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Write $b_n=4^{-n}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$. The given limiting identity and the Cauchy coefficient formula from part (a) give the boundary contour integralThe contour passes through $z=1$, so its value here is an improper integral using the one-sided branch values. Parametrize its open arc by $z=e^{i\theta }$, $0<\theta <2\pi$. Substitution of the analytic branch of a square root calculated in part (b) givesThe factor $e^{-in\theta }$ comes from cancelling $e^{i\theta }$ in $dz$ against $z^{n+1}$; the extra $e^{-i\theta /4}$ comes from the chosen branch. Keeping both phases prevents a change in the half-integer frequency.

Now put $t=\theta /2$. ThenDividing by $i-1$ givesTaking the real part therefore provesAll these improper integrals converge absolutely: the numerators are bounded, while $(\sin t{)}^{-1/2}$ is of order $t^{-1/2}$ at zero and $(\pi -t{)}^{-1/2}$ at $\pi$. Hence the substitutions and passage to the real part are legitimate. The imaginary part also shows that the corresponding sine integral has the same positive value, which checks the phase calculation.

The two terms in the preceding formula are the real part and imaginary part. Their squared sum is the squared complex modulus:Here ${\cosh }^{2}x-{\sinh }^{2}x=1$ follows directly from the exponential function definitions. The hyperbolic sine modulus identity therefore yieldsEquality holds exactly when $\sin y=0$, that is, when $y$ is an integer multiple of $\pi$. The lower bound is uniform in $y$, which will control whole vertical sides of the contour rather than just their endpoints.

The equation requires $\sinh z=\pm i\cosh a$. Write $z=x+iy$, with real $x,y$. Comparing real parts and imaginary parts using part (b)(i) givesIf $x=0$, the second equality would require $|\sin y|=\cosh a>1$, impossible because $a>0$. Thus $\sinh x\ne 0$, forcing $\cos y=0$ andNow $|\sin y|=1$, so $\cosh x=\cosh a$. The hyperbolic cosine is even and is a strictly increasing function on $[0,\infty )$, hence $x=\pm a$. Conversely, substituting these values gives $\sinh (x+iy)=i(-1{)}^k\cosh a$, so every value found does satisfy the original equation. Therefore the complete set is

Use the upper half-disc of radius $R$, with a clockwise semicircular indentation of radius $\epsilon$ above zero. The integrand is holomorphic inside that indented contour, so its total integral is zero. On the large arc,because $a+b>0$, $a-b>0$ and $\mathrm{Im}z\ge 0$. Its integral has modulus at most $\pi /R$ and vanishes as $R\to \infty$.

On the small clockwise arc, $f(z)=b/z+O(1)$, so the integral tends to $-i\pi b$. Therefore the remaining real-axis integral satisfiesThe imaginary part is even and is an ordinary absolutely convergent integral: near zero its integrand tends to $ab$, and at infinity it is $O(x^{-2})$. Taking imaginary parts givesThe complex integral needs the Cauchy principal value because its real part has a $b/x$ singularity; the requested sine-product integral does not. This sine-product integral with inverse-square weight is an application of the upper-half-plane indentation rule.

For $a\ne b$, integratealong the contour consisting of $[-R,R]$ and the upper semicircle, with positive orientation of a contour and $R>\max \{a,b\}$. Its enclosed simple poles are $ia$ and $ib$, with residuesOn the upper semicircle, $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$. The reverse triangle inequality gives $|z^2+a^2|\ge R^2-a^2$ and $|z^2+b^2|\ge R^2-b^2$, so the estimation lemma givesThe integral along the real axis is absolutely convergent: the denominator stays nonzero on bounded real intervals, and the integrand is $O(x^{-4})$ at infinity. The residue theorem therefore givesTake real parts. The real integrand is an even function, so

To include $a=b$, fix $a>0$ and let $b\to a$. For $a/2\le b\le 3a/2$, the absolute value of the integrand is bounded byan integrable function on $[0,\infty )$. The dominated convergence theorem justifies passing the limit under the integral. If $h(t)=e^{-t}/t$, then $h^{\prime }(a)=-e^{-a}(a+1)/a^2$, andHence

There is no such map for the domain printed in the paper. The planar domain $A=\mathbb{C}\setminus [a,b]$ is not simply connected. For example, a sufficiently large positively oriented circle $\gamma$ surrounds $[a,b]$ and lies in $A$, whileThe integrand is holomorphic on $A$. If $A$ were simply connected, the Cauchy integral theorem would force this integral to be zero. A conformal bijection is a homeomorphism and preserves simple connectivity, whereas $H$ is simply connected. Thus a conformal bijection $A\to H$ is impossible.

