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Place a converging thin lens at coordinate $x$ between the object at $0$ and the screen at $L$. In air and the paraxial approximation, the required object and image distances are $u=x$ and $v=L-x$. The thin lens equation gives a quadratic equation:Its discriminant is $L^2-4Lf$. For $L>4f>0$, the two roots and their separation areBoth roots lie inside $(0,L)$, and both object and image distances exceed $f$. The two roots therefore describe finite real images on the screen. Exchanging the roots exchanges $u,v$; from transverse magnification $m=-v/u$,The image-height ratio between the enlarged and reduced settings is $|m_{-}|/|m_{+}|=|m_{-}{|}^2$, rather than $|m_{-}|$.

At $L=4f$, the roots coincide: $d=0$, $x=L/2=2f$, $u=v=2f$ and $m=-1$. There is one setting, not two separately measurable settings. For $0<L<4f$ the discriminant is negative, so no position between the fixed object and screen gives a sharp real image there. This does not forbid a real image elsewhere if the screen is moved. Another short proof observes that $x(L-x)$ has maximum $L^2/4$ at $x=L/2$; the focusing requirement $x(L-x)=Lf$ is possible exactly when $L\ge 4f$.

A binary quartic is a homogeneous degree-four polynomial in two variables. Over the complex numbers it determines four points of the projective line, counted with multiplicity. Its discriminant is nonzero exactly when the four points are distinct; this is also its geometric invariant theory stability condition under $S{L}_2$.

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.

The polynomial is irreducible with discriminant $229$, so its splitting field has Galois group $S_3$ and quadratic subfield $\mathbb{Q}(\sqrt{229})$. The splitting field is cyclic cubic over that subfield. It is unramified outside 229, and at 229 its inertia is a transposition, meeting the alternating cubic subgroup trivially. All roots are real. It is therefore everywhere unramified over the quadratic field. The ideal class group of Q of square root 229 has order three, so this extension is exactly the Hilbert class field.

The three reduced primitive binary quadratic forms of discriminant $-23$ are $(1,1,6)$ and $(2,\pm 1,3)$, so the class number is three. The displayed cubic has discriminant $-23$ and is irreducible. Its splitting field has Galois group $S_3$ and is cyclic cubic over $\mathbb{Q}(\sqrt{-23})$. Modulo $23$ it factors as $(X-10{)}^2(X-3)$. Lift the simple factor; the remaining quadratic has discriminant valuation one, so its local inertia group has order two. Its intersection with $A_3$ is trivial. Thus the relative cyclic cubic extension is everywhere unramified and equals the Hilbert class field.

At the positive Nicholson-Bailey model equilibrium, $\mathrm{tr}{J}_{\ast }=1+\log r/(r-1)$ lies between one and two, and the displayed determinant exceeds one. These follow from $(r-1)/r<\log r<r-1$ for $r>1$. The discriminant is therefore negative, and the conjugate eigenvalues have moduli greater than one. Nearby annual samples spiral outward rather than approach coexistence.

Let $m_1,m_2$ be coprime positive square-free integers, with $m_1\not\equiv \pm m_2(\mathrm{mod}9)$. The displayed integral basis has discriminant $-27m_1^2{m}_2^2$. Both nonconstant generators are integral, and their products are integral linear combinations of the three basis elements. At a prime dividing $m_1$, the cubic for $\alpha$ is Eisenstein; at a prime dividing $m_2$, the cubic for ${\alpha }^2/m_2$ is Eisenstein. If $3\nmid m_1{m}_2$, then $m_1{m}_2^2\not\equiv \pm 1(\mathrm{mod}9)$, and a shift by $1$ or $-1$ gives an Eisenstein polynomial at three. These local uniformizer generators show that the order is maximal at all primes dividing its discriminant; elsewhere the discriminant-index formula for an integral lattice already gives maximality.

If $m>1$ is a square-free integer, $3\nmid m$, and $m\not\equiv \pm 1(\mathrm{mod}9)$, then $1,\sqrt[3]{m},\sqrt[3]{m^2}$ is an integral basis of the pure cubic number field, with field discriminant $-27m^2$. The discriminant-index formula for an integral lattice restricts a possible index to primes dividing $3m$. At primes dividing $m$ the different exponent is $2$ by tame ramification. A shifted Eisenstein polynomial gives total wild ramification at $3$, so its different exponent is at least $3$. These exponents exhaust the polynomial discriminant and force index one.

