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Use the $e^{-i\omega t}$ convention, so passive photon absorption has positive imaginary permittivity. For a single Lorentz oscillator model resonance, write the photon energy as $E=\hslash \omega$ andHere $S$ has units of energy squared, $\Gamma =\hslash \gamma$ is a damping energy, and ${ϵ}_b$ is an approximately constant background. The corresponding real and imaginary parts areThus the real part starts above the background, rises on the low-energy side of the resonance, drops rapidly through ${ϵ}_r^{\prime }={ϵ}_b$ at $2\mathrm{eV}$, and approaches the background from below at larger energy. The imaginary part starts at zero, has a positive photon absorption peak near $2\mathrm{eV}$, and decays above it. For weak damping its peak is close to the resonance, but not exactly fixed there for arbitrary damping. At asymptotically large energy the corrections scale as ${ϵ}_r^{\prime }-{ϵ}_b\sim -S/E^2$ and ${ϵ}_r^{\prime \prime }\sim S\Gamma /E^3$.

The resonance position is fixed at 2 eV; peak height, width and background are not specified. The figure chooses an explicit finite damping and oscillator strength only to illustrate the shape over 0–5 eV. In particular, a negative real permittivity above resonance is possible for suitable strength and damping, but is not forced by the resonance energy alone. Setting damping exactly to zero would give a singular idealized response instead of a finite-width photon absorption peak.

Use physical fields proportional to the real part of $e^{-i\omega t}$. The Drude frequency response isThus the sketch is an even Lorentzian line shape, with maximum ${\sigma }_0^2$ at zero frequency, half that value at $|\omega |={\tau }^{-1}$, and an ${\omega }^{-2}$ tail. The positive-frequency half is shown in the preceding figure. Squaring the modulus differs from plotting either $|\sigma |$ or the real part of $\sigma$.

For a single carrier species, the plasma frequency and electrical conductivity satisfyThereforeThe corresponding relaxation time is $\tau \simeq 3.57\times 10^{-14}\mathrm{s}$. This is an angular frequency; the ordinary frequency is $\boxed{f_{1/2}={\omega }_{1/2}/(2\pi )\simeq 4.46\times 10^{12}\mathrm{Hz}}$. The units of electrical conductivity used in the calculation are ${\mathrm{\Omega }}^{-1}{\mathrm{m}}^{-1}$, consistently with the SI plasma-frequency relation.

Set $D=({\omega }_0^2-{\omega }^2{)}^2+{\gamma }^2{\omega }^2$. The real part and imaginary part of the complex permittivity areFor positive frequency the loss part is nonnegative. The real part starts at $1+{\omega }_p^2/{\omega }_0^2$, rises below resonance, falls steeply through the resonance region, and approaches one from below at high frequency. The imaginary part is a positive absorption peak near ${\omega }_0$, tending to zero on both sides.

For a nonmagnetic medium write its complex refractive index as $N=n_r+i\kappa$, with $N^2={ϵ}_r$ and optical extinction coefficient $\kappa$. In a transparent region with $\kappa \ll n_r$ and ${ϵ}_r^{\prime }>0$,The real part controls phase velocity and optical dispersion; the loss part controls wave absorption and the cycle-averaged dissipated power density $\omega {ϵ}_0{ϵ}_r^{\prime \prime }|E_0{|}^2/2$. Weak damping gives a narrow line but does not make wave absorption weak at its centre. The transparent-medium approximation applies away from the line.

For the Schrödinger equation with a real scalar potential and no vector potential, the probability current isTogether with the probability density $\rho =|\Psi {|}^2$, it satisfies the continuity equation ${\partial }_t\rho +\nabla \cdot \mathbf{j}=0$. Substituting ${\partial }_x\Psi =ik\Psi$ givesA complex wavefunction requires both its real and imaginary parts in the sketch. Choosing the irrelevant constant phase so that $A$ is real, they are $A\cos kx$ and $A\sin kx$, while the probability density is horizontal. It is the spatial phase gradient of a Schrodinger wavefunction ${\partial }_x\arg \Psi =k$ that determines the sign of the probability current. In particular, $e^{ikx}$ and $e^{-ikx}$ have the same real part but opposite probability currents, so the slope of the real part cannot determine the direction.

