real axis, write $z=r{e}^{i\theta }$ with $r>0$ and $-\pi <\theta <\pi$. Then the complex logarithms of $z$ and its complex conjugate are $\log r+i\theta$ and $\log r-i\theta$. Equating real and imaginary parts gives $\log r=\theta$. Consequently the principal-branch locus is the open segment of a logarithmic spiralThe origin is excluded because its complex logarithm does not exist. If the principal value is assigned argument $\pi$ on the negative real axis, a point there equals its own complex conjugate, so its principal complex logarithm would have to equal $i$ times itself. That forces the complex logarithm to be zero and cannot hold there; the endpoints remain excluded.

If the printed equation instead means that some values of the multivalued complex logarithms satisfy it, write those values as $\log r+it$ and $\log r+iu$, with $e^{it}=z/r$ and $e^{iu}=\overline{z}/r$. Equality gives $\log r=-u=t$, hence $u=-t$, which is compatible with conjugation. Every real $t$ is allowed. Under this interpretation the locus is the entire logarithmic spiral $z=e^{(1+i)t}$, $t\in \mathbb{R}$. The PDF does not choose between these branch conventions.

Write $z=x+iy$. Dividing by two givesThe left side is nonnegative, so squaring is reversible only when $x+a\ge 0$. Squaring yields $y^2=4ax$.

If $a>0$, this equation implies $x\ge 0$, so the sign condition holds automatically. The locus is the parabolawith vertex at the origin, focus of a conic $(a,0)$ and directrix of a conic $x=-a$. Indeed the original equation equates distance from the focus of a conic with distance from the directrix of a conic on its right side.

If $a<0$, the squared equation forces $x\le 0$, making $x+a<0$. This violates the sign condition. Thus the locus is empty for $a<0$; squaring without checking that condition would introduce a spurious parabola.

Write $z=x+iy$ and $\alpha =p+iq$. The complex conjugates in the PDF giveSince $(p,q)\ne (0,0)$, the locus is the straight line through the origin perpendicular to the position vector of $\alpha$ in the complex plane. Equivalently,

Let $K$ be the set where the indicator function $f$ equals one. Fix a positive integer $N$. Each permitted length-$N$ decimal prefix $w=(w_1,\dots ,w_N)\in \{2,7{\}}^N$ gives the intervalThere are $2^N$ such intervals, each of length $10^{-N}$, and they cover $K$. Every point of $K$ is strictly inside the interval for its prefix, since its remaining tail lies between $2/(9\cdot 10^N)$ and $7/(9\cdot 10^N)$. Thus none of the interval endpoints belongs to $K$. The intervals are disjoint and separated: their distinct prefix integers cannot be consecutive because their final digits are only $2$ and $7$.

The complement consists of finitely many closed intervals on which $f=0$, of total length $1-2^{N}10^{-N}=1-5^{-N}$. Part (b) therefore already proves Riemann integrability, since this length tends to one.

To compute the Riemann integral directly, make a partition of an interval using $0,1$ and all the endpoints of the $I_w$. On complementary cells the upper Darboux sum contribution is zero, and on the $I_w$ it is at most their total length. Moreover every nonempty cell contains a terminating decimal, which is not in $K$: neither its terminating expansion with a zero tail nor its alternative expansion with a nine tail consists entirely of twos and sevens. Such decimals are dense, so every cell has infimum zero. ThereforeLetting $N\to \infty$ gives the exact answerThis is an example of an indicator of a restricted-digit decimal set: infinitely many, indeed uncountably many, points can support an indicator function whose Riemann integral is zero.

Let $|f|\le M$. If $M=0$ the result is immediate, so take $M>0$. Given $\epsilon >0$, choose $0<\delta <\epsilon /(4M)$ and the finitely many intervals supplied by the hypothesis. On each $[a_i,b_i]$, choose a partition of an interval whose Darboux sum gap is less than $\epsilon /(2n)$, using Riemann integrability there.

Combine all these partitions with $0,1$ and all interval endpoints to obtain a global partition of an interval $P$. On the chosen intervals the gaps add to less than $\epsilon /2$. The complementary intervals have total lengthand on each the difference between the supremum and infimum is at most $2M$. ConsequentlyThe Riemann integrability criterion proves $f$ is Riemann integrable on $[0,1]$. Shared endpoints cause no problem: the Darboux sums use closed intervals, and the complementary interval estimate includes every endpoint value. This proves Riemann integrability from almost full interval coverage.

A Riemann integrable function on a closed interval is bounded and has equal upper and lower Darboux integrals. More explicitly, for a partition of an interval $P:a=x_0<x_1<\cdots <x_m=b$, define the Darboux sumsThen Riemann integrability meansand this common value is the Riemann integral. Equivalently, the Riemann integrability criterion requires that for every $\epsilon >0$ some partition of an interval satisfies $U(f,P)-L(f,P)<\epsilon$. Indeed, the common partition refinement of two partitions increases each lower Darboux sum and decreases each upper Darboux sum, so equality of the two extremal values is equivalent to arbitrarily small gaps.

