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The function $F$ is an odd function, passes through $(0,0)$, and has the horizontal asymptote $F=0$ at both infinities. Its derivative isIt decreases for $|x|>1$ and increases for $|x|<1$, with its lowest value $F(-1)=-1/2$ and highest value $F(1)=1/2$. These features determine the graph of a function shown below.

For $0<ϵ<1/2$, the equation is the quadratic equation $ϵx^2-x+ϵ=0$. The quadratic formula gives exactly two real equation roots:The binomial series gives $\sqrt{1-4{ϵ}^2}=1-2{ϵ}^2-2{ϵ}^4-4{ϵ}^6+O({ϵ}^8)$, hence the two requested asymptotic expansions areTheir product is exactly one. Both are positive, with $x_{-}<1<x_{+}$; there are no negative equation roots because $F(x)<0$ for $x<0$. At $ϵ=1/2$ the two real equation roots coincide at one, and for $ϵ>1/2$ there are none. In the stated limit $ϵ\to 0^{+}$, these two branches are therefore exhaustive.

The quadratic formula gives two algebraic possibilities. To select one consistent holomorphic square root, setwhere each factor is the principal square root. Both arguments lie in the upper half-plane, so neither crosses its branch cut. This choice satisfies $S(\zeta {)}^2={\zeta }^2-4$ and $S(\zeta )\sim \zeta$ at large $\zeta$ in the upper half-plane. The required inverse function isFor $\zeta =iR$, $R>0$, this gives $z=i(R+\sqrt{R^2+4})/2$, whose modulus exceeds one. As $\zeta$ varies continuously in the upper half-plane, this root cannot cross $|z|=1$, since that would make $\zeta$ real. Hence the selected root always lies in the fluid region. The other root is its reciprocal and lies inside the unit disk. Simply writing the principal square root of ${\zeta }^2-4$ would not specify this branch correctly throughout the upper half-plane.

Let $w$ satisfy $\sin w=z$ and put $t=e^{iw}\ne 0$. The exponential definition of the complex sine givesThe quadratic formula therefore gives $t=iz\pm \sqrt{1-z^2}$. Taking the complex logarithm yields the multivalued complex inverse sine formulaThe two quadratic roots have product $-1$, so neither is zero and their logarithms exist as multivalued quantities. If $w_0$ is one local inverse value, the complete set of values is $w_0+2\pi k$ and $\pi -w_0+2\pi k$, with $k\in \mathbb{Z}$.

Branch qualification: the source's $\pm$ formula describes the branches of the complex inverse sine, not two interchangeable expressions for a single chosen inverse branch. At $z=0$, the plus sign with the usual local choices gives $0$, whereas the minus sign with $\log (-1)=i\pi$ gives $\pi$. Thus both signs cannot simultaneously equal the principal inverse at zero. A principal complex inverse sine requires compatible square-root and logarithm branches, normalized to $w_0=0$ at zero and continued on the chosen cut domain.

For the given Möbius transformation, the fixed point equation is $z(z+3)=3z+1$, hence

For the general Möbius transformation, its finite fixed points satisfyWith $c\ne 0$ and nonzero discriminant, the quadratic formula gives the two distinct fixed points of a Möbius transformation:The fixed-point equation for $\alpha$ also gives $b-d\alpha =-\alpha (a-c\alpha )$. ThereforeDividing yields the conjugation of a Möbius transformation with two distinct fixed points:Neither factor in this quotient vanishes, since $(a-c\alpha )(a-c\beta )=ad-bc\ne 0$. The formula extends over the Riemann sphere, including the transformation's pole. For the special transformation and the ordering $\alpha =1,\beta =-1$, $k=1/2$.

Factor the equation asFor the nontrivial roots, the quadratic formula over residues givesThe supplied square root is $14$, and $2^{-1}\equiv 100(\mathrm{mod}199)$. ThereforeIndeed $107^2-107+1=11343=57\cdot 199$, so $107^3+1$ is divisible by $199$, and $107\not\equiv -1$. The other nontrivial root is $(1-14)100\equiv 93$; it satisfies $93^2-93+1=43\cdot 199$. Together with $198\equiv -1$, these are the three cube roots of minus one. The complex-number hint reflects the same quadratic factor: the two roots there are $(1\pm \sqrt{-3})/2$.

Write the complex number as $z=x+iy$. Since $iz=-y+ix$, the modulus of the complex exponential givesThis equals one precisely when $y=0$. The quadratic formula shows that the imaginary part of either root is $\pm \sqrt{4-b^2}/2$. Consequently the roots are real exactly at the two endpoints:Both endpoints satisfy the condition. The repeated roots are respectively $1$ and $-1$; their multiples by $i$ have exponentials on the unit circle.

The quadratic formula givesso both roots of a polynomial have real part $-b/2$. Writing $z=x+iy$, the complex exponential function satisfies $e^z=e^x(\cos y+i\sin y)$, and hence its modulus is $e^x$. ThereforeIntersecting this condition with the allowed parameter interval gives the answer for either root:The endpoint $b=2$ is included: the repeated root of a polynomial is $z=-1$, whose exponential has modulus $e^{-1}<1$.

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.