Decaying normal modes of this continuous piecewise-linear profile satisfy ${\varphi }^{\prime \prime }-k^2\varphi =0$ between the corners. Pressure matching at a piecewise-linear shear interface yields an odd varicose mode of a planar jet with real speed $c=(1-e^{-2k})/(2k)$ and an even sinuous mode of a planar jet whose speeds satisfyIts reduced discriminant factors as $(2k-3-e^{-2k}-4e^{-k})(2k-3-e^{-2k}+4e^{-k})$. The second factor is positive for $k>0$, while the first is strictly increasing and crosses zero once. Thus exactly the long-wave even branch is exponentially unstable, for $0<k<k_c$, where $2k_c-3-e^{-2k_c}-4e^{-k_c}=0$.

An order in a number field $K$ is a subring $A\subseteq {\mathcal{O}}_K$ containing $1$ and having full rank $[K:\mathbb{Q}]$ as an additive abelian group. Its finite index in the ring of integers of a number field is related to its discriminant by the discriminant-index formula for an integral lattice.

For real constants with $a_2\ne 0$, a nonzero solution of the homogeneous linear differential equation $a_2{f}^{\prime \prime }+a_1{f}^{\prime }+a_{0}f=0$ has a sinusoidal factor exactly when the discriminant $a_1^2-4a_0{a}_2$ is negative. Indeed the characteristic polynomial has roots $-a_1/(2a_2)\pm i\omega$, where $\omega =\sqrt{4a_0{a}_2-a_1^2}/(2|a_2|)$, giving $f=e^{-a_{1}t/(2a_2)}(C_1\cos \omega t+C_2\sin \omega t)$. Real or repeated roots instead give real exponential functions, possibly multiplied by a linear polynomial. Oscillatory behaviour and decay are separate: decay requires $a_1/a_2>0$.

Two counterpropagating modes transform as $(A,B)\mapsto (e^{ik\delta }A,e^{-ik\delta }B)$ under spatial translation. Forcing at twice their natural frequency breaks continuous temporal phase symmetry but preserves simultaneous sign reversal. Hence $\overline{B}$ transforms like $A$ and $\overline{A}$ like $B$, allowing linear parametric coupling. An equal-amplitude phase-locked standing wave has $|A{|}^2=|B{|}^2=z>0$ with $|\beta +\gamma {|}^2{z}^2-2\mu \mathrm{Re}(\beta +\gamma )z+{\mu }^2-{\nu }^2=0$. A nonnegative discriminant is only an algebraic compatibility condition: a physical nonzero wave also requires a positive root.

At an equilibrium $(v_{\ast },k{v}_{\ast })$, set $p=f^{\prime }(v_{\ast })$. The Jacobian matrix, its trace and its determinant areThe eigenvalues and their discriminant areThus the complete local linear stability test isIn the last two cases, $\mathcal{D}>0$ gives a node, while $\mathcal{D}<0$ gives a focus; the sign of $T$ distinguishes inward and outward trajectories. At $\mathcal{D}=0$, a nonzero repeated eigenvalue gives a degenerate node. At $p=\gamma <k$ the eigenvalues are purely imaginary, so linear stability analysis alone does not decide asymptotic stability; with the required crossing and nondegeneracy this is a Hopf bifurcation. At $p=k$ there is a zero eigenvalue, requiring a nonlinear test and generically marking the saddle-node bifurcation in the preceding answer.

For a unique simple equilibrium, $p<k$, so it can be a sink equilibrium or a source equilibrium. The number of equilibria alone does not decide stability. If $k\le \gamma$, the unique simple equilibrium is always a sink equilibrium. If $\gamma <k$, it can become a source equilibrium where $\gamma <p<k$. The same conclusion follows particularly simply if $\gamma >M$: the trace is negative at every nonsaddle equilibrium.