A nonzero plane wave on the whole real line is not square integrable. The displayed density is understood with box normalization, generalized normalization, or as a local description of a broad wave packet; it does not define a normalized probability distribution on the whole line.

Parametrize the positively oriented unit circle by $z=e^{i\theta }$. Then $e^{-1/z}=e^{-\cos \theta +i\sin \theta }$ and $dz=i{e}^{i\theta }d\theta$, soEquating the real part and imaginary part with $-2\pi i$ gives

Split the sine into exponentials. The $e^{iz}$ term closes in the upper half-plane and picks up the pole $ia$, giving $2\pi i{e}^{-a}$; the $e^{-iz}$ term closes clockwise below and encloses no pole. Jordan's lemma applies to these separate contours. HenceTo pass to $a\downarrow 0$ without an invalid dominated-convergence argument on an infinite interval, pair positive and negative $x$ first: the imaginary part is odd and the real part is $2{\int }_0^{\infty }x\sin x/(x^2+a^2)dx$. On a bounded interval the limit is harmless, while integration by parts bounds the oscillatory tail uniformly for $0<a\le 1$. This gives the convergent improper Dirichlet integral $\boxed{{\int }_{-\infty }^{\infty }\sin x/xdx=\pi }$.

For the stochastic differential equation $d{X}_s=b(s,X_s)ds+\Sigma (s,X_s)d{B}_s$, the diffusion generator on smooth functions isOne form of the Feynman-Kac formula is the following. Suppose the stochastic differential equation is well posed and nonexplosive, $c\ge 0$ is bounded, the terminal data $g$ and source $h$ are bounded, and $F$ is a classical solution bounded on finite time intervals ofWith the regularity needed for the Itô formula, the parabolic Feynman-Kac formula with a source statesHere ${\mathbb{E}}_{t,x}$ means expectation for the diffusion process started at $X_t=x$. Under hypotheses ensuring existence of a classical solution, this representation also identifies that solution uniquely in the stated bounded class. In the remaining parts, $c=h=0$, so $F(t,x)={\mathbb{E}}_{t,x}g(X_T)$ solves the terminal-value Kolmogorov backward equation $F_t+{\mathcal{L}}_{t}F=0$. Complex terminal data are allowed by applying the real formula to their real part and imaginary part separately.

For $0<\alpha <2$, write $z=r{e}^{i\theta }$ with $0<\theta <\pi$. The chosen power map sends this to $r^{\alpha }{e}^{i\alpha \theta }$, soAt $\alpha =2$ this is the plane slit along the nonnegative real axis. For every $0<\alpha \le 2$, the map is conformal on the upper half-plane.

To check finiteness of the physical exit time, start a Brownian motion $B$ at $i$ and let $T$ be its real-axis exit. The time change of a continuous process in conformal invariance of planar Brownian motion isWe know $T<\infty$ almost surely. The Cauchy exit law from a Brownian half-plane assigns probability zero to $B_T=0$. On such a path, $B_s\ne 0$ throughout the compact interval $[0,T]$, because it is in the open upper half-plane before exit and has nonzero endpoint. Its modulus therefore has a positive minimum and finite maximum. The displayed integrand is bounded on that interval, proving $A_T<\infty$. The image Brownian motion exits $D$ at this finite changed time.

The endpoint $\alpha =0$ printed in the source is degenerate: $z^0=1$, so its image is the singleton $\{1\}$, not an open domain, and the map is not conformal. If exit from that set is defined for Brownian motion started at one, its exit infimum is zero. It cannot be included in a nondegenerate conformal-domain argument.

