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For a Fourier mode, the amplification polynomial isPut $a=1-2\mu$ and $\xi =e^{i\theta /2}\eta$. The roots becomeIf $|a|<1$, the square root is real, both roots have modulus one, and their separation is at least $2\sqrt{1-a^2}>0$. The companion matrix has eigenvectors $({\xi }_{\pm },1{)}^T$, so their uniform separation gives a mesh-independent bound on its powers. By power boundedness of a two-level Fourier scheme and the discrete Fourier transform, this proves stability for $0<\mu <1$.

If $|a|>1$, choose $\theta =\pi$. The roots have product of modulus one and unequal moduli, so one has modulus greater than one. The scheme is unstable.

The boundary cases require more than a root-modulus check. At $\theta =\pi$, $\mu =0$ gives $(\xi -1{)}^2=0$, while $\mu =1$ gives $(\xi +1{)}^2=0$. The companion matrices have nontrivial Jordan blocks and solutions of the form $(A+Bn)(\pm 1{)}^n$. Their linear growth rules out a uniform stability bound. This mode occurs on even periodic grids, which are enough to disprove mesh-uniform stability at the endpoints.

Therefore the stable range is exactlyWithin this range, second-order convergence additionally requires two appropriately accurate starting time levels. The unit-modulus parasitic root near $\theta =0$ does not decay, so poor initialization cannot be repaired by the scheme.

The one-dimensional Time-dependent Schrodinger equation for a wavefunction in a real potential isMultiplying this equation by ${\Psi }^{\ast }$, multiplying its complex conjugate by $\Psi$, and subtracting gives the probability continuity equationIf $\Psi$ and its first derivative decay sufficiently rapidly as $x\to \pm \infty$, then the probability current $j$ vanishes at both ends. Integrating the continuity equation and using the fundamental theorem of calculus yieldsThis conservation of quantum probability is required by the Born rule: once a normalizable wavefunction has total probability one, its time evolution must preserve that normalization.

For the stated Gaussian wave packet, put $\beta =\hslash t/m$. Sinceits squared modulus isThe Gaussian integral then giveswhich is independent of time, as required.

Taking the squared modulus and using ${\cosh }^{2}u=1+{\sinh }^{2}u$ givesThe incident wavevector makes an angle $\psi$ with the downward vertical, so $\sin 2\psi =2km/(k^2+m^2)$. Substitution gives the required formula.

Rouché's theorem states that if $g$ and $h$ are holomorphic functions on a neighbourhood of the closure of a bounded domain andon its positively oriented boundary, then $g$ and $g+h$ have the same number of zeros in the domain, counted with multiplicity.

The Open mapping theorem states that a nonconstant holomorphic function on a domain maps every open subset to an open subset. To prove it, fix $z_0$ and put $w_0=f(z_0)$. By the identity theorem, the zeros of $f-w_0$ are isolated. Choose $r>0$ so that the closed disc $\overline{D(z_0,r)}$ lies in the domain and $z_0$ is the only zero of $f-w_0$ in that disc. ThenWhenever $|w-w_0|<\delta$, the constant $w_0-w$ has modulus smaller than $f-w_0$ on the circle. The Rouche theorem therefore says thathas the same positive number of zeros in the disc as $f-w_0$. Thus every $w\in D(w_0,\delta )$ lies in the image of $f$, proving that the image is open.

If $|f|$ had a local maximum at $z_0$, take a small open disc $D$ on which $|f(z)|\le |f(z_0)|$. The restriction of $f$ to $D$ cannot be constant, since the identity theorem would then make $f$ constant on the whole domain. The complex open mapping theorem says that $f(D)$ is an open neighbourhood of $f(z_0)$, and such a neighbourhood contains points of modulus greater than $|f(z_0)|$, a contradiction. This proves the maximum modulus principle from the complex open mapping theorem.

Now let $\Omega$ be bounded. Its closure is compact, so the continuous function $|f|$ attains a maximum there. If $f$ is nonconstant, the maximum modulus principle excludes an interior maximum; if $f$ is constant, the claimed bound is immediate. Hencewhich is the maximum modulus principle on a bounded domain.