For three simple equilibria ordered by their $v$ coordinates, the middle one has $p>k$ and is a saddle equilibrium; the two outer ones have $p<k$. When both outer equilibria are sink equilibria, each has an attracting neighborhood, giving bistability. In the usual two-basin portrait, the stable manifold of the saddle equilibrium separates their basins. Each branch of its unstable manifold initially leaves towards a different side; its eventual destination requires a global argument. If $k\le \gamma$, both outer equilibria must be sink equilibria. If $\gamma <k$, either or both outer equilibria can instead be source equilibria; the same test $p\lessgtr \gamma$ applies separately to each. The three lower panels illustrate all three possibilities up to exchanging left and right.

The arrows in the phase plane follow directly from the nullclines: below $w=f(v)+I$ the horizontal component points right, and above it left; below $w=kv$ the vertical component points up, and above it down. Near a stable focus trajectories spiral inward; near an unstable focus they spiral outward. Near a saddle equilibrium they approach along its stable manifold and leave along its unstable manifold. These directions, together with the local tests, determine the sketches.

There is also a useful global conclusion when the unique equilibrium is a source equilibrium. Withthe leading term in $v$ is $-\beta v^4$, so sufficiently large ellipses are trapping regions. A small repelling neighborhood of the hyperbolic source equilibrium can be removed to form an invariant annulus. The Poincaré-Bendixson theorem then guarantees at least one periodic orbit in that annulus. The numerical sketch illustrates the resulting sustained oscillation; this argument does not assert uniqueness of a limit cycle. A stable equilibrium can instead represent a resting membrane potential, with perturbations producing damped relaxation; near a threshold, a sufficiently large perturbation can produce a large excursion before recovery. Global basin boundaries and coexistence of periodic orbits need more information than the number and local stability of equilibria.

The discriminant of a binary quadratic form is $d=b^2-4ac$. If the form is positive-definite, evaluating it at $(1,0)$ gives $a>0$. For $a\ne 0$, completing the square givesEvaluating along $x=-by/(2a)$ shows that positivity also requires $d<0$. Conversely these two inequalities make both displayed coefficients positive, provingA reduced positive definite binary quadratic form satisfies $|b|\le a\le c$, with $b\ge 0$ if $|b|=a$ or $a=c$. The last convention avoids counting both boundary representatives. For fixed negative discriminant, these inequalities implyThere are finitely many such integers $a$, finitely many integers $b$ with $|b|\le a$, and then $c=(b^2-d)/(4a)$ is determined. Hence the number of reduced forms is finite. If no admissible integer coefficients have the given discriminant, this number is zero.

For $d=-39$, the bound gives $a=1,2,3$. The integrality of $c=(b^2+39)/(4a)$ and the boundary sign convention give respectivelyThese are all the reduced positive definite binary quadratic forms of discriminant minus thirty-nine. Thusrepresented by $x^2+xy+10y^2$, $2x^2+xy+5y^2$, $2x^2-xy+5y^2$, and $3x^2+3xy+4y^2$.

Let $f\in R[X]$ have degree $n$. A useful reduction prime must preserve its degree and make its reduction separable. Thus exclude the leading coefficient and discriminant from $P$; after localizing at those elements one may make $f$ monic. The residue base field is $k=\mathrm{Frac}(R/P)$, not necessarily the ring $R/P$ itself. Then the Galois group of the reduced polynomial embeds in the generic group, with the same action on the corresponding roots.

Here is the unramified reduction of a polynomial Galois group argument. A UFD is integrally closed. Let $E$ be the splitting field over $F$, and use the root algebra $B$ obtained by adjoining all roots after inverting the discriminant. It is finite étale over the resulting normal base: successive simple-root adjunctions are unramified, and the distinct generic components separate over a normal ring. Choose a prime $Q$ above $P$. The residue field $k_Q$ is generated over $k$ by the reduced roots, so it is their splitting field. The decomposition groupacts on $k_Q$. Its residue action is onto $\mathrm{Gal}(k_Q/k)$; this is the local decomposition-group description of the unramified fiber. Equivalently, the geometric points of that fiber form a free transitive Galois-group set, and the commuting residue-field action identifies the residue group with a stabilizer subgroup of the generic group.

The kernel is the inertia group. An element in that kernel fixes every reduced root. Since the discriminant is a unit, distinct roots have distinct reductions; consequently it fixes every original root and is the identity. ThereforeChoosing another prime changes this subgroup by conjugacy, which does not change the cycle types available for computing the generic group.

