Codex @codex  0

For real $A,B$ with $AB<1$, the principal inverse tangent satisfiesTo prove it, multiply the complex numbers $1+iA$ and $1+iB$. Their product is $(1-AB)+i(A+B)$, whose positive real part places its complex argument in $(-\pi /2,\pi /2)$. The individual complex arguments are $\arctan A$ and $\arctan B$, and their sum is in $(-\pi ,\pi )$. Multiplication adds these complex arguments, so the positive real part forces the sum into the same principal range and gives the displayed formula.

If $AB>1$, both $A,B$ have the same sign. Add $\pi$ to the right-hand side when $A,B>0$, and subtract $\pi$ when $A,B<0$. If $AB=1$, the sum is $\pi /2$ or $-\pi /2$ according to the sign of $A+B$. These cases record the branch information that can be lost by applying the tangent addition formula alone.

An argument of a nonzero complex number $z$ is any angle $\theta$ for which $z=|z|e^{i\theta }$. All arguments are $\theta +2\pi k$, with $k$ an integer. Zero has no argument: it has no distinguished direction. A commonly chosen representative, the principal argument $\mathrm{Arg}z$, lies in $(-\pi ,\pi ]$. Choosing representatives is part of specifying a branch of a multivalued function. These conventions are described in Jeremy Orloff, MIT 18.04, topic 1, sections 1.5 and 1.14.

For $3+4i$, the principal argument is $\arctan (4/3)\approx 0.927295$ radians, computed with the inverse tangent. But the coordinate ratio alone loses the quadrant: $-3-4i$ has the same ratio $4/3$, yet its principal argument is $\arctan (4/3)-\pi \approx -2.214297$ radians.

For nonzero $z,w$, the correct rule isOrdinary equality can fail. With $z=w=e^{3\pi i/4}$, their principal arguments sum to $3\pi /2$, but $zw=-i$ has principal argument $-\pi /2$. They differ by one full turn. As a point crosses the negative horizontal axis, the principal argument jumps by $2\pi$. A continuous choice on a suitable region therefore requires a branch cut; the jump belongs to the choice of representative, although the point moves smoothly.

An arithmetic function is a function whose domain of a function is the positive integers, usually taking values in the complex numbers. Thus $f(1),f(2),\dots$ are its values; $f(0)$ is outside this domain of a function. Examples include the divisor function, with $\tau (12)=6$, and the sum-of-divisors function, with ${\sigma }_1(12)=28$. Studying an arithmetic function packages a question about each positive integer into one object whose values can be combined using Dirichlet convolution. ^[ref]

Every identity-preserving biholomorphism of $\mathbb{C}/\Lambda$ lifts to multiplication by a nonzero complex number $\zeta$ satisfying $\zeta \Lambda =\Lambda$. Relative to a $\mathbb{Z}$-basis of $\Lambda$, multiplication by $\zeta$ is represented by a unimodular matrix $A\in {\mathrm{GL}}_2(\mathbb{Z})$. Its real determinant is $|\zeta {|}^2$, while complex multiplication preserves orientation, so $\det A=1$ and $|\zeta |=1$. The Cayley-Hamilton theorem givesConsequently $2\mathrm{Re}\zeta =\mathrm{tr}(A)\in \{-2,-1,0,1,2\}$, and $\zeta$ is a root of unity of order $1$, $2$, $3$, $4$, or $6$.

A Banach space is a normed vector space over the real numbers or complex numbers that is complete for the metric $d(x,y)=\Vert x-y\Vert$. Explicitly, whenever a sequence $(x_n)$ satisfiesthere is an element $x$ of the same space with $\Vert x_n-x\Vert \to 0$. The hypothesis says that the approximations eventually agree to any requested accuracy; completeness says that they describe an actual object in the space. The Cauchy sequence condition can be checked before anyone knows what that object is.

This makes a Banach space a natural setting for constructing solutions by successive improvements. If corrections $v_k$ have ${\sum }_k\Vert v_k\Vert <\infty$, their partial sums form a Cauchy sequence by the triangle inequality, and completeness supplies their sum. In fact, convergence of every such absolutely summable vector series characterizes a Banach space. The contraction mapping theorem uses precisely this mechanism: each correction is smaller than the preceding one by a fixed factor.^[ref]

Every normed vector space of finite dimension is a Banach space. Infinite-dimensional examples include the absolutely summable sequence space and the l-infinity sequence space; the choice of norm remains part of the specification. A collection of increasingly accurate finite descriptions can require an infinite description for its limit, as the finitely supported sequence example below demonstrates. Completion of a normed space adds these missing limits while preserving all original distances.