Here is the martingale argument for maximal comparisons, including the starting-value issue. For $P=f+ig$, the Cauchy-Riemann equations and Itô's formula show thatare orthogonal continuous local martingales with the same quadratic variation. Localize before large disks and before exit. The Burkholder-Davis-Gundy inequalities then givewith a constant depending only on $p$; Fatou's lemma removes the localization. These are centered martingales. For the uncentered processes this impliessince $|f(i^{\alpha })|\le \sup |f(Z_t)|$.

The source normalizes $P(0)=0$ at a boundary point, rather than the value of $g$ at the starting point. Consequently its displayed constant cannot be uniform in $\alpha$ with only $p$ dependence. To see this, take $P(z)=iz$ and $0<\alpha <\min (1/2,1/p)$. Then $f=-\mathrm{Im}z$ and $g=\mathrm{Re}z$. Inside this narrow wedge,Both $p$th moments are finite and the right one is positive; finiteness follows from the wedge radial-maximum tail of order $R^{-1/\alpha }$, justified below. The source inequality would force $C_p\ge {\cot }^p(\alpha \pi )$, unbounded as $\alpha \downarrow 0$. At $\alpha =0$, the immediate-exit interpretation gives the direct contradiction $1\le C_{p}0$ for the same polynomial.

For a fixed nondegenerate wedge, an uncentered inequality does hold with a constant allowed to depend on $\alpha$ as well as $p$. One rigorous outline for the polynomial class in this question is as follows. Separation of variables in a sector of radius $R$ gives, for exit through its circular arc, the expansionThus at the specified starting point the radial-maximum tail is comparable to $R^{-1/\alpha }$. Hitting any fixed interior subarc has that same power, with positive leading coefficient. If a nonzero polynomial has degree $d$, its real part has magnitude at least $c{R}^d$ on some fixed interior subarc for sufficiently large $R$. Therefore its maximal real-part $p$th moment is infinite whenever $dp\ge 1/\alpha$. In that case the desired extended-real inequality is automatic.

The remaining polynomials have $d<1/(\alpha p)$, so belong to a finite-dimensional vector space space. On that spaceis a genuine norm: if it is zero, Brownian motion's positive probability of reaching any interior neighborhood forces $\mathrm{Re}P=0$ throughout $D$, so analyticity makes $P$ a purely imaginary constant, and $P(0)=0$ makes it zero. The corresponding imaginary-part functional is bounded on this finite-dimensional vector space normed space; its finiteness follows from $dp<1/\alpha$ and the radial tail bound. equivalence of norms in finite dimensions givesThis establishes the fixed-domain interpretation and identifies exactly why the universal centered martingale inequality cannot silently be applied to the printed uncentered boundary normalization. It is the distinction in starting-value obstruction in wedge conjugate maximal inequalities.

The velocity coupling changes the momenta and the undifferentiated field terms. Variation with respect to ${\varphi }_2$ gives the equation for ${\varphi }_1$:so the two Euler-Lagrange field equations areThey describe damped and antidamped fields from an antisymmetric velocity coupling. With the convention ${\varphi }_j=A_j{e}^{i(k_{j}x-{\omega }_{j}t)}$, their characteristic equations areThus the dispersion relations and modes areFor each spatial Fourier mode, the general solution is the linear combination of its two temporal branches; spatial superposition gives the general field. Take real parts or suitable complex conjugate pairs to obtain the required real fields. At the repeated-root threshold, the independent second temporal solution contains an extra factor $t$, as detailed next.

For the modal matrix $M(s)$, $s\ge 0$, its trace is always negative:Its determinant isThe proposed inequality is exactly $\beta \kappa \eta <\alpha \kappa +1$. It makes the coefficient of $s$ positive, while the leading and constant coefficients are already positive. Therefore $D(s)>0$ for every $s\ge 0$, independent of which discrete wave numbers the domain allows.

A real two-by-two matrix with negative trace and positive determinant has both eigenvalues with negative real part: if they are real they have positive product and negative sum, and if nonreal their common real part is half the trace. Every mode therefore decays, proving linear stability for all domain sizes under the stated sufficient condition.