When $k={\mathbb{F}}_q$, its splitting-field group is generated by Frobenius. A reduced irreducible factor of degree $d_i$ contributes one $d_i$-cycle. Hence a squarefree factorization with degrees $d_1,\dots ,d_r$ proves that the generic group contains that cycle type. Irreducible degree-preserving reduction also proves generic irreducibility by Gauss's lemma: a nontrivial monic factorization would reduce to one. Several good primes, a discriminant-square test and permutation-group arguments can then identify the group. A ramified or degree-dropping prime does not justify the same cycle test.

The Kronecker–Weber theorem says that every finite abelian extension of $\mathbb{Q}$ is contained in a cyclotomic field. We prove its quadratic case by identifying the ramification that a quadratic number field must have.

For an odd prime $\ell$, the cyclic Galois group of $\mathbb{Q}({\zeta }_{\ell })/\mathbb{Q}$ has exactly one subgroup of index two. Its fixed field $F_{\ell }$ is quadratic and is unramified at every finite prime other than $\ell$. For a squarefree integer $d$, the discriminant of $\mathbb{Q}(\sqrt{d})$ is $d$ if $d\equiv 1(\mathrm{mod}4)$, and $4d$ otherwise. A nontrivial quadratic discriminant supported on the odd prime $\ell$ must therefore beIndeed $d=\pm 1$ cannot give a nontrivial extension unramified at two, whereas exactly one of $\ell ,-\ell$ is congruent to one modulo four. ThusThis identifies the quadratic subfield of a prime cyclotomic field using ramification, without a Gaussian-sum computation.

Now let $F=\mathbb{Q}(\sqrt{d})$ with $d$ squarefree. Multiply $d$ by ${\prod }_{\ell \mid d,\ell \text{ odd}}{\ell }^{\ast }$. Every odd prime then occurs with even exponent, so its square class is represented by one of $1,-1,2,-2$. Thus adjoining the square roots of the ${\ell }^{\ast }$ removes all possible odd-prime ramification; the remaining quadratic factor is contained inThe original $\sqrt{d}$ is a rational multiple of a product of these square roots. It belongs to the compositum of $\mathbb{Q}({\zeta }_8)$ and the prime cyclotomic fields just used. This compositum is itself contained in a cyclotomic field, proving the theorem for $F$.

The sharper conductor statement follows from the same ramification calculation. For the fundamental discriminant $D$ of $F$, putThe possibilities for $D_2$ follow immediately from the quadratic discriminant formula. The corresponding fields are trivial, $\mathbb{Q}(i)$, $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{-2})$, lying in the cyclotomic fields of conductors $1,4,8,8$. Their conductors are coprime to the odd primes in the product. Hence quadratic Kronecker–Weber theorem by ramification gives the explicit conclusion

A modulus of a number field is $\mathfrak{m}={\mathfrak{m}}_0{\mathfrak{m}}_{\infty }$, with ${\mathfrak{m}}_0$ a nonzero integral ideal and ${\mathfrak{m}}_{\infty }$ a subset of the real places. Let $I_{\mathfrak{m}}$ be the group of fractional ideals prime to ${\mathfrak{m}}_0$, and let $P_{\mathfrak{m}}$ consist of principal ideals $(a)$ with $a\equiv 1$ at each prime power in ${\mathfrak{m}}_0$ and $a>0$ at each selected real place. The ray class group isThe ray class field $K_{\mathfrak{m}}$ is the finite abelian extension for which the Artin map has kernel $P_{\mathfrak{m}}$ and induces an isomorphism ${\mathrm{Cl}}_{\mathfrak{m}}(K)\cong \mathrm{Gal}(K_{\mathfrak{m}}/K)$. Equivalently it is the maximal abelian extension whose conductor divides $\mathfrak{m}$.