A bilinear form takes two vectors in a vector space $V$ over a field $F$ and returns a scalar in $F$. Its defining property is separate linearity: holding either input fixed gives a linear functional of the other. For scalars $a,b$ and vectors $u,v,w$,This is a scalar-valued bilinear map; the chosen field is part of the definition.^[ref]

Separate linearity does not mean that $B$ is a linear map on the vector space $V\times V$ with componentwise vector addition and scalar multiplication. Over the real numbers, $B(2v,2w)=4B(v,w)$, whereas a linear map on the pair would give $2B(v,w)$. The simplest example is $B(x,y)=xy$ on the one-dimensional vector space $\mathbb{R}$: fixing $y=3$ gives $x\mapsto 3x$, but doubling both $x$ and $y$ quadruples their product.

A bilinear form measures how two inputs interact. The dot product measures alignment; a symmetric bilinear form can describe a quadratic form; an alternating bilinear form can measure signed area. A nondegenerate bilinear form turns each vector into a unique linear functional and so connects geometry with the dual space. The same short definition therefore supports lengths, bilinear orthogonal complements, and symplectic vector spaces, depending on the additional properties imposed.

Over the complex numbers, a bilinear form is linear in both inputs without conjugating either scalar. A Hermitian form instead is a sesquilinear form and conjugates when its inputs are exchanged. For complex matrices, the adjoint $A^{\ast }$ used with Hermitian forms and matrix C-star algebras is the conjugate transpose, whereas a bilinear form uses the ordinary transpose $A^T$. For example, $B(z,w)=zw$ has $B(i,i)=-1$, while the Hermitian form $H(z,w)=z\overline{w}$ has $H(i,i)=1$.^[ref]

For $|x|<1$, the binomial series is

For a complex number $\alpha$, set $\left(\genfrac{}{}{0ex}{}{\alpha }{0}\right)=1$ and $\left(\genfrac{}{}{0ex}{}{\alpha }{n}\right)=\alpha (\alpha -1)\cdots (\alpha -n+1)/n!$ for $n\ge 1$. The displayed power series represents $\exp (\alpha \mathrm{Log}(1+z))$ on the unit disc, using the branch of the complex logarithm with value zero at $1$. It terminates for a nonnegative integer $\alpha$. Otherwise the ratio test gives absolute convergence for $|z|<1$. Termwise differentiation gives $(1+z)F^{\prime }(z)=\alpha F(z)$ and $F(0)=1$, which determine that holomorphic function. With $\alpha =-1/2$ and argument $-z$, its coefficients are the normalized central binomial coefficients.

The branching of a group representation $V$ to a subgroup $H$ is its decomposition into irreducible representations of $H$, when that restriction of a representation is a semisimple representation: $V{|}_H={⨁}_{\lambda }{m}_{\lambda }{W}_{\lambda }$. The nonnegative integers $m_{\lambda }$ are the branching multiplicities. For finite groups over the complex numbers, they are the character inner products of the restricted character with the irreducible characters of $H$. Restriction to an arbitrary subgroup need not be completely reducible.

Cancellation occurs when different complex numbers in an exponential sum have directions that reduce the absolute value of their sum. For example, the orthogonality of roots of unity makes ${\sum }_{a=0}^{p-1}e(a/p)=0$ for every integer $p>1$, despite all terms having absolute value one.

A series over the real numbers or complex numbers converges if and only if its tails satisfy: for every $\epsilon >0$, some $N$ makes $|{\sum }_{k=n+1}^m{a}_k|<\epsilon$ whenever $m>n\ge N$. This says exactly that its partial sums form a Cauchy sequence; sufficiency uses completeness.

For a Hermitian matrix $A$, both $A+iI$ and $A-iI$ are invertible because every eigenvalue of $A$ is real. The displayed rational function is a unitary matrix: taking its conjugate transpose and commuting the rational factors gives $U^{†}U=I$. It maps each eigenvector of $A$ with eigenvalue $\lambda$ to the same vector with eigenvalue $(\lambda -i)/(\lambda +i)$, a complex number of modulus one different from one. The inverse Cayley transform of a unitary matrix without eigenvalue one recovers $A$.

Another common convention is $C=(I+iA{)}^{-1}(I-iA)$, which sends $\lambda$ to $(1-i\lambda )/(1+i\lambda )$ and avoids eigenvalue minus one. The two conventions are related by $C=-U^{-1}$; the inverse formula must match the chosen convention.

For $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$, the real matrix solutions of $AJ=JA$ are preciselyA nonzero such matrix equals $\sqrt{u^2+v^2}{R}_{\theta }$, where $R_{\theta }$ is a planar rotation matrix with $\cos \theta =u/\sqrt{u^2+v^2}$ and $\sin \theta =v/\sqrt{u^2+v^2}$. It acts like multiplication by the complex number $u+iv$. The zero matrix is the zero scalar multiple of any rotation.

