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In an orthonormal computational basis, the squared norm is the sum of the squared moduli of the coefficients:Thus the vector is normalized and represents a pure state of a qubit. The imaginary part contributes positively to the squared modulus; squaring the coefficient without complex conjugation would give the wrong norm.

Use time dependence $e^{-i\omega t}$ and assume the dopants supply $N$ mobile electrons per unit volume. The Drude model equation isThe free-carrier electric polarization is ${\mathbf{P}}_c=-Ne\mathbf{x}$, so, after adding the bound-electron background,ThereforeThe symbol ${\omega }_p$ here is the background-screened plasma frequency; the unscreened carrier frequency is $\sqrt{{ϵ}_{\infty }}{\omega }_p$. For positive $\gamma$, the imaginary part of ${ϵ}_r$ is positive in this time convention, consistent with absorption. Incomplete dopant ionization would replace the nominal doping density by the actual mobile carrier density.

The line vortex is mapped toIts imaginary part is positive because $c$ lies in the fluid region. A conformal map preserves the circulation of the local line vortex, so its strength in the mapped complex plane is still $\Gamma$.

Again $|e^{-ax}\sin (bx)|\le e^{-ax}$ proves Lebesgue integrability. Taking the imaginary part of the absolutely convergent complex integral in part i gives

Take the imaginary part of the eighth-power binomial expansion in De Moivre's theorem. With $s=\sin \theta$ and $c=\cos \theta$,Substitute $c^2=1-s^2$ and collect powers:Therefore the requested leading coefficient isThe other coefficients, if retained as an algebraic check, are $b=192,c=-80,d=8$. No division by $s$ or $c$ is needed, so the identity remains valid when either vanishes.

Set $q=e^{i\theta }/2$. The required sine sum is the imaginary part of a finite geometric series:The denominator is $(5-4\cos \theta )/4$. The numerator's imaginary part isDividing and multiplying numerator and denominator by $2^N$ provesFor real $\theta$, $5-4\cos \theta \ge 1$, so there is no exceptional angle requiring a limiting interpretation. At $N=1$ the expression reduces to $\sin \theta /2$.

Rationalizing the denominator givesHence its real part, imaginary part, complex modulus, and principal argument are respectivelyThe quadrant matters: the real part is positive and the imaginary part negative. For the exponential expression, Euler's formula gives $e^{2+i\pi /4}=e^2(1+i)/\sqrt{2}$, so the same four quantities areGeneral complex arguments are obtained from each stated principal argument by adding $2\pi k$, $k\in \mathbb{Z}$.

Take the imaginary part of De Moivre's theorem at exponent five. Writing $s=\sin \theta$ and $c=\cos \theta$ givesFor real $x\in [-1,1]$, put $x=\sin \theta$. The equation becomes $\sin 5\theta =1$, so $\theta =\pi /10+2\pi k/5$. The distinct sine values are $1$, $(\sqrt{5}-1)/4$, and $-(1+\sqrt{5})/4$. To establish completeness over the complex numbers as well, factor the polynomial:Thus all polynomial roots, including their multiplicities, areThe repeated sine values account for the double roots; there are no additional nonreal roots hidden by the trigonometric parametrization.

Squaring first is simpler than expanding the tenth powers. Since $(1-i{)}^2=-2i$ and $(1+i{)}^2=2i$, while $i^4=1$, the requested Cartesian forms areThus the real part is zero in each case, and the imaginary parts are $-32$ and $32$. The results are complex conjugates, as also follows because the two inputs are complex conjugates and the fifth-power polynomial has real coefficients.

Put $z=x+iy$. The expression whose complex argument is constrained becomesIts principal argument is $\pi /2$ exactly when its real part is zero and its imaginary part is strictly positive. HenceThe locus consists of the upper-right half of the upper hyperbola branch, $y=\sqrt{x^2+1}$ with $x>0$, and the lower-left half of the lower branch, $y=-\sqrt{x^2+1}$ with $x<0$. The vertices $z=\pm i$ are excluded because $z^2+1=0$ there and zero has no complex argument. The remaining halves have negative imaginary part and give argument $-\pi /2$, not $\pi /2$. The right panel of the original illustration marks the excluded vertices with open circles.

Multiply numerator and denominator by the denominator's complex conjugate:Thus the real part is $\boxed{7/41}$ and the imaginary part is $\boxed{22/41}$.