Finally let $\Omega =\{z:\mathrm{Re}z>1\}$ and suppose $|f|\le K$ throughout $\overline{\Omega }$. Fix $z_0\in \Omega$. The principal complex logarithm is holomorphic on the right half-plane, so for each positive integer $n$ the functionis holomorphic on $\Omega$ and continuous on its closure. Given $\epsilon >0$, choose $R>|z_0|$ so large thatApply the bounded-domain result to $g_n$ on $\Omega \cap D(0,R)$. On the vertical part of the boundary, $|z|\ge 1$ and hence $|g_n(z)|\le M$. On the circular part,It follows thatLetting $n\to \infty$ and then $\epsilon \to 0$ gives $|f(z_0)|\le M$. Since $z_0$ was arbitrary, this proves the bounded half-plane maximum principle.

The boundedness assumption is necessary. The functionis holomorphic on $\Omega$ and continuous on its closure, and $|f(1+iy)|=e$ on the boundary, but $|f(x)|=e^x$ is unbounded for real $x>1$. Thus the boundary estimate does not control an unbounded holomorphic function on this unbounded domain.

Letbe reflection in the unit circle, and represent a Möbius transformation $g$ byA direct calculation shows that $JgJ$ is represented byThus $gJ=Jg$ exactly when $M$ and $\tau (M)$ differ by a nonzero scalar. Applying the conjugate-linear involution $\tau$ twice shows that this scalar has modulus one. Rescaling $M$ by a suitable complex scalar then makes $M=\tau (M)$. Consequently all commuting maps, and only those maps, have the formThis is the Möbius maps commuting with reflection in the unit circle classification.

For such a map,It follows that $|g(z)|<1$ whenever $|z|<1$ precisely whenThe inverse has the same property, so these and only these maps preserve the unit disc.

For a slowly varying envelope $E=\psi e^{-ikx}$, the paraxial approximation to the Helmholtz equation isBecause $n=1+\mu W$ and ${\mu }^2\ll 1$, passage through a sufficiently thin phase screen producesIts modulus is one at the screen exit. Beyond the screen, $n=1$, so the parabolic wave equation is $E_x=i{E}_{zz}/(2k)$. A Taylor expansion in propagation distance givesSincewe findIt follows thator, equivalently, $|E{|}^2=1-x{\varphi }^{\prime \prime }/k+O(x^2)$. Thus random phase curvature produces local focusing and defocusing: free-space diffraction converts phase fluctuations into amplitude fluctuations immediately after the screen.

The reflected P-wave is propagating only while its normal wavenumber is real, equivalently while $|\sin \varphi |\le 1$. Sinceit is evanescent whenPutThen its spatial factor iswhich decays as $z\to -\infty$. The reflected SV amplitude becomesThe numerator and denominator have equal complex modulus, soThus evanescent mode conversion changes the reflected SV phase but not its amplitude, as described by Unit-modulus SV reflection with evanescent P conversion.

The travelling-wave substitution $\xi =x-Ut$ givesWriteIts real part satisfiesFor $U>0$, the imaginary part also givesWith $\alpha =\sqrt{1-2U^2}$ and a translation chosen so that the notch is centred at zero,Its limiting modulus is $\sqrt{{\alpha }^2+2U^2}=1=d$, as required. The same expression at $U=0$ is obtained directly from the second-order equation.

The special orthogonal group isIfits first column is a unit vector, say $(\cos \theta ,\sin \theta {)}^T$. The second column is the unique unit vector perpendicular to it that gives positive determinant, namely $(-\sin \theta ,\cos \theta {)}^T$. Thusso $A$ is rotation through $\theta$ about the origin.

For $A\in SO(3)$, its real characteristic polynomial of odd degree has a real eigenvalue. Every eigenvalue of an orthogonal matrix has modulus one, so every real eigenvalue is $\pm 1$. Nonreal eigenvalues occur in conjugate pairs whose product is one, while $\det A=1$; if all eigenvalues are real, their product likewise forces at least one to be $+1$. Hence $A$ fixes a nonzero vector $u$. The plane $u^{\perp }$ is $A$-invariant, and the restriction to it is an orientation-preserving orthogonal map. By the $SO(2)$ result it is a planar rotation. Therefore $A$ is a rotation about the axis $\mathbb{R}u$.

An algebraic number is an algebraic integer when it is a root of a monic polynomial in $\mathbb{Z}[X]$.

If $\alpha$ and $\beta$ are algebraic integers, the ring $\mathbb{Z}[\alpha ,\beta ]$ is a finitely generated $\mathbb{Z}$-module. Multiplication by $\alpha \beta$ preserves this module, so the algebraic integer module criterion shows that $\alpha \beta$ is an algebraic integer.