Set ${\omega }_1=u$, ${\omega }_2=\Omega +v$, ${\omega }_3=w$ and retain terms linear in the perturbations in Euler equations for a torque-free rigid body. Using $I_3=I_1+I_2$ givesConsequently $\ddot{u}={\beta }^2{\Omega }^{2}u$, where $\beta =\sqrt{(a^2-b^2)/(a^2+b^2)}>0$. For $\Omega >0$ the growing solution isThere is also a decaying mode with the opposite exponent. For negative $\Omega$, select the exponent $+\beta |\Omega |$ as the growing one. This is the intermediate axis theorem: steady rotation about the intermediate principal inertia axis is unstable whenever $\Omega \ne 0$.

Take the sheet to lie in the $x_1{x}_2$ plane, with lengths $a,b$ in those directions. Its surface density is $M/(ab)$. Reflection symmetry about the coordinate axes makes the off-diagonal entries of the inertia tensor zero. Its principal moments of inertia areand similarly $I_2=M{a}^2/12$. Since $I_3$ integrates $x_1^2+x_2^2$, the perpendicular axis theorem gives

The given vectors satisfyand $y_k,y_{k+1}$ are independent, so their span is invariant under $A$. On that plane the characteristic polynomial is ${\lambda }^2-2\lambda +2$. ThereforeIndeed $A{v}_{\pm }=y_{k+2}-(1\mp i)y_{k+1}=(1\pm i)v_{\pm }$. No third eigenvalue is determined by these data.

The displayed $x_k=(1,1,1)$ has norm $\sqrt{3}$, although a large-$k$ normalized iterate should have norm one. This harmless source normalization is resolved by dividing all three displayed vectors by $\sqrt{3}$; the recurrence, eigenvalues and eigenvector directions are unchanged.

Since $A$ is diagonalizable over the complex numbers, write $x^{(0)}={\sum }_i{a}_i{v}_i$. Provided iteration does not encounter zero, its direction is that ofWith one strictly dominant eigenvalue and $a_n\ne 0$, all other contributions are smaller by factors $(|{\lambda }_i|/|{\lambda }_n|{)}^k$. The dominant eigenvalue is real, since nonreal eigenvalues of a real matrix occur in conjugate pairs of equal modulus. The normalized power method therefore converges to a signed unit eigenvector if ${\lambda }_n>0$, and alternates between opposite unit eigenvectors if ${\lambda }_n<0$. A quotient such as $(x^{(k)}{)}^{T}A{x}^{(k)}$ approaches ${\lambda }_n$ in either case. If $a_n=0$, replace it by the largest active modulus; if $A{x}^{(k)}=0$, the prescribed iteration is undefined.

With two dominant eigenvalues of equal modulus and both components active, the iterates approach their real invariant plane rather than one direction. Distinct real equal-modulus eigenvalues are $\rho ,-\rho$ and produce a limiting two-cycle. A conjugate pair $\rho e^{\pm i\theta }$ produces a rotation in suitable coordinates on that plane: the normalized directions are periodic when the angle is rational relative to $2\pi$ and otherwise generally keep rotating without convergence. If the two eigenvalues are equal and real, the initial projection is itself one eigenvector in their repeated eigenspace, and converges or alternates as in the single-value case; one orbit cannot recover every independent eigenvector of that eigenspace.

For distinct dominant values, retain unnormalized consecutive vectors $y,Ay,A^{2}y$ in the nearly invariant plane and solveThe roots of ${\lambda }^2-s\lambda +p$ estimate the two eigenvalues. Corresponding eigenvectors are $Ay-{\lambda }_{2}y$ and $Ay-{\lambda }_{1}y$, since applying $A$ and using the recurrence multiplies them by ${\lambda }_1$ and ${\lambda }_2$. This is dominant eigenpair extraction from a two-step Krylov recurrence. Equivalent extraction uses the $2\times 2$ matrix of $A$ on the computed plane. It needs two independent iterates and nonzero projections onto both distinct eigenvectors.

The thin-layer condition isIntegrate the differential equation once: $(f^{\prime \prime }+f{f}^{\prime }{)}^{\prime }=0$. Since $f(0)=f^{\prime \prime }(0)=0$, its integration constant is zero. A second integration gives $f^{\prime }+f^2/2=b$. The outward-jet solution with $b>0$ and $f(0)=0$ isUsing the given fourth-power integral in the normalization gives $1=4k^4[4/(3k)]=16k^3/3$, so $k=(3/16{)}^{1/3}$. Thus the similarity profile of an axisymmetric radial viscous free jet isIt is symmetric, outward, decays in $z$, and has exactly the conserved flux $F$. It describes the region beyond the small injection slit where the stated boundary-layer approximation applies; its formal $r=0$ singularity is outside that approximation.