Write $z=x+iy$ with real $x,y$. The complex hyperbolic sine is $\sinh z=\sinh x\cos y+i\cosh x\sin y$. Its zero real part requires $\sinh x\cos y=0$. If $\cos y=0$, the magnitude of the imaginary part would be $\cosh x\ge 1$, which cannot equal $1/2$. Therefore $x=0$ and $\sin y=1/2$. All solutions areOn the Argand diagram these are two interlaced arithmetic sequences on the imaginary axis; both repeat with imaginary period $2\pi i$.

Using the complex forcing method for a real differential equation, $\mathrm{Im}[(4x+2i)e^{ix}]=4x\sin x+2\cos x$, a suitable complex forcing is $f(x)=(4x+2i)e^{ix}$. Solve $u^{\prime \prime }+2u=f$ and take its imaginary part. Writing $u=p(x)e^{ix}$ reduces the equation toA linear polynomial $p=ax+b$ gives $a=4$ and $b=-6i$. ThusFor a direct check, $(4x\sin x{)}^{\prime \prime }+2(4x\sin x)=8\cos x+4x\sin x$, while $(-6\cos x{)}^{\prime \prime }+2(-6\cos x)=-6\cos x$. Their sum is the required forcing. Adding $A\cos (\sqrt{2}x)+B\sin (\sqrt{2}x)$ gives the full real solution if needed.

By De Moivre's theorem, $(\cos \theta +i\sin \theta {)}^3=\cos 3\theta +i\sin 3\theta$. Expanding the cube and equating its real part givesThe last form uses ${\sin }^2\theta +{\cos }^2\theta =1$; comparing the imaginary parts similarly gives the corresponding sine identity.

For a genuinely second-order linear ordinary differential equation, assume $a_2\ne 0$. Substituting the exponential function $f=e^{rt}$ into the homogeneous linear differential equation gives $a_2{r}^2+a_{1}r+a_0=0$. Oscillations arise precisely when its two roots of a polynomial have nonzero imaginary parts, namelyIn that case, writing $\omega =\sqrt{4a_0{a}_2-a_1^2}/(2|a_2|)>0$, the real complementary function isEvery nonzero such solution has an oscillatory factor; the oscillation may decay, grow, or keep constant amplitude according to the sign of $a_1/a_2$. This is the oscillatory-root criterion for a real second-order equation; decay is not a prerequisite for oscillation.

For completeness, if $a_2=0$ and $a_1\ne 0$, only real exponential functions occur. If $a_2=a_1=0$ and $a_0\ne 0$, only $f=0$ is allowed. If all three coefficients vanish, every twice-differentiable function is allowed, including oscillatory ones; the second-order criterion does not apply to this degenerate identity.

Using Euler's formula, the difference of the two complex exponentials is $2i\sin (\pi /4)=i\sqrt{2}$. HenceThese are the real part and imaginary part, respectively.

The polar form is $1+i\sqrt{3}=2e^{i\pi /3}$. By De Moivre's theorem,Therefore the real part and imaginary part are

Multiply numerator and denominator by the denominator's complex conjugate:Thus the real part and imaginary part are

The degree-five row of the wavevector selection rule for equivariant monomials in part (a) has two vectors other than $(1,0,0,0)$. They correspond exactly toThe vector $(1,0,0,0)$ gives only regular fifth-order terms $A_1|A_j{|}^2|A_k{|}^2$. Omitting those as required, and applying the same two square-symmetry generators, the truncated normal form of a dynamical system isHere $e,f$ are real because rotation by $\pi$ acts by complex conjugation. These are equations for steady patterns, so all four complex right-hand sides must vanish; constancy of the two invariant angles alone would only establish a relative equilibrium and could allow translation drift.

Take a nonzero equal-magnitude state, $|A_j|=R>0$, and set $z=R^2$, $S=c_1+c_2+c_3+c_4$, $\rho =\sigma +Sz$. Dividing each equation by its nonzero amplitude gives, in order,Taking imaginary parts, adding and subtracting pairs, yieldsThus, unless $e=f=0$, each invariant phase is zero or $\pi$ modulo $2\pi$. Equality of the real parts further givesThis identifies the equal-amplitude states of an eight-mode square pattern completely. For generic $e\ne f$, the two nonzero phase types, with their amplitude equations, areRepresentatives are respectively $(A_1,A_2,A_3,A_4)=R(1,1,1,1)$ and $R(-1,1,-1,1)$. Spatial translations change individual phases but leave the indicated phase pair fixed. Reflection acts on the invariant angles as $({\chi }_1,{\chi }_2)\mapsto (-{\chi }_1,{\chi }_2)$, and interchange of $x,y$ as $({\chi }_1,{\chi }_2)\mapsto (-{\chi }_2,-{\chi }_1)$. Thus the two rows are not equivalent under square symmetry and translations. Every positive root of the corresponding polynomial gives an amplitude branch of that type; if there is no positive root, that type is absent at those parameter values.

