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For real $\lambda$, the numerator and denominator have equal complex modulus:The denominator is always nonzero. HenceThus the real axis maps into the unit circle in the complex plane.

Write $z=x+iy$. Squaring the nonnegative complex modulus on each side givesThe linear terms cancel, leaving $3(x^2+y^2)=12$. Therefore the locus isIts real and imaginary axis intersections are $\pm 2$ and $\pm 2i$.

The original PDF places a complex modulus around the quotient; those bars are essential. Its denominator excludes $z=-i$. Using $|z^{\ast }-i|=|z+i|$, the case $\lambda =1$ becomes $|z-i|=|z+i|$. In Cartesian coordinates,This locus is the real axis, the perpendicular bisector of the points $i$ and $-i$ in the complex plane. The excluded denominator point does not lie on it.

A quotient with nonzero denominator has zero complex modulus exactly when its numerator is zero. ThusAt that point $z^{\ast }-i=-2i\ne 0$, so the solution is admissible.

The same completed square gives $|z+b{|}^2=|b{|}^2-c$. A nondegenerate circle therefore occurs exactly whenFor completeness, $c>|b{|}^2$ gives an empty locus because a squared complex modulus cannot be negative.

For arbitrary complex $b$ and real $c$, complete the complex modulus square:The locus is a single point exactly when

The complex modulus is a nonnegative real number. Equality therefore requires $z-1$ itself to be real and nonnegative. Its imaginary part gives $y=0$, and its real part then gives $x-1\ge 0$. Conversely every such point satisfies the equality. Hence the complete solution set isGeometrically this is the closed ray on the real axis starting at $z=1$.

In polar coordinates, $x=r\cos \theta$, $y=r\sin \theta$, with $r=\sqrt{x^2+y^2}$. Thus the polar form of a complex number isHere $r=|z|\ge 0$ is its complex modulus; for $z\ne 0$, the complex argument $\theta$ is determined modulo $2\pi$. At $z=0$ the angle is undefined, although every value of $\theta$ gives the same point when $r=0$.

The complex modulus $|z-(-1-i)|$ is the Euclidean distance from $z$ to $-1-i$. Thus the locus isa circle with centre $(-1,-1)$ and radius $8$. Its extreme horizontal coordinates are $-9$ and $7$, and its extreme vertical coordinates are also $-9$ and $7$.

A squared complex modulus is nonnegative. Conversely, every nonnegative right-hand side of $|z+b{|}^2=|b{|}^2-c$ can be attained. The necessary and sufficient condition is therefore

Completing the square directly with the complex modulus givesFor $|b{|}^2>c$, this is the circle with centre $-b$ and radius $\sqrt{|b{|}^2-c}$. Equality gives just the point $-b$, and $|b{|}^2<c$ gives the empty locus. In particular, the nondegenerate answer is

The two indistinguishable paths to the same final state must be added as probability amplitudes:Taking the squared modulus gives the time-dependent decay rate, with the common final-state phase-space normalization absorbed into the probability amplitudes:The cross term is the interference between direct decay and decay after flavour conversion. For reference, diagonal propagation gives $g_{\pm }=(e_{\ell }\pm e_h)/2$, where $e_j=e^{-i{m}_{j}t-{\Gamma }_{j}t/2}$ and $t$ is proper time. The unfactored expression remains usable if $A_f=0$, when the ratio $\lambda$ itself is undefined.

Put $r=p/(E+m)$ and assume $p>0$. The common plane wave has unit modulus. The component norm isFor unit spinor probability density choosewhere $E^2=p^2+m^2$ was used. This is spinor normalization, not square normalization of a plane wave over infinite space; a box or wavepacket gives the corresponding spatial normalization. If the convention is instead ${\psi }^{†}\psi =2E$, use $N=\sqrt{(E+m)/10}$.

For momentum along $+z$, the positive-helicity components are the first and third, while the second and fourth have negative helicity. HenceBoth are independent of speed. In the source's chiral block order the exact left-chiral weight is $1/2+3p/(10E)$, tending to $4/5$ at high energy; helicity and chirality probabilities agree only in that limit.

Phase matching makes wave amplitudes emitted at different depths add coherently. Consider collinear plane waves in a uniform transparent material of length $L$, with fixed polarization of waves and nonzero selected nonlinear optical susceptibility. Assume negligible pump depletion, negligible absorption, an initially absent harmonic, and a wide beam for which diffraction is negligible. The pump's wave amplitude is then constant throughout the interaction.