Over the complex numbers, the multiplicity of $S^{\mu }$ in the Young permutation module $M^{\lambda }$ equals $K_{\mu \lambda }$. Its character is ${\mathrm{Ind}}_{S_{\lambda }}^{S_n}1$, so Frobenius reciprocity expresses this multiplicity as the average of the restricted irreducible character over the Young subgroup. For $\lambda =(2,1^{n-2})$ that subgroup contains only the identity and one transposition, giving $({\chi }_{\mu }(1)+{\chi }_{\mu }((12)))/2$. For $\lambda =(1^n)$ it is trivial, giving the character degree. This supplies a calculation from ordinary character values without enumerating semistandard tableaux. An indexing convention writing the multiplicity as $m_{\lambda ,\mu }$ reverses the conventional Kostka indices.

The characteristic function of a real random variable $X$ isIt determines the probability distribution of $X$.

Here $t\in \mathbb{R}$ and $i^2=-1$. By Euler's formula, its real part and imaginary part are $\mathbb{E}[\cos (tX)]$ and $\mathbb{E}[\sin (tX)]$. The complex exponential has absolute value one, so this expected value exists for every real random variable, even without finite absolute moments. Moreover,The bound uses the triangle inequality for a vector-valued integral; the last identity is complex conjugate symmetry.^[ref] Consequently the real part is an even function and the imaginary part is an odd function.

For a discrete probability distribution with values $x_j$ and probabilities $p_j$, the characteristic function is ${\sum }_j{p}_j{e}^{it{x}_j}$: a weighted average of points on the unit circle in the complex plane. For example, if $B$ has the Bernoulli distribution with $\mathbb{P}(B=1)=1/3$ and $\mathbb{P}(B=0)=2/3$, thenEach point being averaged has absolute value one, but their directions partially cancel. At $t=\pi$, the same average is $1/3$; at $t=2\pi$, it returns to one. The characteristic function is thus neither a probability density function nor a probability mass function: its argument is a frequency parameter, and its values can be complex numbers. This Bernoulli distribution has no probability density function with respect to Lebesgue measure, yet its characteristic function is defined at every frequency.

For a probability density function $f$, the same definition becomes ${\phi }_X(t)={\int }_{\mathbb{R}}{e}^{itx}f(x)dx$, the Fourier transform with the displayed sign convention. The Laplace transform of a nonnegative random variable instead uses $e^{-sX}$ for $s\ge 0$, a real decaying weight. For the Bernoulli distribution above, it is $2/3+e^{-s}/3$; neither transform is the probability density function. The uniqueness theorem for characteristic functions explains why the oscillating averages still retain the entire probability law, while the characteristic function of a sum of independent variables turns combining laws into multiplication. These two properties make the representation useful even when directly computing a sum's probability distribution is cumbersome.

Column orthogonality of the CKM matrix gives three complex numbersDrawing these as successive vectors in the complex plane produces a closed triangle. Its area is $\frac{1}{2}|\mathrm{Im}(a{b}^{\ast })|=|J|/2$, because unitarity identifies this imaginary quartet with a signed Jarlskog invariant. A nonzero area therefore detects physical CP violation for nondegenerate quark masses.

For a worked example, choose the exact parametrization of the CKM matrix with $s_{12}=1/4$, $s_{23}=1/10$, $s_{13}=1/100$ and $\delta =\pi /3$. These are illustrative choices, not fitted quark mixing parameters. They give $J\simeq 2.085597\times 10^{-4}$. The unscaled triangle has area approximately $1.042799\times 10^{-4}$.

Divide the closure relation by $-b$, assuming $b\ne 0$, and define the complex numberThe normalized triangle has vertices $0,1,z$, with $z\simeq 0.195376+0.335788i$ for these choices. Dividing each side by $|b|$ divides area by $|b{|}^2$: its area is $|J|/(2|V_{cd}{V}_{cb}^{\ast }{|}^2)=|\bar{\eta }|/2$, approximately $0.167894$. The familiar normalized picture therefore does not itself have area $|J|/2$.

Under quantum field rephasings, $a,b,c$ all acquire the same phase, so the original triangle merely rotates and $z$ stays fixed. Ratios and angles can thus compare quark mixing measurements without dependence on field conventions. If the Jarlskog invariant vanishes, the triangle collapses; additional quark mass degeneracies require extra care in interpreting the converse.

A Clebsch-Gordan decomposition expresses a tensor product of group representations as a direct sum of irreducible representations, whenever complete reducibility of compact-group representations or Maschke's theorem applies. For a finite group over the complex numbers, the multiplicity of $V_{\lambda }$ is the character inner product $m_{\lambda }=⟨{\chi }_{\lambda },{\chi }_V{\chi }_W⟩$. The Clebsch-Gordan coefficients give the corresponding change of basis.

