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An orientation-reversing local conformal map of planar domains. Equivalently, its complex conjugation is a holomorphic function with nonzero derivative. Complex conjugation $z\mapsto \bar{z}$ is the basic example. The metric pullback uses the modulus of its local scale factor, just as for an orientation-preserving conformal map.

An antiunitary operator between complex Hilbert spaces is a surjective antilinear map satisfying the displayed inner product identity. It preserves norms but conjugates scalars. In a chosen orthonormal basis it is $UK$, with $U$ a unitary operator and $K$ coefficientwise complex conjugation. Conversely any such $UK$ is antiunitary. Antilinearity by itself does not preserve norms.

For a linear, fixed-frequency beam splitter, choose power-normalized complex amplitudes $a_1,a_2$ at its input optical ports and $b_1,b_2$ at its output optical ports: $|a_j{|}^2$ and $|b_j{|}^2$ are powers. The scalar description treats one specified electromagnetic polarization and matching transverse profiles; other fields require additional components. With the geometrical ordering in the diagram,Here $t,t^{\prime }$ are transmission complex amplitudes per unit input and $r,r^{\prime }$ are reflection complex amplitudes per unit input. In this power normalization, a single input gives $T=|t{|}^2$ and $R=|r{|}^2$. Raw electric field amplitudes in different media generally need additional power-normalization factors.

If these two output optical ports account for all outgoing power and the device has no loss or gain, $S$ is a unitary matrix:The dagger is the Hermitian transpose and a star is complex conjugation. To prove the condition, substitute $\mathbf{b}=S\mathbf{a}$ into conservation of total power for every input vector: ${\mathbf{a}}^{†}(S^{†}S-I)\mathbf{a}=0$. Inputs $(1,0)$ and $(0,1)$ fix the diagonal entries; $(1,1)$ and $(1,i)$ fix the real and imaginary parts of the off-diagonal entry. A lossy two-output description instead has $S^{†}S\le I$ for a passive device.^[ref]

One consistent ideal symmetric 50:50 choice isEach reflection carries $+\pi /2$ relative wave phase in this convention. The columns have unit vector norm and zero inner product, directly verifying that $S$ is a unitary matrix. Choosing a different phase convention for complex amplitudes at the optical ports replaces $S$ by $D_{\mathrm{out}}S{D}_{\mathrm{in}}^{-1}$, where the $D$ are diagonal matrices whose entries have modulus one; the input complex amplitudes must be transformed consistently. Observable powers are unchanged. The displayed $i$ is a convention for this ideal model, not a universal reflection rule for every coating or incidence geometry.

For two Fourier elastic wave fields with the same real elastic stiffness tensor, its major symmetry cancels the cross-strain terms in the divergence displayed above. The force-balance equations supply the remaining volume terms, and the divergence theorem gives their boundary integral of reciprocal tractions. Taking one field to be a complex conjugate gives energy-work identities and weighted modal orthogonality. The bilinear identity uses no complex conjugation until that field is chosen explicitly.

For a complex representation of a finite group, the eigenvalues of each represented element are roots of unity. They have absolute value one, so inversion equals complex conjugation on each eigenvalue. Taking traces gives the identity.

For a Dirac field, the charge-conjugation matrix implements ${\psi }^c=C{\bar{\psi }}^T$. In the standard gamma matrix representation one can take $C=i{\gamma }^2{\gamma }^0$. An overall phase is conventional. The transpose intertwining relation is different from the entrywise complex conjugation involved in an antiunitary operator.

A CM-field is a totally imaginary number field that is a quadratic extension of a totally real number field. Its nontrivial automorphism over that real subfield acts as complex conjugation in every complex embedding. Imaginary quadratic number fields are examples. An arbitrary totally imaginary number field need not have such a real subfield of index two.

Let $V_{\infty }$ be the norm limit of local principal units in the $p$-power cyclotomic field tower, for odd $p$. The Coleman power series $f_u$ gives the Frobenius-corrected logarithm $p^{-1}\log (f_u^p/\varphi f_u)$, whose inverse Amice transform is supported on ${\mathbb{Z}}_p^{\times }$. This is the middle map in the displayed exact sequence; the last map is the first moment $\mu \mapsto \int xd\mu$. Its kernel consists of norm-compatible $p$-power roots of unity. The logarithmic difference equation has one resonant degree, namely degree one in $\log (1+T)$, accounting for the last term. Complex conjugation acts by $-1$ on both copies of ${\mathbb{Z}}_p(1)$. Taking its even part therefore identifies the local principal-unit limit in the real tower with ${\mathbb{Z}}_p[[{\mathbb{Z}}_p^{\times }/\{\pm 1\}]]$.

For the quadratic polynomial $p(x)=2x^2-6x+5$, the discriminant is $(-6{)}^2-4(2)(5)=-4$, so its roots of a polynomial are the complex conjugatesTheir sum is $3=-(-6)/2$, and their product is $(9+1)/4=5/2$, matching the constant coefficient divided by the leading coefficient. More revealingly, $r_{+}{r}_{-}=r_{+}\overline{r_{+}}=|r_{+}{|}^2$: the product is the square of a complex modulus. Coefficients in the real numbers therefore need not imply polynomial roots in the real numbers.