Let $S(x,y,z)=(-x,-y,z)$. In the first two components of the vector field, replacing $x,y$ by their negatives changes the sign of the whole component; in the third, the product $xy$ is unchanged. Thus $F(Sv)=SF(v)$, establishing the symmetry, with solutions mapped to solutions by $S$.

At an equilibrium point, positivity of $a$ makes $y=x$. The remaining equations give $x(2c-a-z)=0$ and $z=x^2/b$. Hencewhere the two nonzero equilibrium points exist when $0<a<2c$. They merge with the origin at $a_p=2c$, in a pitchfork bifurcation.

At the origin the Jacobian matrix and its eigenvalues areAt $a=2c$ these are $0,-c,-b$. To see the local pitchfork bifurcation dynamics more explicitly put $\delta =2c-a$, $y=x+v$, and use $z=x^2/b+\cdots$ on the center manifold. The equations becomeAt $\delta =0$, invariance at cubic order gives $v=-x^3/(bc)+\cdots$, hence the reduced equation isIt has the supercritical pitchfork bifurcation form as $\delta$ increases, with the two branches $x\simeq \pm \sqrt{b\delta }$ attracting near onset while the origin becomes unstable.

For $0<a<2c$, the determinant of the planar block is $a(a-2c)<0$, so the origin has one positive and two negative eigenvalues and is a saddle equilibrium. At $a=c$, the planar eigenvalues are explicitly $\pm c$. For $a>2c$, the planar trace $c-a$ is negative and its determinant positive, so the origin is asymptotically stable. To distinguish its type putFor $2c<a<a_{\ast }$ both planar eigenvalues are real negative, giving a stable node. At $a=a_{\ast }$ they coincide at $-c/\sqrt{3}$; the planar block is not a scalar matrix, so it is a defective stable node. For $a>a_{\ast }$ they are nonreal with negative real part, giving a stable spiral in the planar block together with the stable third direction. This refines the requested node-versus-saddle classification.

The origin cannot have a Hopf bifurcation: a planar imaginary pair would require trace zero, namely $a=c$, but the determinant there is $-c^2<0$, so the pair is real and of opposite signs.

For either nonzero equilibrium point put $s=\pm \sqrt{b(2c-a)}$. The Jacobian matrix isIts characteristic polynomial isA nonzero imaginary root of an equation $i\omega$ requires, by separating real and imaginary parts, ${\omega }^2=B$ and $AB=C$. Equivalentlyso the candidate Hopf bifurcation values areAt such an admissible value the factorization is $p(\lambda )=(\lambda +A)({\lambda }^2+bc)$, giving the pair $\pm i\sqrt{bc}$ and a negative third eigenvalue $-A$, since $A=C/B>0$ on the nonzero branches.

The mathematical model parameter restrictions matter. A real candidate requires $b/c\le 17/8$, and the nonzero equilibrium points also require $0<a_h<2c$. The upper-sign value is admissible for $0<b/c<17/8$; the lower-sign value additionally requires $b/c>1$. For fixed $b,c$, differentiation of the characteristic equation at the imaginary root of an equation gives the eigenvalue crossing speedThus the transverse eigenvalue crossing condition holds except at the coincident-root case $b/c=17/8$, $a_h=3c/4$. At a transverse eigenvalue crossing, a generic nonzero first Lyapunov coefficient gives a Hopf bifurcation. The displayed radical is the spectral threshold; it does not justify an unconditional generic-Hopf claim for all positive mathematical model parameters, nor does it by itself exclude a degenerate Hopf bifurcation point. For example $b>17c/8$ yields no real candidate at all. The cubic Routh-Hurwitz stability criterion gives local Lyapunov stability on the side with $A>0$ and $AB-C>0$, where

For the modified equation, the planar block at the origin is $\left(\begin{array}{cc}-a& a\\ c& c\end{array}\right)$, with determinant $-2ac<0$ for the stipulated positive mathematical model parameters. Thus the origin is always a saddle equilibrium for nonzero $a,c$ in that mathematical model parameter range. The nonzero equilibrium points now satisfy $y=x$, $z=2c$ and $x^2=2bc$, so they areFor $b,c>0$ these branches stay a positive distance from the origin and do not merge with it as $a$ varies. The origin also has no zero eigenvalue for $a,c>0$. Therefore the pitchfork bifurcation is absent in the modified system.