The explicit construction also pinpoints the missing point. DefineThis Möbius transformation sends the removed line segment to the nonpositive real axis, together with infinity. More precisely, its image on the stated planar domain issince $T(z)=1$ has no finite solution when $a\ne b$. Conversely, its inverse $z=(bw-a)/(w-1)$ is defined for every $w$ in this set and cannot lie on $[a,b]$, proving this image description exactly.

Take the principal square root of $T(z)$, characterized by positive real part, and then multiply by $i$. The result is the conformal mapEach stage is injective and onto the indicated image; the square root maps the slit plane to the right half-plane, with the point $1$ omitted here. Hence this is a conformal bijection onto the punctured upper half-plane. Its inverse isIf the source is instead $(\mathbb{C}\cup \{\infty \})\setminus [a,b]$ on the Riemann sphere, the same map extends with $F(\infty )=i$ and becomes a conformal bijection onto all of $H$. Indeed, in the local coordinate $u=1/z$ it has expansion $F=i+\frac{i}{2}(b-a)u+O(u^2)$, whose derivative at $u=0$ is nonzero. This gives the sphere version of the requested construction without silently changing the printed planar domain.

In charge-free, current-free vacuum, Maxwell's equations areTaking the curl of the third equation and using the fourth, the curl of the curl identity, and the zero divergence of the electric field givesThe magnetic flux density obeys the same wave equation. For the plane wave factor $e^{i(\omega t-kx)}$, the spatial Laplacian contributes $-k^2$ and the second time derivative contributes $-{\omega }^2$. Thusfor propagation in the positive $x$ direction. The zero-divergence conditions require $E_{0x}=B_{0x}=0$, while the curl equations require ${\mathbf{B}}_0=(k/\omega )\hat{\mathbf{x}}\times {\mathbf{E}}_0$. These conditions show that the proposed plane waves satisfy all four Maxwell's equations, rather than only their derived wave equations. Physical electric fields and magnetic fields are the real parts of the complex expressions.

For electrical resistance $R_p>0$ in a parallel circuit with inductance $L>0$, add electrical admittances and invert:Conjugate impedance matching to the series RC impedance requiresIt is possible at a finite positive angular frequency exactly when $C{R}_p^2>L$, in which caseThe expression for $R$ follows by substituting this ${\omega }^2$ into the real part of $Z$. If $C{R}_p^2\le L$, no such finite matching angular frequency exists. Under literal equality, both passive inductive configurations fail by the sign of their imaginary parts.

Use the $e^{i\omega t}$ phasor convention and interpret impedance matching as conjugate matching for the maximum power transfer theorem. The load electrical impedance of the series circuit isFor $Z=A+i\omega B$, the condition $Z_{RC}=Z^{\ast }$ equates the real parts and imaginary parts:These formulas require $B>0$, $C>0$ and a passive realization with $A\ge 0$, evaluated at the matching angular frequency; if $A,B$ depend on $\omega$, the equation for $\omega$ is implicit.

There is a convention ambiguity in the word “matched”. Literal equality $Z_{RC}=Z$, as required for equal bridge arms in part (b), instead gives $R=A$ and ${\omega }^{2}BC=-1$, requiring $B<0$. The inductive cases below use conjugate matching; literal equality is impossible for either passive inductive configuration at a finite positive angular frequency because its electrical reactance is positive while that of the series RC impedance is negative.