Let $U(r)$ be the radial velocity scale. Advection-viscosity balance gives $U^2/r\sim \nu U/{\delta }^2$, while the conserved momentum flux gives $F\sim U^{2}r\delta$. Eliminating $U$ yields $\delta /r\sim ({\nu }^2/F{)}^{1/3}$, a constant, so $\delta =Cr$ and $U\propto r^{-1}$.

The streamfunction automatically satisfies incompressibility. With the proposed form it giveswhere $u_r,u_z$ are the two velocity components. Differentiation in the momentum equation cancels the terms proportional to $\eta f^{\prime }{f}^{\prime \prime }$ and leavesChoose $A{C}^2=\nu$. The momentum integral is $F=A^{2}C\int f^{\prime 2}d\eta$, so choosing $A^{2}C=F$ makes its normalization one. ThereforeReflection symmetry gives $u_z=0$ and ${\partial }_z{u}_r=0$ at zero. Taking the streamfunction's odd representative makes $f(0)=0$, and symmetry gives $f^{\prime \prime }(0)=0$; far-field decay gives $f^{\prime }\to 0$. The factor $f$ multiplying $f^{\prime \prime }$ is present in the original PDF but missing from the TeX aid; the direct substitution confirms the PDF equation.

Axisymmetric incompressibility is ${\partial }_r(r{u}_r)+r{\partial }_z{u}_z=0$. Multiply the radial momentum equation by $r$ and use this to put advection in conservative form:Integrating through the layer givesThe boundary terms vanish because radial velocity and its shear decay into the surrounding fluid; an entrainment velocity need not itself vanish as long as its product with $u_r$ does. Hence

For the difference of two solutions the equation is $w_{tt}+w_t-\Delta w=0$ with zero initial data. The same shrinking-ball energy now obeysAgain $E(0)=0$ forces the difference to vanish at every chosen spacetime point. Therefore the two $C^2$ solutions agree for every $t>0$. Damping contributes an additional negative energy term and does not weaken the local uniqueness proof.

Let $w$ be the difference between a $C^2$ solution and the formula in part (ii). It satisfies $w_{tt}-\Delta w=0$ and zero initial data. Fix $(T,x_0)$ and $\epsilon >0$, and put $B_t=B(x_0,T+\epsilon -t)$. Define the local energyDifferentiating on the shrinking ball and integrating by parts givessince $2w_t{\partial }_{n}w\le w_t^2+({\partial }_{n}w{)}^2\le w_t^2+|\nabla w{|}^2$. Initially $E=0$, hence $w_t=\nabla w=0$ inside each ball. In particular $w(t,x_0)$ is constant from its initial value zero, so $w(T,x_0)=0$. Since the point was arbitrary, every $C^2$ solution equals the Kirchhoff solution. This shrinking cone energy argument uses only energy on compact sets, so it does not assume global decay of the competing solution.

Rotate the polar axis to $\xi$. Since $W_t$ is a finite sphere measure, its Fourier transform iswith value $t$ at $\xi =0$. Fourier transforming the wave equation gives ${\hat{u}}_{tt}+|\xi {|}^2\hat{u}=0$ with initial displacement zero and velocity $\hat{g}$. Its solution is this multiplier times $\hat{g}$, so inversion gives $u=W_t\ast g$. Thus the Kirchhoff formula isThis extends to arbitrary smooth $g$, without any growth restriction: for each finite spacetime neighborhood, multiply $g$ by a compactly supported smooth cutoff equal to one on every sphere occurring in that neighborhood. The formula is unchanged there, so its verified differential equation remains valid locally. Its initial values follow from $u(t,x)=tg(x)+o(t)$ and differentiation under the sphere integral. Since the neighborhood was arbitrary, it is a smooth solution everywhere for $t\ge 0$.

Choose the Fourier transform conventionIt maps the Schwartz space to itself. For a tempered distribution $u$, define the transform by bilinear distributional duality:For an ordinary integrable function this agrees with the first formula by interchanging its integrals. The absence of complex conjugation is the distribution-pairing convention.

Use the Runge exhaustion of a slit disk, with the principal argument $-\pi <\arg z<\pi$:These are nested compact subsets of $\Omega$. Their complements are connected: the inner disk and exterior region join through the omitted sector around the negative real radius. Every compact subset of $\Omega$ has positive distance from zero, the unit circle and the slit, so it is contained in $K_m$ for all sufficiently large $m$.

Since $f$ is holomorphic on $\Omega$, it is holomorphic on a neighborhood of each $K_m$. The polynomial Runge theorem gives $P_m$ with ${\sup }_{K_m}|P_m-f|<1/m$. Given any compact $K\subset \Omega$, eventually $K\subseteq K_m$, so the same estimate holds on $K$. ThereforeThe nested exhaustion makes this one sequence work simultaneously on all compact subsets.