The pump electric field has spatial wave phase $k_{\omega }z$; its quadratic source has spatial wave phase $2k_{\omega }z$. The freely propagating harmonic has wave number $k_{2\omega }$. Define the second-harmonic phase mismatch byFor the chosen complex-field convention, the outgoing harmonic wave amplitude, up to an overall constant and propagation phase, is proportional toChoosing the conjugate complex-field convention reverses the exponent's sign without changing the wave intensity. Taking the squared complex modulus givesHere $C$ is fixed for the chosen modes and material; its dimensions accommodate the stated wave intensity convention. The quadratic pump dependence follows because the source wave amplitude is proportional to $E_0^2$, whereas $I_{\omega }$ is proportional to $E_0^2$. The sinc function comes from summing contributions along the whole material, just as a finite aperture produces diffraction through addition of complex amplitudes.

At $\Delta k=0$, the integral is $L$ and the harmonic wave intensity grows as $L^2$. At $\Delta kL=2\pi$, all those contributions cancel. This cancellation needs no absorption: their wave phases wind through a full turn. At high conversion, the constant-pump assumption fails, so the $L^2$ law cannot continue indefinitely. Tight focusing, finite optical pulses, absorption, and transverse separation of modes also require a more complete model.

Optical dispersion generally makes $n_{2\omega }\ne n_{\omega }$. In a suitable crystal, birefringence allows modes with different polarization of waves to experience different refractive indices, making their selected values equal. The appropriate crystal direction and allowed nonlinear optical susceptibility component must both work.^[ref]

For a worked calculation, choose $L_0=1\mathrm{mm}$ and $\Delta k=\pi \mathrm{m}{\mathrm{m}}^{-1}$. At $L=L_0$, the harmonic wave intensity is $4/{\pi }^2\simeq 0.4053$ of its perfectly matched value. At $L=2L_0$ it vanishes in this model. More material can therefore reduce the output. If instead $\Delta k=0$, doubling $L$ multiplies output by four. The two curves retain that common normalization, so their different peak heights are meaningful.

For a polynomial $p(z)=c{\prod }_j(z-a_j)$ with $c\ne 0$, no polynomial root at zero, and $R>0$ different from every $|a_j|$, Jensen's formula gives the displayed expression. The polynomial roots are repeated according to their multiplicity of a zero: an exterior root contributes $\log |a_j|$, whereas an interior root contributes $\log R$.

For an explicit example, take the polynomialIts two simple zeros are $1/4$ and $-1/2$. The circle log-mean of a holomorphic function is thereforeAt $R=0.2$ the open disc contains no zeros of a function, at $R=0.4$ it contains $1/4$ only, and at $R=0.75$ it contains both. These three circles avoid the polynomial roots; the radii $1/4$ and $1/2$ are excluded from the displayed version.

The corresponding values of the circle log-mean of a holomorphic function and the circle geometric mean of a holomorphic function areFor example, at $R=0.4$ the contribution of the sole interior zero of a function is $\log (0.4/0.25)=\log (8/5)$; adding it to $\log (3/8)$ gives $\log (3/5)$. The second polynomial root still affects $f(0)$, but contributes no additional interior-zero term at this radius.

Fix a detected position and polarization of waves. Write the electric field operator as $\hat{E}(t)={\hat{E}}^{(+)}(t)+{\hat{E}}^{(-)}(t)$, where the positive-frequency part contains annihilation operators and ${\hat{E}}^{(-)}=({\hat{E}}^{(+)}{)}^{†}$. An ideal absorbing photodetector responds towith density matrix $\rho$. The diagonal gives the single-count rate up to the detector coupling. Multiple absorptions require higher, normal-ordered field products. Roy J. Glauber used these correlations to develop the quantum account of optical coherence.^[ref]

The ordering follows from a detection amplitude. If absorption takes an initial pure state $|i⟩$ to an unobserved final state $|f⟩$, its field factor is $⟨f|{\hat{E}}^{(+)}|i⟩$. Squaring its complex modulus and summing over a complete orthonormal basis of final states gives $⟨i|{\hat{E}}^{(-)}{\hat{E}}^{(+)}|i⟩$. Averaging over initial preparations replaces this expression by a trace. Thus a detector that absorbs cannot count the Fock vacuum, although the vacuum has quantum fluctuations.

For one selected normal mode, ${\hat{E}}^{(+)}(t)=\mathcal{E}u(t)a$. All Fock states with nonzero occupation, coherent states, and thermal light in this same mode havewhere both denominators are nonzero. Its magnitude is one even when the expected value of the field vanishes. Perfect first-order optical coherence therefore does not identify a coherent state. Pair detection supplies additional information.

For real $\alpha \ne 0,\gamma$ and complex $\beta \ne 0$, completing the square in $\alpha |z{|}^2-\beta \bar{z}-\bar{\beta }z+\gamma =0$ gives center $\beta /\alpha$ and radius $\sqrt{|\beta {|}^2-\alpha \gamma }/|\alpha |$, when the radicand is nonnegative. The displayed map is reflection in the line through $0$ and $\beta$: it preserves the complex modulus, fixes the center, and squares to the identity. Hence it maps the circle onto itself, including a radius-zero degeneration.