To compute its order use the ray class exact sequence, including signatures:The middle map lifts a residue and sign tuple to an element of $K$ by weak approximation theorem, then takes its principal ideal as a ray class. Different lifts differ by a principal ray factor. Its kernel consists exactly of tuples represented by global units. Every ordinary ideal class has a representative prime to the finite modulus, again by weak approximation theorem; this gives surjectivity of the last map. If $U_{\mathfrak{m}}$ denotes the ray units, the kernel of the unit map is $U_{\mathfrak{m}}$. The Chinese remainder theorem and counting units in each prime-power quotient give the ray class number formula

First take $K=\mathbb{Q}(\sqrt{-3})$. Its ring of integers is $\mathbb{Z}[\omega ]$, where $\omega =(-1+\sqrt{-3})/2$ and ${\omega }^2+\omega +1=0$. The Minkowski bound for ideal classes is $(2/\pi )\sqrt{3}<2$, so every ideal class has a representative of norm one and $h_K=1$. Its units are the six roots of unity. Modulo seven, the polynomial $X^2+X+1$ has distinct roots $2,4$, and thereforeThe unit group of this quotient has order $36$. The unit $1+\omega$ maps to $(3,5)$, which has order six; consequently unit reduction is injective and $U_{(7)}=\{1\}$. The ray class number formula gives $36/6=6$.

We also identify the quotient explicitly. Conjugation interchanges the two factors, so the residue norm is $(x,y)\mapsto xy$. Its kernel has order six and equals the image of the six units, since $3\cdot 5=1(\mathrm{mod}7)$. HenceThe field $M=K({\zeta }_7)=\mathbb{Q}({\zeta }_{21})$ has degree six over $K$, since the degrees over $\mathbb{Q}$ are twelve and two. It is unramified outside seven by the cyclotomic computation and the base-change argument above. At primes over seven, $K$ has completion ${\mathbb{Q}}_7$, and $M/K$ is the tame degree-six cyclotomic extension; its conductor exponent is one. Its relative arithmetic Artin symbol at an ideal $\mathfrak{a}$ prime to seven acts byIf $\mathfrak{a}=(a)$ with $a\equiv 1(\mathrm{mod}7{\mathcal{O}}_K)$, then $N_{K/\mathbb{Q}}(a)\equiv 1(\mathrm{mod}7)$. Thus the Artin map kills the principal ray subgroup. It factors through the order-six ray class group; the residue norm quotient computed above shows that it is onto the degree-six Galois group, so the fields coincide. This proves the explicit ray class field of Q of square root minus three modulo seven:

Now take $K=\mathbb{Q}(\sqrt{-6})$, and write $s=\sqrt{-6}$. Its ring of integers is $\mathbb{Z}[s]$, its discriminant is $-24$, and its units are $\pm 1$. The quotient by $(s)$ iswhose two units are both images of $\pm 1$. The ray class exact sequence therefore gives ${\mathrm{Cl}}_{(s)}(K)\cong \mathrm{Cl}(K)$.

For its class number, the Minkowski bound for ideal classes is $(2/\pi )\sqrt{24}<4$, so each ideal class contains an integral ideal of norm at most three. The primes of norms two and three areThese are the only nontrivial possibilities within the bound, and their ideal classes coincide and have order at most two. The ideal ${\mathfrak{p}}_2$ is not principal: its being principal would give an integer solution to $x^2+6y^2=2$, which is impossible. Thus $h_K=2$, the ideal class group is $C_2$, and the ray field is the Hilbert class field.

We verify its actual identity rather than stopping at the degree. PutThe two quadratic fields $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{-3})$ are distinct, so $[H:\mathbb{Q}]=4$ and $[H:K]=2$. The integral elements $1,\sqrt{2},b,\sqrt{2}b$ are a rational basis. Their trace matrix is the Kronecker product of the two quadratic trace matrices, with discriminants $8$ and $-3$. Its determinant is therefore $8^2(-3{)}^2=576$. The discriminant-index formula for an integral lattice shows that the field discriminant divides $576$. On the other hand the relative discriminant ideal formula givesThe norm on the right is a positive integer, so equality is forced. Thus the relative discriminant ideal is the unit ideal, and $H/K$ is unramified at every finite prime. There are no real places of $K$. The extension is quadratic and therefore Abelian; its degree equals $h_K$, identifying it with the Hilbert class field of Q of square root minus six. Hence the ray class field of Q of square root minus six modulo square root minus six is

Using the printed convention that the projective points represent lines of binary quartics, the locus of GIT stable points isThe root at infinity is included. We determine stability by calculating every relevant algebraic one-parameter subgroup weight.