For any polynomial with coefficients in the real numbers, $p(\overline{z})=\overline{p(z)}$. Thus a nonreal root of a polynomial brings its complex conjugate along with the same multiplicity of a root. This follows by applying complex conjugation to the polynomial expression and its factors. It explains why sums of complex conjugate terms can produce coefficients in the real numbers.

A complex-valued function is complex differentiable at $z_0$ ifexists. The same limit must hold for every approach, including curved paths. Equivalently,The error divided by $|h|$ tends to zero. Thus the derivative is a single complex multiplier, not an arbitrary real linear map of the plane. If the real coordinate map is differentiable, complex differentiability is equivalent to its satisfying the Cauchy-Riemann equations at that point.

The simplest failure is complex conjugation, $f(z)=\overline{z}$. At every $z_0$ its increment quotient is $\overline{h}/h$. For real $h=t\ne 0$ it is $1$, while for $h=it$ it is $-1$. More generally, $h=t{e}^{i\theta }$ gives $e^{-2i\theta }$. Although complex conjugation is a smooth real reflection and has every real directional derivative, these incompatible quotients rule out a complex derivative everywhere.

Agreement on two axes is necessary, but does not by itself establish a limit over the whole complex plane. The Cauchy-Riemann equations need the real differentiability condition above, or a sufficient regularity condition such as continuous partial derivatives near the point. There is a second distinction: $f(z)=|z{|}^2$ has complex derivative zero at zero because its quotient there is $\overline{h}\to 0$, but its Cauchy-Riemann equations fail at every nonzero point. It is complex differentiable at zero without being a holomorphic function on any neighbourhood of zero.

An invertible matrix over the complex numbers preserving the symmetric bilinear form $v^{T}w$: $A^{T}A=I$. The transpose here does not include complex conjugation, unlike the defining condition for a unitary matrix. For example, $\left(\begin{array}{cc}\cosh t& i\sinh t\\ -i\sinh t& \cosh t\end{array}\right)$ is complex orthogonal for real $t$, but is not unitary when $t\ne 0$.

For a normalized pure two-qubit state written in a magic basis, concurrence is $\mathcal{C}=|{\sum }_j{b}_j^2|$, while $\det {\rho }_A={\mathcal{C}}^2/4$. This bilinear sum uses no complex conjugation inside its square; replacing it by ${\sum }_j|b_j{|}^2$ would incorrectly give maximal concurrence for every state.

The conjugate Dolbeault operator is the type-$(1,0)$ component of the exterior derivative: $\partial :{\mathcal{A}}^{p,q}\to {\mathcal{A}}^{p+1,q}$. Complex conjugation interchanges $\partial$ and $\bar{\partial }$. On a complex manifold, $d=\partial +\bar{\partial }$, ${\partial }^2={\bar{\partial }}^2=0$, and $\partial \bar{\partial }+\bar{\partial }\partial =0$.

If $f$ is holomorphic on an open set $U$, its conjugate reflection $f^{\ast }$ is holomorphic on $U^{\ast }=\{z:\overline{z}\in U\}$, with $(f^{\ast }{)}^{\prime }(z)=\overline{f^{\prime }(\overline{z})}$. The identity follows by conjugating the difference quotient with increment $\overline{h}$; the two conjugations restore complex differentiability.

If $U$ is invariant under complex conjugation and is connected, and $f$ is real at a set of real points accumulating inside $U$, then $f=f^{\ast }$ by the identity theorem. In particular, an entire function real at every $1/n$ has this symmetry. This is a symmetry statement on an already defined domain; Schwarz reflection also constructs extensions across a boundary.

For $|{\Phi }_d⟩=d^{-1/2}{\sum }_i|ii⟩$ and normalized $|u⟩={\sum }_i{u}_i|i⟩$, contraction of the first factor gives $d^{-1/2}{\sum }_i\overline{u_i}|i⟩$. Hence a rank-one projective measurement outcome $u$ occurs with probability $1/d$ and leaves the other subsystem in the coordinatewise complex conjugate state $|\overline{u}⟩$. The complex conjugation refers to the paired bases defining $|{\Phi }_d⟩$; it must not be omitted for a complex measurement vector. Conditioning is essential: the unconditional reduced density matrix remains $I/d$.

Write $e^u\varphi (u)=\int \psi (t)e^{-iut}dt$. The double integral $\iint \psi (t)\psi (t^{\prime })(1+it)(1+i{t}^{\prime })/(2+i(t+t^{\prime }))dtd{t}^{\prime }$ equals the displayed constant. Express the denominator as a Laplace integral and each differentiated Fourier factor as $-{\varphi }^{\prime }(u)$, then use Fubini's theorem. For a real cutoff the resulting square is positive even though the Fourier transform factors have no complex conjugation. The Cauchy-Schwarz inequality and $\varphi (0)=1$, $\varphi (1/3)=0$ give $c_{\varphi }\ge 3$.