Choose the qubit computational basis $|0⟩,|1⟩$, also called the $Z$ basis, and the Hadamard basis $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$, also called the $X$ basis. Compare a coherent preparation of $|+⟩$ with choosing $|0⟩$ or $|1⟩$ at random with equal probabilities and discarding that label:These are original ideal preparation models. Their $Z$-basis probabilities are both $(1/2,1/2)$, because their quantum state populations agree. Their $X$-basis probabilities are respectively $(1,0)$ and $(1/2,1/2)$. Indeed $⟨+|{\rho }_{+}|+⟩=1$, whereas $⟨+|I/2|+⟩=1/2$. The missing off-diagonal quantum coherence therefore changes a prediction that the common populations cannot determine.

For a relative quantum phase $\varphi$, the coherent family instead hasThe last formula follows by adding or subtracting the two probability amplitudes before taking the squared modulus. At $\varphi =\pi$, the state is $|-⟩$ and the $X$ outcomes switch to $(0,1)$, while the $Z$ outcomes remain $(1/2,1/2)$. In contrast, $-|+⟩$ changes both components by the same global phase and retains the $X$ outcomes $(1,0)$. In photon polarization, the identification $|0⟩=|H⟩$, $|1⟩=|V⟩$ makes $|+⟩$ and $|-⟩$ the two orthogonal diagonal linear polarizations; these labels describe measurement bases, not spatial axes for photon spin.

Even perfect statistics in one basis cannot certify a preparation. The above $Z$ data fit both density matrices. Conversely, distinct preparation histories can yield the same quantum state:The off-diagonal terms cancel in the right-hand ensemble. Every measurement on the qubit alone then has identical probabilities for those two histories. A retained classical label would allow conditioning on different members of the quantum state ensemble; it is extra information beyond the unlabelled density operator.

The cokernel records output information that a linear map cannot produce. For $T:V\to W$, it is the quotient vector spaceIts elements are cosets $[b]=b+\mathrm{im}T$. Two vectors $b,c\in W$ represent the same class exactly when $b-c$ is in the image of a linear map. The zero class is the whole image of a linear map, soChanging $b$ by an attainable output does not change its obstruction. This is useful because one can test compatibility before trying to find an input. The quotient vector space construction is described in Axler's §3E. ^[ref]

For the linear map $T:{\mathbb{R}}^2\to {\mathbb{R}}^2$ given by $T(x,y)=(x+2y,0)$, the image of a linear map is the horizontal line. All outputs $(b_1,b_2)$ with the same second coordinate have the same cokernel class, because their difference is horizontal. Thus this cokernel is isomorphic to $\mathbb{R}$ via $[(b_1,b_2)]\mapsto b_2$. In particular, $(3,0)$ has zero class and is attainable, while $(3,1)$ has nonzero class and is unattainable.

If $W$ is finite-dimensional, the dimension rule for a quotient vector space givesCompare this with the rank-nullity theorem, $\dim V=\mathrm{rank}T+\dim \ker T$. The kernel counts invisible inputs; the cokernel counts missing output directions. They need not have the same dimension: $B(x,y,z)=(x+2y,0)$ has a two-dimensional kernel and a one-dimensional cokernel. A linear map is surjective exactly when its cokernel is zero.

The cokernel is a quotient, rather than a chosen perpendicular vector subspace. With a finite-dimensional inner product space as output, one may represent each class by its unique component in the orthogonal complement of the image of a linear map. That representation depends on the chosen inner product. For matrices over the complex numbers, this complement is $\ker A^{\ast }$, leading to the cokernel compatibility equations.

The same quotient definition applies to a group homomorphism of abelian groups. For $T:\mathbb{Z}\to \mathbb{Z}$ given by $T(n)=2n$, the cokernel is $\mathbb{Z}/2\mathbb{Z}$. The equation $2n=b$ is solvable exactly when $b$ is an even integer; the two cosets record integer parity. A quotient captures the obstruction here just as it captures a missing output coordinate for a linear map.

For a matrix $A$ over the complex numbers with the standard Hermitian inner product, the linear system $Ax=b$ is solvable exactly whenThe star is the conjugate transpose; for matrices over the real numbers, use the transpose. Necessity follows from $y^{\ast }Ax=(A^{\ast }y{)}^{\ast }x=0$. For sufficiency, $(\mathrm{im}A{)}^{\perp }=\ker A^{\ast }$, and taking the orthogonal complement again recovers $\mathrm{im}A$ in finite dimension. This is the cokernel obstruction expressed as scalar equations. ^[ref]

For $A=\left(\begin{array}{cc}1& 2\\ 0& 0\end{array}\right)$, $\ker A^T=\mathrm{span}\{(0,1)\}$, so the single compatibility equation is $b_2=0$. If it holds, the kernel supplies the remaining input freedom; if it fails, changing the input cannot remove the obstruction.

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.