Let the source's top terminal have voltage phasor $V_0$ and its bottom terminal zero. Denote the left and right junctions by $L$ and $R$. The two left resistors are equal, so at balance $V_L=V_0/2$. Zero electric current through the ammeter requires $V_R=V_L$, and therefore equal upper and lower right-arm electrical impedances:Write $x=\omega R_x{C}_x$. Separating real parts and imaginary parts givesThe second equation implies $1+x^2={\omega }^2{R}_x^{2}C{C}_x$. Subtract $x^2={\omega }^2{R}_x^2{C}_x^2$ to obtainA finite balance with positive components requires $0<C_x<C$, and then $0<R_v<R_x$. This equal-ratio RC bridge uses equality of arm electrical impedances, independent of the conjugate-matching convention in part (a).

Detector convention. The stated negligible-electrical impedance ammeter is an ideal short circuit, whose actual terminal voltage is zero at every angular frequency. The nontrivial bridge output usually intended is the open-circuit junction difference, obtained with the detector removed or replaced by a high-input impedance voltmeter. Define its polarity byFor a detector of finite electrical impedance $Z_A$, its terminal voltage is insteadHere $Z_{\mathrm{th}}$ is the Thévenin equivalent electrical impedance, found by shorting the ideal source. This exhibits why the ideal-ammeter limit $Z_A\to 0$ has zero terminal voltage even when the bridge is unbalanced.

time average of harmonic power. Expand the two real parts:The terms proportional to $e^{\pm 2i\omega t}$ average to zero over complete periods, while $F{u}^{\ast }$ is time-independent. For real peak amplitudes this is $F_0{u}_0\cos \varphi /2$.

String dynamics. A short element of length $dx$ experiences transverse force $T[{\psi }_x(x+dx)-{\psi }_x(x)]$. Newton's second law therefore gives $\rho dx{\psi }_{tt}=Tdx{\psi }_{xx}$, orThe approximation is a flexible, lossless string with small slope and constant tension; $\rho$ is its linear mass density.

energy transport. Across a cut toward increasing $x$, the transverse force is $F_y=-T{\psi }_x$ and the transverse velocity is $u={\psi }_t$. For a rightgoing wave these have complex amplitudes $ikTA$ and $i\omega A$. HenceThis real mechanical impedance is force divided by transverse velocity for a travelling wave, with units $\mathrm{kg}{\mathrm{s}}^{-1}$.

Matching and power ratios. At an ordinary massless junction without an external transverse force, displacement is continuous and transverse tension is balanced: ${\psi }_1={\psi }_2$, $T_1{\psi }_{1x}=T_2{\psi }_{2x}$. For the same tension, the slope is therefore continuous even if density changes. With real positive impedances,Their sum equals $P_i$; the transmitted power includes the impedance ratio and is not simply $t_{\psi }^2{P}_i$.

Response and resonance. The terms $m\ddot{x}$, $b\dot{x}$ and $kx$ describe inertia, linear viscous resistance and the spring's restoring force; the right side is the applied harmonic force. Substituting the complex trial motion and taking real parts givesThe steady response excludes the decaying homogeneous transient. Its magnitude is $[(k-m{\omega }^2{)}^2+b^2{\omega }^2{]}^{-1/2}$. Differentiating the denominator gives $2\omega [-2m(k-m{\omega }^2)+b^2]=0$, soA nonzero-frequency amplitude maximum exists only for $b^2<2km$; otherwise the maximum is at zero frequency. Increasing damping lowers and broadens the peak and shifts it downward. Every curve starts at $|R(0)|=1/k$ and tends to $1/(m{\omega }^2)$ at high frequency.

Series circuit. The PDF circuit has the resistor, inductor and capacitor in one loop. With current $I=\dot{q}$, Kirchhoff's voltage law gives $V=L\dot{I}+RI+q/C$, henceThus $m,b,k$ correspond to $L,R,1/C$. The resistor's rms voltage isIt is maximized exactly at ${\omega }_{\max }={\omega }_0=1/\sqrt{LC}$ for $R>0$. The charge-amplitude resonance is lower. Expanding its square root for $R^{2}C/L\ll 1$ givesThe two maxima refer to different observables: resistor voltage follows current, whereas charge follows displacement.