The ring of integers of a quadratic field givessince $2\equiv 2(\mathrm{mod}4)$. For a direct verification, if $x=a+b\sqrt{2}$ with $a,b\in \mathbb{Q}$ is integral, then its trace $2a$ and norm $a^2-2b^2$ are integers. Writing $2a=m\in \mathbb{Z}$ and reducing the norm condition in lowest terms shows first that $b$ can have denominator at most $2$; a denominator $2$ with odd numerator would require $m^2\equiv 2(\mathrm{mod}4)$, which is impossible. Thus $b\in \mathbb{Z}$, and then the norm condition forces $m$ even, so $a\in \mathbb{Z}$. Conversely, every $a+b\sqrt{2}$ with $a,b\in \mathbb{Z}$ is integral because $\sqrt{2}$ is integral and algebraic integers form a ring.

Now let $f$ have degree $d$ and roots ${\alpha }_1,\dots ,{\alpha }_d$. Sinceand $|{\alpha }_1|>1$, the integer $|a_d|$ cannot be $2$; hence $|a_d|=1$. Thus $f$ is monic up to sign and ${\alpha }_1$ is an algebraic integer.

Let $B$ be the product of all roots with modulus greater than one, counting multiplicity. Complex roots occur in conjugate pairs, so $B$ is real up to the signs of the real roots andAs a product of algebraic integers, $B$ is an algebraic integer; the equality forces $B=\pm 2$. Thereforeis the absolute value of the product of the remaining conjugates, is rational, and is an algebraic integer. It is consequently a nonzero rational integer. Every remaining factor has modulus at most one, so this integer is at most one. It must equal one, and henceThis is the mahler-measure norm argument at measure two.

Let the intercepted ciphertexts beSince $\gcd (e_i,e_j)=1$, the extended Euclidean algorithm gives integers $a,b$ with $a{e}_i+b{e}_j=1$. The common-modulus RSA attack recoversNegative powers are evaluated using modular inverses. If an inverse does not exist, its greatest common divisor with $N$ already factors the modulus.

Let $D=\mathrm{diag}(A)$. The Jacobi method isso $H=I-D^{-1}A$ and $v=D^{-1}b$. If $A$ is strictly diagonally dominant, every Gershgorin disc of $H$ is centered at zero and has radiusEvery eigenvalue therefore has modulus below one, so $\rho (H)<1$ and part (i) proves convergence.

The product $S(z)=f(z)f(1-z)$ is entire. From $f(z+1)=zf(z)$ and $f(1-z)=-zf(-z)$,Its modulus is therefore periodic with period one. The assumed bound for $f$ on $0\le \mathrm{Re}z\le 1$, together with the reflected bound for $f(1-z)$, bounds $S$ on that strip and hence on all of $\mathbb{C}$. By Liouville's theorem $S$ is constant, while $S(z+1)=-S(z)$ forces that constant to be zero.

If $f$ were nonzero at one point, it would be nonzero on a neighborhood, so $f(1-z)=0$ on an open set; the identity theorem would then give $f\equiv 0$, a contradiction to that choice. Therefore $f\equiv 0$ and $F=\Gamma$.

The power method starts from $q_0\ne 0$ and iteratesIf the eigenvalue of largest modulus is separated by a spectral gap and the initial vector is nonorthogonal to a corresponding eigenvector, the iterates converge in direction to that eigenvector.

Inverse iteration with shift $s$ instead solvesIt applies the power method to $(A-sI{)}^{-1}$ and therefore finds an eigenvector whose eigenvalue is closest to $s$.

The vertices satisfyAfter removing the root $z=0$, the product of the other roots has modulus equal to the constant term of $((z+1{)}^N-1)/z$, namely $N$. Hence the product of the $N-1$ chord lengths from one vertex of a unit regular polygon is $N$. Multiplying this identity over all $N$ vertices counts every chord twice, so the product of all chord lengths is $N^{N/2}$. Scaling the circle by $R$ scales each of its $N(N-1)/2$ chords by $R$, giving

The eigenvalues of the recurrence matrix have modulus $e^{-\kappa T}<1$, so transients decay and the coefficients approach the fixed pointThusFor $T=k\pi$ and $\kappa k\ll 1$, this isEven multiples deliver impulses in phase and produce resonance; odd multiples alternate the phase and largely cancel.