A dot product measures how much two Euclidean vectors point along each other. This article uses real Euclidean vectors in ${\mathbb{R}}^n$, with the usual Cartesian basis: multiply matching coordinates and add,The result is a scalar, while $\Vert \mathbf{a}\Vert$ is the Euclidean norm. This real convention has no complex conjugation; for vectors over the complex numbers, use the convention specified in inner product.

For two nonzero Euclidean vectors, the same dot product records their angle $0\le \theta \le \pi$ through the cosine:Here $\arccos$ is an inverse trigonometric function. The Cauchy-Schwarz inequality guarantees that the normalized ratio lies in $[-1,1]$. The zero vector has dot product zero with every vector, but its angle is undefined: the ratio would divide by zero. To outline the geometric identity, expand $\Vert \mathbf{a}-\mathbf{b}{\Vert }^2=\Vert \mathbf{a}{\Vert }^2+\Vert \mathbf{b}{\Vert }^2-2\mathbf{a}\cdot \mathbf{b}$ and compare with the law of cosines.^[ref]

Take $\mathbf{a}=(3,4)$ and $\mathbf{c}=(2,0)$. Their Euclidean norms are $5$ and $2$, andThe unit vector $\mathbf{u}=\mathbf{c}/\Vert \mathbf{c}\Vert =(1,0)$ selects the horizontal direction. The scalar projection $\mathbf{a}\cdot \mathbf{u}=3$ says how far $\mathbf{a}$ reaches along that direction; its vector projection is $3\mathbf{u}=(3,0)$. Thus $\mathbf{a}=(3,0)+(0,4)$ splits into an aligned part and an orthogonal projection residual.

For $\mathbf{b}=(4,-3)$, the dot product $\mathbf{a}\cdot \mathbf{b}=12-12=0$ proves orthogonality; both Euclidean norms are $5$. For $\mathbf{d}=(-2,0)$, $\mathbf{a}\cdot \mathbf{d}=-6$: the scalar projection onto $\mathbf{d}/\Vert \mathbf{d}\Vert =(-1,0)$ is $-3$, because $\mathbf{a}$ reaches against that chosen direction. Its angle with $\mathbf{d}$ is $126.8699^{\circ }$.

Keeping $\mathbf{a}=(3,4)$ fixed and choosing $\mathbf{b}$ of Euclidean norm $2$ gives $\mathbf{a}\cdot \mathbf{b}=10\cos \theta$. At angles $0^{\circ },60^{\circ },90^{\circ },120^{\circ },180^{\circ }$, the dot products are $10,5,0,-5,-10$. Positive values mean partial alignment, zero means orthogonality, and negative values mean a component against the other direction.

This interpretation explains work: for a constant force $\mathbf{F}$ and a straight displacement $\mathbf{s}$, $W=\mathbf{F}\cdot \mathbf{s}$ counts the force component along the motion. A sideways force contributes zero work, and an opposing force contributes negative work.^[ref]

The dual pairing evaluates a linear functional on a vector:It is bilinear: it is linear in each argument separately. Even over the complex numbers, this algebraic evaluation has no complex conjugation. An inner product instead has two vectors as inputs and, over the complex numbers, is sesquilinear. For $\phi (x,y)=2x-y$ and $v=(3,1)$, the dual pairing equals five. Its value is independent of the basis used to describe its two inputs.

Linearizing a uniform Kerr-fibre background couples opposite frequency sidebands. With $i{A}_z-{\beta }_2{A}_{TT}/2+\gamma |A{|}^{2}A=0$, the displayed dispersion relation permits exponential growth precisely when ${\beta }_2\gamma <0$ and $0<{\Omega }^2<-4\gamma P/{\beta }_2$. For positive $\gamma$, this is anomalous group velocity dispersion. complex conjugation requires the paired complex-mode ansatz on the unstable branch. amplitude and power growth exponents differ by a factor of two.

On the unit sphere, use $\zeta =(x+iy)/(1-z)$ away from the north pole and $\eta =(x-iy)/(1+z)$ away from the south pole. Their overlap relation $\eta =1/\zeta$ is a holomorphic map, giving a holomorphic atlas for the Riemann sphere. The opposite sign of $iy$ in the second manifold chart ensures a holomorphic rather than antiholomorphic transition.

Writing the sphere coordinates as $(X,Y,Z)$ avoids confusing their height with the complex coordinate $z$. The two manifold charts cover the whole unit sphere:On their overlap, $X^2+Y^2=1-Z^2$ givesThe coordinate transformation and its inverse are holomorphic, so these manifold charts make the sphere a Riemann surface. The sign in $X-iY$ is essential. Geometric projection from the south pole with the same planar axis labels would give $v=(X+iY)/(1+Z)=1/\overline{z}$. The complex conjugation in that transition prevents it from being holomorphic. For $z=i$, the correct second coordinate is $w=-i$: positive $Y$ corresponds to negative imaginary part in this manifold chart.