The free electron model replaces the microscopic ionic potential and interacting electronic motion by independent mobile electrons in a uniform neutralizing background, with approximately parabolic dispersion $ϵ={\hslash }^2{k}^2/(2m^{\ast })$. Its quantum version fills states according to Fermi-Dirac statistics and the Pauli exclusion principle. Scattering is represented by a momentum relaxation time $\tau$, rather than derived from the model's ideal free trajectories. This model explains a Fermi energy much larger than room-temperature thermal energy, an electronic heat capacity proportional to temperature at low temperature, and a Drude electrical conductivity. With appropriate scattering assumptions it also yields the Wiedemann-Franz law. It cannot by itself explain band gaps, detailed band anisotropy, strong correlations, or the microscopic temperature dependence of $\tau$.

The inertial term is mass times the rate of change of mean drift velocity. The electric force is $eE$, taking the printed $e$ as the signed electron charge, and $-\gamma ⟨v⟩=-m^{\ast }⟨v⟩/\tau$ describes loss of directed momentum to scattering. electrons retain large microscopic velocities; this damping is of their mean directed motion. In a steady DC field,Both charge factors are negative for electrons, so the conductivity is positive. If $e$ instead denotes a positive charge magnitude, the force and current equations each require a minus sign and give the same result.

For AC response choose physical fields as real parts of phasors proportional to $e^{-i\omega t}$. Substitution givesThusThe real part decreases from ${\sigma }_0$ to zero. The imaginary part starts at zero, reaches ${\sigma }_0/2$ at $\omega \tau =1$, and falls as $1/\omega$. The real part falls as $1/{\omega }^2$ at high frequency. With an $e^{+i\omega t}$ convention, only the sign of the imaginary part changes. Positive $\mathrm{Re}\sigma$ gives positive cycle-averaged dissipation $⟨jE⟩=\mathrm{Re}\sigma |\hat{E}{|}^2/2$.

The supplied DC conductivity determines the transport time:Inertia matters when $\omega \tau \sim 1$, giving ${\omega }_c\simeq 4.02\times 10^{13}\mathrm{rad}{\mathrm{s}}^{-1}$, or ${\nu }_c={\omega }_c/(2\pi )\simeq 6.40\times 10^{12}\mathrm{Hz}$. Quoting the former number as hertz would introduce an erroneous factor of $2\pi$.

Distributed circuit model. A transmission line must be treated as a distributed system when its propagation time is appreciable compared with the signal timescale. Divide a uniform two-conductor line into elements of length $dx$. Each has series resistance $R^{\prime }dx$ and inductance $L^{\prime }dx$, and shunt electrical conductance $G^{\prime }dx$ and capacitance $C^{\prime }dx$. The primes denote quantities per unit length, with units $\mathrm{\Omega }{\mathrm{m}}^{-1}$, $\mathrm{H}{\mathrm{m}}^{-1}$, $\mathrm{S}{\mathrm{m}}^{-1}$ and $\mathrm{F}{\mathrm{m}}^{-1}$, respectively.

Kirchhoff's voltage and current laws give the Telegrapher's equations:For a lossless line, $R^{\prime }=G^{\prime }=0$, differentiation gives $V_{xx}=L^{\prime }{C}^{\prime }{V}_{tt}$ and the same wave equation for $I$. The propagation speed and characteristic impedance areA general lossless signal has $V=V^{+}(t-x/u)+V^{-}(t+x/u)$ and $I=[V^{+}(t-x/u)-V^{-}(t+x/u)]/Z_0$. Thus characteristic impedance is the voltage/current ratio of one travelling wave, not simply the resistance of the wires. It can be real even when the line dissipates no energy, because energy is carried away by propagation.

For sinusoidal quantities with convention $e^{i\omega t}$, the general lossy-line formulas areChoose the square-root branches for forward attenuation and positive forward power. The real part of ${\gamma }_{\mathrm{prop}}$ is attenuation per unit length; its imaginary part is the phase wavenumber. For small losses, attenuation is approximately $R^{\prime }/(2Z_0)+G^{\prime }{Z}_0/2$, using the lossless real $Z_0$ in this estimate.

Fields and geometry. Two separated conductors support a transverse electromagnetic mode, with electric and magnetic fields transverse to the propagation axis. For a homogeneous dielectric and negligible conductor penetration, both examples below have $L^{\prime }{C}^{\prime }=\mu ϵ$ and hence $u=1/\sqrt{\mu ϵ}$. The distributed circuit model follows from the field energy and charges in each slice.