For a nonzero, time-independent vector $\hat{\mathbf{u}}$, substitute $\mathbf{u}(t)=\hat{\mathbf{u}}{e}^{i\omega t}$. Taking two time derivatives givesCancel the nonzero exponential to obtain the eigenvalue problemThus an eigenvector specifies a normal mode shape and its negative eigenvalue specifies the square of its angular frequency, $\omega =\sqrt{-\lambda }$. Real displacements are obtained by taking the real part, or by combining the two temporal solutions as $\hat{\mathbf{u}}(a\cos \omega t+b\sin \omega t)$.

For $0\le r<1$, this Poisson kernel for the unit disk has the uniformly convergent Fourier seriesThe identity follows by taking the real part of the geometric series $1+2{\sum }_{n\ge 1}(r{e}^{i\theta }{)}^n$. For integrable $2\pi$-periodic boundary data $f$, its Poisson extension iswhere $a_n,b_n$ are the usual Fourier coefficients. Each term is a harmonic function in the open disk. For continuous boundary data, the extension solves the Dirichlet problem with those boundary values.

A nonzero complex number has polar formwhere $r$ is its modulus and $\theta$ is an argument. The equality uses Euler's formula; sine and cosine recover the imaginary part and real part from polar coordinates. For $z=3+4i$, take $r=5$ and $\theta =\arctan (4/3)\approx 0.927295$ radians, so $\cos \theta =3/5$ and $\sin \theta =4/5$.

If $w=s{e}^{i\varphi }\ne 0$, thenThus multiplication multiplies moduli and adds arguments modulo $2\pi$; division divides moduli and subtracts arguments modulo $2\pi$. Since $1+i=\sqrt{2}{e}^{i\pi /4}$, multiplying $3+4i$ by $1+i$ gives $-1+7i$, of modulus $5\sqrt{2}$ and argument $\theta +\pi /4\approx 1.712693$ radians. One formula simultaneously describes its new size and direction.

Repeated multiplication now becomes especially simple: $(r{e}^{i\theta }{)}^n=r^n{e}^{in\theta }$ for a positive integer $n$. This explains the evenly spaced arguments of the roots of a complex number. For example, $(3+4i{)}^2=-7+24i$ has modulus $25$ and principal argument $2\theta \approx 1.854590$ radians. The Cartesian calculation and the polar form provide independent checks on both its location and its size. Choose polar form when the task concerns multiplication, division, or repeated turns; use real parts and imaginary parts for addition.

Let $a>0$, $t>0$ and $z=t+ia/t$. The real and imaginary coordinates obey $XY=a$, a hyperbola. Put $s=t^2-a^2/t^2$, which is strictly increasing. Then $z^2=s+2ia$ traces a horizontal line, and $z^4=s^2-4a^2+4ias$ traces a parabola. Eliminating $s$ gives the displayed quadratic relation between real part and imaginary part. The value of $t$ fixes an orientation and determines which segment is traversed; the full real parabola is obtained as $t$ ranges over $(0,\infty )$.

The principal cube root is $z^{1/3}=\exp (\mathrm{Log}z/3)$ using the principal complex logarithm. In the right half-plane, its complex argument lies between $-\pi /6$ and $\pi /6$. Consequently the roots of $r^3=-z$ consist of one root with negative real part, $-z^{1/3}$, and two with positive real part, $z^{1/3}{e}^{\pm i\pi /3}$.