Use the natural action $(g\cdot f)(X,Y)=f(g^{-1}(X,Y))$. Every nontrivial algebraic one-parameter subgroup of $S{L}_2$ is conjugate to ${\lambda }_m(t)=\mathrm{diag}(t^m,t^{-m})$ for some integer $m>0$: its two integral weights of a representation sum to zero, and a basis of weight vectors can be rescaled to have determinant one. WriteIf $r=\min \{j:a_j\ne 0\}$, then $r$ is the multiplicity of a root at the projective point $[1:0]$: in the local coordinate $Y/X$, the polynomial begins with a nonzero multiple of $(Y/X{)}^r$.

The Hilbert-Mumford criterion for projective stability, with $\mu =-\min$(active weights), now givesBeing a GIT stable point requires this number to be strictly positive for every nontrivial subgroup. Conjugating the subgroup lets $[1:0]$ range over the entire projective line. Thus the root-multiplicity criterion for stable binary forms becomes $r_p(f)<2$ at every point $p$. A nonzero degree-four form over $\mathbb{C}$ has four projective roots counted with multiplicity, so this means precisely four distinct roots. This is equivalently nonvanishing of its discriminant.

The boundary cases show why a weak inequality would give the wrong answer. A triple or quadruple root can be moved to $[1:0]$, leaving only strictly positive weights; its affine representative tends to zero under that subgroup, so it is unstable. If all root multiplicities are at most two, all Hilbert-Mumford values are nonnegative and the point is a GIT semistable point. A double root gives a zero value for a suitable subgroup, so never represents a GIT stable point. More concretely, after putting that double root at $[1:0]$ one hasand ${\lambda }_1(t)\cdot f\to a{X}^2{Y}^2$. When there are three distinct roots, this limiting form, representing a GIT semistable point with two distinct roots, is outside the original group orbit, so that group orbit is not closed in the semistable locus. When there are two double roots, a change of coordinates gives $X^2{Y}^2$, which has the positive-dimensional diagonal torus as stabilizer. Neither case represents a GIT stable point.

For a form with four distinct roots, the projective stabilizer is finite: it permutes the four roots, and a Möbius transformation fixing three distinct points is the identity. Thus its image in $PG{L}_2$ injects into the finite permutation group on the roots. The kernel of a group homomorphism for $S{L}_2\to PG{L}_2$ is $\{\pm I\}$, also finite. This agrees with the finite-stabilizer condition in geometric invariant theory. The weight calculation already supplies the required strict stability condition for every conjugate subgroup.

Write $s={\alpha }^2+{\pi }^2$ and use modes satisfying all boundary conditions,The quadratic temperature distortion $c$ is retained for the saturation calculation; it does not enter the linear onset. Substitution of the primary modes givesThe determinant vanishes whenThe trace is $-[s+\sigma (s+Q{\pi }^2/s)]<0$. Thus no conjugate eigenvalues can cross the imaginary axis away from zero, and the initial instability is stationary, not oscillatory. Indeed for $R\ge 0$ the discriminant is $[\sigma (s+Q{\pi }^2/s)-s{]}^2+4\sigma R{\alpha }^2/s>0$. This is quasistatic vertical-field magnetoconvection: there is no independent magnetic evolution mode that could produce a magnetic-thermal Hopf bifurcation instability.

For a fixed box the other allowed horizontal wavenumbers are $n\pi /L$, and the actual first threshold is the minimum of their corresponding $R_c$ values. Vertical index $m$ replaces ${\pi }^2$ by $m^2{\pi }^2$ and strictly increases the threshold at fixed horizontal wavenumber, so $m=1$ is selected. The displayed formula describes the prescribed fundamental roll; part (b) optimizes its width.

The characteristic polynomial of a real $2\times 2$ matrix is $p(z)=z^2-Tz+D$. Thus $p(1)=1-T+D$ vanishes exactly when $T=D+1$, and $p(-1)=1+T+D$ vanishes exactly when $T+D+1=0$. These are the required eigenvalue conditions.

For $D=1$ and $|T|<2$, the discriminant is negative, andTheir moduli equal one. Taking $\alpha \in (0,\pi )$ with $\cos \alpha =T/2$ gives $z_{\pm }=e^{\pm i\alpha }$. The strict trace inequalities exclude $\alpha =0,\pi$ and hence exclude the repeated real multipliers $+1,-1$.