For a coaxial cable with inner radius $a$ and inner surface of the outer conductor at $b$, Gauss's law and Ampère's law giveThe electric field is radial and the magnetic field circles the axis. Integration of voltage and field energy givesThe equal and opposite return current on the outer conductor confines the ideal external field to the annulus.

For wide parallel plates of width $w$ separated by $d\ll w$, neglect fringing. The electric field crosses the gap, $E\simeq V/d$, while the magnetic field lies in the gap parallel to the plates and perpendicular to propagation, $H\simeq I/w$. ThereforeFor a forward wave, $\mathbf{E}\times \mathbf{H}$ points along the line. Integrating this Poynting vector over either cross-section gives instantaneous power $VI$. For complex peak amplitudes, average power is $\mathrm{Re}(V{I}^{\ast })/2$, or $|V^{+}{|}^2/(2Z_0)$ for a single forward wave on a lossless line. The conductors guide an energy flux predominantly in the dielectric between them.

Terminations and reflection. At a load $Z_L$, require $V_L=Z_L{I}_L$. If $V^{+}$ and $V^{-}$ are the incident and reflected voltage amplitudes there, the total current is $(V^{+}-V^{-})/Z_0$. ThusA matched load, $Z_L=Z_0$, absorbs the incident power with no reflection. An open circuit has ${\Gamma }_V=+1$, a current node and a doubled voltage at the end. A short circuit has ${\Gamma }_V=-1$, a voltage node and a doubled current there. A purely reactive load on a lossless line has $|{\Gamma }_V|=1$ and accepts no average power. For real $Z_0$, the reflected power fraction is $|{\Gamma }_V{|}^2$ and the accepted fraction is $1-|{\Gamma }_V{|}^2$.

Incident and reflected waves produce a standing-wave pattern with voltage standing wave ratio $(1+|{\Gamma }_V|)/(1-|{\Gamma }_V|)$. The input impedance of a lossless length $\ell$ isThis gives $i{Z}_0\tan (k\ell )$ for a short and $-i{Z}_0\cot (k\ell )$ for an open. A quarter-wave section gives $Z_{\mathrm{in}}=Z_0^2/Z_L$, enabling impedance transformation. After a source launches a pulse, a distant termination affects it only after the reflected pulse returns; the return time is $2\ell /u$. Real conductor resistance, dielectric loss and departures from a TEM field profile limit the ideal model at high frequency.

For free parallel RLC voltage decay, measure $V=V_A-V_B$ and branch currents from A to B. ThenThere is no external current, so Kirchhoff's current law gives $I_C+I_R+I_L=0$. Differentiating yieldsFor $V\propto e^{st}$ the characteristic equation is $C{s}^2+s/R+1/L=0$. Its underdamped roots giveThe supplied inequality makes $\omega$ real. A real physical voltage has the general form $V=A{e}^{-\gamma t}\cos (\omega t+\varphi )$, the real part of the printed complex expression. A general complex-valued solution uses both roots, $e^{-\gamma t}(C_{+}{e}^{i\omega t}+C_{-}{e}^{-i\omega t})$. The inductor's initial current is fixed consistently by the undifferentiated current balance; it is not an arbitrary additional voltage-mode constant.

The force law separates into two identical harmonic oscillator equations:For positive $m,k$, the general real solution isSince $\mathrm{Re}(e^{i\alpha })=\cos \alpha$ by Euler's formula, these are exactly the stated real parts of complex exponentials. The amplitudes $x_0,y_0$ and phases ${\varphi }_1,{\varphi }_2$ are fixed by the four initial position and velocity components. This is two-dimensional isotropic harmonic oscillator flow: the same angular frequency acts in every direction.

Let $b$ denote buoyancy in a stably stratified Boussinesq approximation. Geostrophic balance and hydrostatic balance imply the thermal wind relation $f_0{U}_z=-{\bar{b}}_y$. Consequently the specified vertical shear requiresFor northern-hemisphere $f_0>0$ and $\Lambda >0$, the mean potential temperature decreases northward and increases upward: ${\bar{\Theta }}_y=-({\Theta }_0/g)f_0\Lambda$, ${\bar{\Theta }}_z=({\Theta }_0/g)N^2$. The corresponding mean isentropes slope upward toward the north, with $dz/dy=f_0\Lambda /N^2$. Stable potential-temperature stratification does not require the actual temperature to increase upward; the dry adiabatic lapse also enters that comparison.