For a contour integral with factor $e^{{\zeta }^4/4+z\zeta }$, the quartic term dominates the linear term on rays at large radius. Writing $\zeta =r{e}^{i\theta }$ gives real part $r^4\cos (4\theta )/4$ for the leading exponent. There are four decay sectors, centered at $\theta =\pi /4+j\pi /2$ and each of angular width $\pi /4$. On a closed subsector the integrand and all its $z$ derivatives decay faster than any exponential in $r$, locally uniformly in $z$. Contours joining two such sectors therefore allow differentiation under the integral sign and have vanishing infinite endpoint terms. Distinct geometric contours still need an independence proof; their initial moments often give a Vandermonde matrix calculation.

A leading fixed-discharge, level-surface approximation to a shallow erodible-bed flow gives perturbation equations $C_x=4U-C$ and $U_t+3\beta U_x=-4U+C$. The displayed Fourier growth rate has negative real part for every nonzero wave number, independently of bedload parameter $\beta$. Bedload changes the propagation speed but not this damping. The zero-wave-number mode is neutral, so absence of growing modes should not be confused with strict asymptotic decay of every spatially uniform perturbation.

For equal exterior potential energies and constant mass, the exact transmission probability isBelow the barrier, the hyperbolic sine is positive for $a>0$, so $0<T<1$. At $E=V_0$, the interior wavefunction is linear in $x$; either neighbouring formula has the displayed finite limit, since $\sinh z/z$ and $\sin z/z$ tend to one. Above the barrier, the interior wavefunction oscillates: the quantum reflection probability generally remains nonzero, but $qa=n\pi$ for a positive integer $n$ gives perfect transmission probability $T=1$ through constructive wave interference. These are scattering resonances, not below-barrier quantum tunnelling.

At half height, $E=V_0/2$, we have $k=\kappa$ and $T={\mathrm{sech}}^2(\kappa a)$, consistent with finite square barrier transmission at half barrier height after replacing this width $a$ by that example's width $2a$. In the illustration, the width is four times $\ell =\hslash /\sqrt{2m{V}_0}$ and $E/V_0=1/2$, giving $T\simeq 0.01388$. The plotted wavefunction comes from solving the four interface equations; the transmission probability curve uses the closed formula.

Take a straight hollow rectangular electromagnetic waveguide with vacuum interior $0<x<a$, $0<y<b$ and axis $z$, where $a>b>0$. Its four walls are ideal perfect conductors. The time-dependent Maxwell equations require zero tangential electric field and zero normal oscillating magnetic field at these walls. Real metal has finite electrical conductivity, so real guides lose electromagnetic energy by Joule heating; the perfect-wall model isolates confinement and wave dispersion from that loss.

The simplest propagating transverse electric mode of a rectangular waveguide is ${\mathrm{TE}}_{10}$. “Transverse electric” means $E_z=0$, not that the magnetic field is also wholly transverse. With real peak scale $E_0$, its complex amplitudes, including the common axial factor, can be writtenThe physical fields are the real parts after multiplication by $e^{-i\omega t}$. In particular, $E_y$ is zero at $x=0,a$, maximal at $x=a/2$, and independent of $y$. At $y=0,b$ it is normal to the metal, so it need not vanish there. The ${\mathrm{TE}}_{10}$ electric field crosses the narrow gap, while its strength varies across the broad dimension.

For an original example choose $a=30\mathrm{mm}$, $b=12\mathrm{mm}$ and $f=1.25f_c$, where $f_c=c/(2a)$. With $c=299792458m/s$, this givesHere ${\lambda }_0=c/f$ is vacuum wavelength and ${\lambda }_g=2\pi /\beta$ is guided wavelength. The next cutoff is ${\mathrm{TE}}_{20}$ at $2f_c$; ${\mathrm{TE}}_{01}$ starts at $(a/b)f_c=2.5f_c$. Thus only ${\mathrm{TE}}_{10}$ can propagate at the chosen frequency, although a source or discontinuity can additionally excite local evanescent waves. These are calculated values, not specifications or measurements of a commercial guide.