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Measuring the coset state $|g_0+K⟩$ in the group shift operator eigenbasis samples uniformly from the annihilator of a subgroup of a finite abelian group $K^{\perp }$. Indeed,so the character-sum cancellation lemma gives probability $|K|/|G|$ on $K^{\perp }$ and zero elsewhere. The modulus of $\chi (g_0)$ is one, so the offset has no effect on the distribution.

For a sufficiently smooth local map fixing the origin, a local invariant centre manifold is tangent to the generalized eigenspaces with eigenvalues of modulus one. Writing the map as $(u,v)\mapsto (F(u,v),G(u,v))$ and the manifold as $v=h(u)$ gives the invariance equation $h(F(u,h(u)))=G(u,h(u))$. Finite Taylor series coefficients can be found from this equation. If all transverse eigenvalues have modulus less than one, local asymptotic stability reduces to that of the map on the centre manifold.

A complex number can encode both the wave amplitude and phase of a harmonic disturbance: $a(x,t)=\mathrm{Re}\{\hat{a}{e}^{i(kx-\omega t)}\}$. Its modulus is the real oscillation amplitude and its complex argument fixes the wave phase.

A complex number has the form $x+iy$, with conjugate $x-iy$ and modulus $\sqrt{x^2+y^2}$.

An eigenvalue is dominant when its modulus is greater than or equal to that of every other eigenvalue. It is uniquely dominant when the inequality is strict.

An eigenphase of a unitary operator is a real number $\theta$ modulo one specifying an eigenvalue $e^{2\pi i\theta }$. Every unitary eigenvalue has modulus one. Quantum phase estimation records the eigenphase corresponding to an input eigenvector; a superposition retains the corresponding labels coherently before measurement.

A global phase multiplies an entire quantum state by one complex number of modulus one. It does not change any measurement probability, unlike a relative phase between components.

A square nonnegative matrix $A$ is irreducible if for each pair $(i,j)$ there is an integer $k\ge 0$ with $(A^k{)}_{ij}>0$. This says that every index can reach every other through positive entries. The Perron–Frobenius theorem then gives a positive leading eigenvector and an algebraically simple eigenvalue. Irreducibility alone permits other eigenvalues of the same modulus: the cyclic permutation matrix $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ has eigenvalues $1,-1$.

For real $x,y$, the complex exponential function has $e^{x+iy}=e^x(\cos y+i\sin y)$. Since the trigonometric factor lies on the unit circle,Thus the real part of the exponent controls its modulus, while the imaginary part controls its direction on the complex plane.

Write a nonzero complex number as $w=r{e}^{i\theta }$. Matching modulus and complex argument gives $r^4=2\sqrt{2}$ and $4\theta =\pi /4+2\pi k$. Thus the four distinct fourth roots areThese angles differ by $\pi /2$, so the values are distinct. Every further integer $k$ repeats one of them; this is multiplication by the fourth roots of unity.

For a real line $ux+vy+w=0$, with $(u,v)\ne (0,0)$, choose $B=(u-iv)/2$ and $C=w$. Direct multiplication gives $Bz+\overline{B}\overline{z}=ux+vy$, so every line has the required form with $B\ne 0$.

For a circle of centre $c$ and radius $\rho >0$, expand its modulus equation:Thus the required circle parameters are $\boxed{B=-c,C=|c{|}^2-{\rho }^2}$. Conversely, completing the square in the circle form gives $|z+B{|}^2=|B{|}^2-C$, which describes a genuine circle when $|B{|}^2-C>0$. The complex conjugate terms make both equations real-valued.

Let $r=e^{i\theta }\ne 1$. The geometric series from any starting index is bounded:Using summation by parts,Since the coefficients are nonincreasing and positive, the modulus is at most $K(a_N+a_m-a_N)=K{a}_m\to 0$. The Cauchy criterion proves convergence for every $\theta \notin 2\pi \mathbb{Z}$. This is a direct proof of the Dirichlet test in this setting.

For $a_n=1/n$, the imaginary parts of the convergent partial sums give convergence of the sine series at those angles. If $\theta \in 2\pi \mathbb{Z}$, every sine term is zero. Hence the sine series converges for all real angles.

The characteristic polynomials of a linear multistep method areThe consistency of a numerical method conditions hold for every real $a$:Both roots of a polynomial of $\rho$, namely $1$ and $-1$, are simple and have modulus one. The root condition for a multistep method therefore gives zero-stability for every $a$. The Dahlquist equivalence theorem now gives convergence for every fixed $a\in \mathbb{R}$. As usual, this means convergence of a numerical method on each fixed finite interval for a sufficiently regular ordinary differential equation with a Lipschitz continuous vector field and starting values tending to the exact starting values. When $a\ne 0$, the implicit time-stepping method update is locally uniquely solvable for sufficiently small $h$, for example by a contraction mapping if $|ha|L<1$. Zero-stability does not assert that a large fixed step is suitable for a stiff differential equation.

Apply the Dahlquist test equation and put $z=h\lambda$. Every recurrence mode must satisfyThe amplification polynomial of a multistep method, rather than just its root near one, determines absolute stability. Put $B=2a-1$ and use the Cayley transform between the half-plane and diskFor $\zeta \ne -1$, the characteristic equation becomesIf $B>0$, the denominator cannot vanish at a root of the characteristic equation: $1+B{w}^2=0$ would force $w=0$, a contradiction. Hence $\mathrm{Re}z<0$ forces $\mathrm{Re}w<0$, so all amplification roots have modulus less than one. On the imaginary axis the roots have modulus one and are simple: the transformed quadratic has discriminant $4(1-B{z}^2)>0$ for imaginary $z$. At $z=0$ they are the two simple roots $\pm 1$. Also $\zeta =-1$ cannot be a root when $B>0$ and $z\ne 0$, and the leading coefficient $1-az$ cannot vanish in the closed left half-plane.

If $a<1/2$, the root near $-1$ isFor small negative real $z$ it lies below $-1$, violating absolute stability. The endpoint $a=1/2$ deserves separate treatment:One root is always $-1$; the other is the trapezoidal rule multiplier $(1+z/2)/(1-z/2)$. They are distinct for every finite $z$ in the left half-plane. Consequently, with absolute stability understood as the bounded root condition for a multistep method,There is a convention at this reducible endpoint: if A-stability is defined to require every unreduced recurrence mode to decay for $\mathrm{Re}z<0$, the answer is $a>1/2$, since the $(-1{)}^n$ mode persists at $a=1/2$. Canceling the common factor $\zeta +1$ gives the A-stable trapezoidal rule, but cancellation removes an actual starting-error mode of the original two-step recurrence. The fourth-order member $a=1/3$ is outside either A-stability range.