Single-slit diffraction is the spreading of light transmitted by one slit. Consider an ideal rectangular slit of width $a$, illuminated normally by a uniform plane wave with a single wavelength $\lambda$ in the surrounding medium. This monochromatic wave has uniform wave amplitude and equal incident wave phases across the slit, providing the required optical coherence. In the scalar Fraunhofer diffraction model, the normalized angular wave intensity isHere $\theta$ is measured from the forward normal in the plane across the slit width. The sinc function uses the unnormalized convention $\mathrm{sinc}u=\sin u/u$. The limit makes the central value one. This ideal envelope neglects additional direction-dependent propagation and detector factors; near the forward direction it describes the normalized screen irradiance. ^[ref].

The noncentral zeros satisfyThe restriction follows from the range of the sine of a real angle. The first minima have $\sin \theta =\pm \lambda /a$ when $\lambda /a\le 1$. If $a<\lambda$, this model has no noncentral zero in the forward hemisphere; equality puts the first zeros at grazing directions, outside the small-angle approximation and off any finite perpendicular screen. ^[ref].

For $0<\lambda /a<1$, define the central width as the separation between the two first minima. Its angular width and its width on a perpendicular screen areThe angular approximation uses radians and requires $\lambda /a\ll 1$; the screen approximation uses $y\approx L\theta$. The tangent relation is exact straight-line geometry, even though the underlying Fraunhofer diffraction field is approximate. This width is distinct from the full width at half maximum.

For $\lambda =500\mathrm{nm}$, $a=0.1\mathrm{mm}$ and $L=1\mathrm{m}$, the positive first minimum isThe dimensionless far-field parameter is $a^2/(\lambda L)=0.02$. Doubling the slit width to $a=0.2\mathrm{mm}$ gives $W\approx 5.00002\mathrm{mm}$ and parameter $0.08$; both curves below use the Fraunhofer diffraction approximation. Choosing $L=2\mathrm{m}$ halves those parameters and doubles both screen widths.

Each curve is normalized to its own $I_0$; the arrows measure between first minima. The calculation uses $\theta =\arctan (y/L)$ before evaluating the sinc function, rather than replacing $\sin \theta$ by $y/L$. The lower diagram shows the exact screen geometry with the vertical scale enlarged. Normalization removes the change in peak brightness when the slit width changes.

The wave interference explanation is especially simple: divide the slit into equal narrow strips and add their complex amplitudes, rather than their wave intensities. With wavenumber $k=2\pi /\lambda$, the continuous slit givesIntegrating the complex exponential gives the endpoint difference; its continuous value at zero is one. Taking the squared complex modulus gives the stated wave intensity. This is the Fourier transform of an interval indicator expressed in angular coordinates. Feynman's continuous-source argument obtains the same pattern by letting the number of sources tend to infinity. ^[ref].

At the positive first minimum, the wave phases span exactly $2\pi$ across the slit. Every strip in one half has a partner $a/2$ away, with equal wave amplitude and phase difference $\pi$, so the pairs cancel. At $\theta =0$, all the wave phases align instead. ^[ref].

The drawing uses $N=8$ equally weighted midpoint samples $x_j/a=-1/2+(j+1/2)/8$, with $j=0,\dots ,7$. At the first minimum their complex amplitudes are a common rotation of the eight roots of unity, so their head-to-tail polygon closes exactly. This finite sum illustrates cancellation; the continuous integral, rather than eight point sources, defines the slit pattern at general angles. The circle shows the continuous range of wave phases; coloured arrows show the eight samples.

A narrower aperture requires a larger angle to accumulate the cancelling phase difference, so narrowing the slit widens the beam. This is a useful limit on geometrical optics: confining light more tightly at an opening increases its subsequent angular spread. At fixed incident irradiance and slit height, reducing $a$ also reduces transmitted power, exposing a practical balance between diffraction-limited resolution and light throughput. In a diffraction grating, this single-slit diffraction envelope modulates the wave interference between different slits.

Let a power series ${\sum }_{n\ge 0}{a}_n{z}^n$ have finite radius of convergence $R$. If its terms were bounded in complex modulus at a point $z$ with $|z|>R$, say $|a_n{z}^n|\le M$, choose $w$ with $R<|w|<|z|$. ThenThe comparison test for series with a convergent geometric series would give absolute convergence at $w$, contradicting its position outside the convergence disc. Hence the terms are unbounded at every such $z$. The argument also works when $R=0$ and gives a stronger obstruction than the usual term test for divergence.