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For a possibly unstable normal mode, $c_i\ne 0$ ensures $V=U-c$ never vanishes, so choose one continuous branch of $V^a$. Substitute $\hat{w}=V^{a}q$ into the Taylor–Goldstein equation and collect terms:Multiplication by $V^{2a}$ puts its first two terms in divergence form:Multiply by the complex conjugate $\overline{q}$ and apply integration by parts. Since $q(\pm L)=0$, the boundary term vanishes and the power-transformed Taylor–Goldstein energy identity isThe weight is $V^{2a}$, not $|V{|}^{2a}$: preserving its complex phase is essential for the subsequent stability proofs.

Since $z\overline{z}=|z{|}^2=x^2+y^2$, the complex conjugate givesTherefore

Factor the expression using the complex conjugate:It is real, so

The one-dimensional Time-dependent Schrödinger 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.

The Dirac field Lagrangian isChoose the Majorana spinorSubstitution shows that its two chiral kinetic terms agree after integration by parts and thatThus ${\mathcal{L}}_D[{\psi }_M]$ equals twice the displayed one-Weyl-field Lagrangian, up to a total derivative.

The lower chiral component of the Dirac equation isPutting $u_R=i{\sigma }^2{u}_L^{\ast }$ gives exactly $i{\bar{\sigma }}^{\mu }{\partial }_{\mu }{u}_L-im{\sigma }^2{u}_L^{\ast }=0$; the upper component is its complex conjugate.

Write $m=|m|e^{i\theta }$. Parity reverses the spatial coordinates and must exchange the two chiral Weyl spinors. One convenient choice isThe identities ${\sigma }^0={\bar{\sigma }}^0$ and ${\sigma }^i=-{\bar{\sigma }}^i$ exchange the two kinetic terms after $\mathbf{x}\mapsto -\mathbf{x}$. The mass bilinear $m{\bar{\psi }}_R{\psi }_L$ maps to its complex conjugate $m^{\ast }{\bar{\psi }}_L{\psi }_R$, so the two mass terms are exchanged and the action is invariant. More generally, independent unit phases $a,b$ may multiply the two transformations provided $b^{\ast }a=m^{\ast }/m$.

For $0<\lambda <1$, the preceding change of variables and one integration by parts giveSplit the last improper integral at one and add and subtract one on $(\lambda ,1)$. Since ${\int }_{\lambda }^{1}dx/x=-\log \lambda$,

The definition also gives the complex conjugate relationConsequently, for $-1<\lambda <0$,

For functions $(f_{ϵ}{)}_{ϵ\in \{0,1{\}}^3}$ on ${\mathbb{F}}_p^n$, the Gowers inner product iswhere $\mathcal{C}$ is the complex conjugate operator. The Gowers uniformity norm isThe Gowers-Cauchy-Schwarz inequality states

The symmetric form of the functional equation of the Riemann zeta function isThe complex conjugation identity $\zeta (\overline{s})=\overline{\zeta (s)}$ and the functional equation show that every Nontrivial zero of the Riemann zeta function $\rho$ is accompanied by $\overline{\rho }$, $1-\rho$, and $1-\overline{\rho }$.

Let $\rho =\beta +i\gamma$ be the given nonreal zero. If $\beta >1/2$, use $\rho$ itself; if $\beta <1/2$, use $1-\rho$, whose real part is $1-\beta >1/2$. The resulting zero cannot have real part greater than one, by the stated zero-free half-plane. It therefore has real part in $(1/2,1]$.

Write $z=x+iy$, with $y\ne 0$ because the triangle is nondegenerate. Let $w=u+iv$ be its circumcenter. Equality of the complex modulus values $|w|=|w-1|$ gives $u=1/2$. Equality of $|w|=|w-z|$, after expanding the squares, givesThus the real part and imaginary part determineUsing the complex conjugate $\bar{z}=x-iy$, we also haveThereforeThe nonzero denominator is exactly the nondegeneracy condition. If $z$ is real, the three distinct collinear points do not have a finite circumcenter.

The product rule gives $y^{\prime }=x^n{z}^{\prime }+n{x}^{n-1}z$ and $y^{\prime \prime }=x^n{z}^{\prime \prime }+2n{x}^{n-1}{z}^{\prime }+n(n-1)x^{n-2}z$. On an interval avoiding $x=0$, substitution and division by $x^{n+2}$ giveFor a constant-coefficient differential equation, the coefficient of $z^{\prime }$ forces $2n+4=0$. At $n=-2$, the remaining coefficient $n^2+3n+2$ also vanishes. ThusThe characteristic equation of a constant-coefficient differential equation is $r^2+2r+2=0$, with complex conjugate roots of a polynomial $-1\pm i$. Its real solutions $e^{-x}\cos x$ and $e^{-x}\sin x$ are linearly independent. Therefore the general solution isThe constants can be chosen independently on the two sides of the singular point. The power substitution reducing a differential equation to constant coefficients here is invertible wherever $x\ne 0$.

Multiply the numerator and denominator by the complex conjugate $1+i$: $(1+i)/(1-i)=(1+i{)}^2/2=i$, whose square is $-1$. Thus

Multiplying by the complex conjugate of the denominator gives $(1+i)/(i-1)=-i$. Cubing yields

The quadratic formula gives $z=-b\pm \sqrt{b^2-1}$. For $-1\le b\le 1$, write this asEach root of a polynomial has modulus $1$, since $b^2+(1-b^2)=1$. As $b$ increases from $-1$ to $1$, the upper root traverses the upper unit semicircle from $1$ to $-1$, and the lower root traverses the lower semicircle in the opposite angular direction. At $b=-1$ the roots coincide at $1$; at $b=1$ they coincide at $-1$. For intermediate $b$ they are complex conjugates.

Multiply numerator and denominator by the complex conjugate of $1+z$. For $z\ne -1$ this givesThus its real part and imaginary part are respectively

Multiply numerator and denominator by the complex conjugate of $1-i$:Thus the real part and imaginary part are respectively

Multiply by the complex conjugate of the denominator:Thus

Multiply numerator and denominator by the complex conjugate $5+4i$:The real part and imaginary part are therefore

Rationalize with the complex conjugate $4-3i$:Thus the real part and imaginary part are

Multiply numerator and denominator by the complex conjugate of the denominator:Squaring gives $(-1-3i{)}^2/4=(-8+6i)/4$. Thus the real part and imaginary part are