There are two coefficient degeneracies to include. When $e=f\ne 0$, the mixed pairs $(0,\pi )$ and $(\pi ,0)$ are allowed and are equivalent under square symmetry. They form one additional phase type, with amplitude equation $\boxed{\sigma +Sz=0}$; $iR(1,1,1,1)$ is a representative. When $e=f=0$, all invariant phases are unrestricted, and the same cubic amplitude equation gives a continuous family modulo translations and the finite square-group action. Even if $e+f=0$ but $e\ne f$, only the original two phase types remain: the equality of their amplitude equations does not remove the phase constraints. Finally, the all-zero state is always a separate steady state and has no defined polar phases.

The spatial reflection $x\mapsto -x$ acts on the two complex amplitudes as $(A,B)\mapsto (A,-B)$, because the first spatial eigenfunction is even and the second is odd. The first equation must therefore be even in $B,\overline{B}$, and the second odd. For a nonresonant double Hopf bifurcation, the temporal phases give independent actions $A\mapsto e^{i{\phi }_1}A$, $B\mapsto e^{i{\phi }_2}B$ in the cubic normal form of a dynamical system. A cubic monomial in the first equation must have net $A$ phase weight one and net $B$ phase weight zero. Only $A|A{|}^2,A|B{|}^2$ satisfy this; the second equation similarly has $B|B{|}^2,B|A{|}^2$. Parametrizing the independent cross-saturation coefficients givesAll coefficients can be complex: their real parts determine amplitude growth and their imaginary parts frequency shifts. These are cubic truncations of a family of reduced vector fields on the four-real-dimensional extended centre manifold, with the two unfolding parameters entering ${\nu }_1,{\nu }_2$.

The nonresonance assumption must be made explicit. Merely having ${\omega }_1\ne {\omega }_2$ does not imply this form: for ${\omega }_1=2{\omega }_2$, the quadratic terms $B^2$ in $\dot{A}$ and $A\overline{B}$ in $\dot{B}$ satisfy both temporal resonance and spatial parity. Thus the initial form is the generic nonresonant double-Hopf form, rather than a statement for every possible unequal frequency pair.

At the one-to-one reflection-symmetric one-to-one Hopf resonance, only the common temporal phase action remains. Cubic terms must then have total phase weight one. Spatial reflection retains, in the first equation, the three terms $A|A{|}^2,A|B{|}^2,B^2\overline{A}$; the second retains $B|B{|}^2,B|A{|}^2,A^2\overline{B}$. Common phase covariance excludes all quadratic terms. The additional resonant terms are consequently $-{\gamma }_1{B}^2\overline{A}$ and $-{\gamma }_2{A}^2\overline{B}$, with independent complex coefficients. This is the reflection-symmetric one-to-one Hopf resonance.

In the specified equal-coefficient specialization, put $r=R^2$. A nonzero pure-$A$ periodic branch exists for $\mathrm{Re}{\nu }_1>0$ and hasIts radial eigenvalue is $-2{\alpha }_{R}r<0$, and its phase is neutral. For a transverse $B$ perturbation the linear equation isIts time dependence can be removed exactly by $B=C{e}^{i\Omega t}$. Using ${\nu }_1-\alpha r=i\Omega$ and ${\nu }_2={\nu }_1-d$ givesWriting $C=u+iv$, $d=d_R+i{d}_I$, $\beta ={\beta }_R+i{\beta }_I$ yields the constant real stability matrixLet $Q=\mathrm{Re}(\beta \overline{d})={\beta }_R{d}_R+{\beta }_I{d}_I$. The transverse characteristic polynomial isBoth transverse eigenvalues have negative real part exactly whenThese hold at $r=0$ because $d_R>0$. Thus a sufficiently small pure branch is transversely stable; it is stable modulo its neutral temporal phase until one of the two inequalities fails. The secondary steady and Hopf thresholds of a resonant pure mode are derived in the two parts below. They refer to accessible points on the pure branch; in a multi-parameter problem $d$ may also vary, so a proposed parameter path must actually cross the corresponding surface.