In the rigid-lid constant-shear baroclinic model, perturbation quasi-geostrophic potential vorticity isThe mean interior potential-vorticity gradient vanishes on this constant-$f_0$, constant-$N$, linear-shear basic state. Linear advection of that perturbation therefore becomes $({\partial }_t+\Lambda z{\partial }_x)q^{\prime }=0$, the supplied interior equation. It retains slow balanced motion and omits friction, planetary-vorticity gradients and nonlinear disturbance advection.

Substitution of ${\psi }^{\prime }=\Psi (z)\cos (ly)e^{ik(x-ct)}$ givesA possible branch has zero interior perturbation potential vorticity, soBoth supplied applications have $f_0>0$, so this equals the source's expression. For an unstable mode the nonzero imaginary part of the phase speed prevents a real critical level $\Lambda z=c$; for real speeds the displayed solution remains a valid zero-interior-PV branch away from, and smoothly through, any critical level. Taking the real part gives the physical disturbance.

The rigid lids enforce $w^{\prime }=0$. Linearized adiabatic buoyancy conservation isUse the geostrophic conventions $v^{\prime }={\psi }_x^{\prime }$ and $b^{\prime }=f_0{\psi }_z^{\prime }$. At a lid this gives $({\partial }_t+\Lambda z{\partial }_x){\psi }_z^{\prime }-\Lambda {\psi }_x^{\prime }=0$, henceThese are the no-normal-flow condition expressed through boundary buoyancy advection, not conditions of zero horizontal velocity. Boundary temperature anomalies can interact even though this solution has no interior perturbation PV, allowing baroclinic instability to extract mean available potential energy.

For $l=0$ put $L_D=NH/f_0$, $\eta =k{L}_D$ with $k>0$, and write the discriminant asThe second factor is positive for every $\eta >0$, since $x>\tanh x$. The first changes sign at the positive solution of $x=\coth x$, $x=\eta /2$. Numerically $x_c=1.19968$, so ${\eta }_c=2.39936$. The growing branch exists for $0<k{L}_D<{\eta }_c$, givingEquality is neutral, so this is an infimum wavelength, rather than an attained shortest growing mode.

In the unstable interval,Both phase-speed branches have the same real part there; the imaginary parts have opposite signs. At and beyond the short-wave cutoff the growth rate is zero and the two real speeds areThey split continuously from $1/2$ and approach the boundary-flow speeds zero and $\Lambda H$ at large $\eta$. As $\eta \to 0$, $D=-{\eta }^2/12+O({\eta }^4)$, so $\sigma \to 0$ linearly with $k$, despite a finite limiting $c_i=\Lambda H/\sqrt{12}$. A zonally uniform disturbance has no exponential growth from this mechanism.

The maximum growth minimizes $D$. Its derivative isThe positive interior minimum is at ${\eta }_m\simeq 1.606$, yieldingThe supplied phase-speed formula therefore predicts fastest growth at a wavelength of order four deformation lengths, not at the short-wave cutoff.

For nonzero meridional wavenumber,At least one growing zonally varying mode can exist only ifThe upper value is a supremum approached as $k\to 0$, where the growth itself tends to zero. Thus not every permitted meridional wavenumber has the same fastest-growing zonal scale; nonzero $l$ restricts the available band of $k$ and weakens the growth factor $k/\sqrt{k^2+l^2}$ at fixed $\eta$.

Using $N=\sqrt{N^2}$, not the tabulated squared value itself, givesThe planetary radii put these into angular context: $l_{\sup }R$ is about 280 for Earth and 149 for Jupiter, corresponding to minimum meridional angular wavelengths about $1.28^{\circ }$ and $2.42^{\circ }$, respectively. They are reciprocal-length wave numbers; the dimensionless angular counts should not be substituted for $l$ in the Cartesian dispersion relation. If a channel admits a half-wave across its width, the corresponding minimum width is $\pi /l_{\sup }$ rather than a full wavelength.