The factor theorem supplies the factor $x-3$. Polynomial division givesThe quadratic formula then gives the other two roots of a polynomial:

The quadratic formula gives the two polynomial roots $(-3\pm \sqrt{5})/2$. Its leading coefficient is positive, so the quadratic polynomial is strictly negative between them. Therefore

The quadratic formula gives $t=-1\pm \sqrt{2}$. The tangent function is positive in the first quadrant, so the positive polynomial root is selected:

Applying the quadratic formula to the characteristic polynomial gives the two eigenvalues

Put $d=\sqrt{a^2-b^2-c^2}>0$ and $h=b-ic$. The denominator of the complex rational function becomesThe quadratic formula gives its two zeros, hence the two poles:Their product is $(b+ic)/(b-ic)$, so $|z_{\mathrm{in}}{z}_{\mathrm{out}}|=1$. To locate them one also needs to exclude the possibility of both lying on the unit circle. Here their moduli are explicit:Both are simple poles: $P^{\prime }(z)=hz+a$ takes the nonzero values $d$ and $-d$ at them. In particular their residues are $1/d$ and $-1/d$.

For the real integral, use the change of variables $z=e^{i\theta }$. As $\theta$ runs from $-\pi$ to $\pi$, this traverses the unit circle once counterclockwise, andThus $I=(1/i){\oint }_{|z|=1}f(z)dz$. Only $z_{\mathrm{in}}$ is enclosed. The residue theorem, or Cauchy integral formula after subtracting its simple-pole part, givesThe real denominator is positive because $a>\sqrt{b^2+c^2}$. As an independent check, write $b\cos \theta +c\sin \theta =\sqrt{b^2+c^2}\cos (\theta -\varphi )$; a period shift reduces the answer to the existing reciprocal affine sine integral.

Solving $\mu x(1-x)=1$ by the quadratic formula givesFor $\mu >4$, $0<x_{-}<1/2<x_{+}<1$. On $L=[0,x_{-}]$ the derivative $f^{\prime }(x)=\mu (1-2x)$ is strictly positive, while on $R=[x_{+},1]$ it is strictly negative. Thus the restrictions are strictly monotonic functions. Their endpoint values areBy continuity and strict monotonicity, each restriction maps its interval bijectively onto $[0,1]$, with opposite orientations. The corresponding inverse functions are

Setting $a(x-h{)}^2+k=0$ gives $(x-h{)}^2=-k/a$. A positive $-k/a$ gives two distinct roots of a polynomial in the real numbers; zero gives a repeated root of a polynomial at $h$; a negative value gives a pair of complex conjugate roots of a polynomial. For the downward-opening example, the two roots of a polynomial are $2\pm \sqrt{3/2}$, approximately $0.775255$ and $3.224745$.

The discriminant $\Delta =b^2-4ac$ packages the same calculation, becauseSince $4a^2>0$, the sign of $\Delta$ determines which of the three cases occurs. This derives the quadratic formula by a polynomial identity, and explains its structure: first shift to $h$, then take a square root. For the upward-opening example, $(x+3{)}^2=-4$ gives the complex conjugate roots of a polynomial $-3\pm 2i$.

A solubility product constrains a product of thermodynamic activities; the available inventories determine which combinations of concentrations are possible. Their intersection determines the molar solubility. Consider an original, abstract AB solid, with the preceding one-to-one dissolution chemical equation. Stipulate $c^{\circ }=1\mathrm{M}$, ${\gamma }_{\mathrm{A}}={\gamma }_{\mathrm{B}}=1$ and $K_{\mathrm{sp},c}=10^{-4}{\mathrm{M}}^2$ at fixed temperature and pressure. Thus $K_{\mathrm{sp}}=10^{-4}$ is dimensionless. These are chosen model parameters, not data for an identified material.

Initially, a completely dissociated soluble XB salt supplies B anions at molar concentration $c_0\ge 0$ and X cations at the same molar concentration; A is absent. All three ions have unit electric charge magnitude. X remains a spectator ion, and only AB can form a solid. Assume constant volume, no additional chemical reactions, and negligible contributions from self-ionization of water. Dissolving AB raises both reacting-ion concentrations by the same $s$:Charge neutrality is satisfied: $c_{\mathrm{A}}+c_{\mathrm{X}}-c_{\mathrm{B}}=s+c_0-(s+c_0)=0$. Adding B alone without its accompanying cation would omit the initial electric charge balance.

In the plane of the two reacting-ion concentrations, this inventory constraint is the line $c_{\mathrm{B}}=c_{\mathrm{A}}+c_0$. Saturation supplies the hyperbola $c_{\mathrm{B}}=K_{\mathrm{sp},c}/c_{\mathrm{A}}$ for $c_{\mathrm{A}}>0$. Their positive intersection solves a quadratic equation:The quadratic formula supplies two roots of a polynomial; only the positive one is a physical concentration. Uniqueness also follows because $s(s+c_0)$ strictly increases for $s\ge 0$, starting at zero and growing without bound. The second expression, obtained by multiplying by the conjugate, avoids subtraction of nearly equal numbers at large $c_0$.

For pure water, $c_0=0$ gives $s_{\ast }=\sqrt{K_{\mathrm{sp},c}}=0.010\mathrm{M}$. For $c_0=0.030\mathrm{M}$,Their product is $10^{-4}{\mathrm{M}}^2$, although neither concentration equals the pure-water value. This is the useful geometrical distinction: an inventory line and an equilibrium constant specify different constraints.

Check the solid inventory before imposing saturation. If the initially added AB has amount of substance $n_0$ in volume $V$, the saturated branch requires $n_0>V{s}_{\ast }$ for excess solid to remain. For example, choose $n_0=0.0020\mathrm{mol}$ and $V=0.100\mathrm{L}$. In pure water, $0.0010\mathrm{mol}$ dissolves and the same amount of substance remains as solid. At $c_0=0.030\mathrm{M}$, approximately $0.0003027756\mathrm{mol}$ dissolves, leaving $0.0016972244\mathrm{mol}$ as solid. In this model, an insufficient supply instead givesIf $n_0/V<s_{\ast }$, all AB dissolves and the ion product is below the solubility product. At $n_0/V=s_{\ast }$, the last solid disappears exactly at saturation.