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Yes: for $k=2$ the stated inequality guarantees zero-stability. Part (ii) puts every polynomial root in the closed unit disc. The remaining root condition for a multistep method is that every root on the unit circle has multiplicity of a root one.

If this failed for a quadratic polynomial, it would have a repeated root $\xi$ with $|\xi |=1$, andConsequently $|{\alpha }_1|=2|{\alpha }_2|$ and $|{\alpha }_0|=|{\alpha }_2|$, contradictingbecause ${\alpha }_2\ne 0$. Thus all roots on the unit circle are simple andThis conclusion does not assert consistency of a numerical method: part (i) rules out consistency of a numerical method whenever the coefficient inequality is strict. Zero-stability and consistency of a numerical method are separate requirements.

The potential plane is the strip $0<\psi <1$. The wall maps to its lower boundary, with upstream $\varphi \to -\infty$, the corner at $w=0$, and downstream $\varphi \to +\infty$. The free surface maps to its upper boundary.

Write $q=w^{\prime }=u-iv$. Far upstream $q=1$; far downstream the flow points vertically down and $q=i$. On the two wall rays, $q$ runs along the positive real and positive imaginary axes. The reentrant fluid corner has angle $3\pi /2$, so its potential behaves locally as $(z-B{)}^{2/3}$ and its speed diverges like $|z-B{|}^{-1/3}$. Thus the corner corresponds to $q=\infty$, not $q=0$. The free surface has $|q|=1$, giving the hodograph domainIt is the exterior of a quarter unit disc in the first quadrant.

Now $s=q^2$ maps this domain to the part of the upper half-plane outside the unit disc. The maptakes it to the first quadrant: its real and imaginary parts are positive because $|s|>1$ and $\Im s>0$. Finally $\eta =t^2$ maps that quadrant to the upper half-plane. The corresponding strip map is $\eta =e^{\pi w}$. Both give upstream $\eta =0$, corner $\eta =1$, and downstream $\eta =\infty$, so the three boundary normalizations fix the remaining real Möbius freedom. ConsequentlyOn the free surface $q=e^{i\theta }$, $t=i\tan \theta$, so $\eta$ is negative and $\Im w=1$ as required. On the wall $\eta >0$, giving $\Im w=0$.

The printed upper-tail identity cannot be a probability formula: it vanishes at zero and increases in $x$. Its correction isEquivalently the cumulative distribution function is $1/2+{\pi }^{-1}\arctan x$. Differentiating that cumulative function gives the densityThe normalization and sign can also be checked directly by conformal invariance of planar Brownian motion: the Cayley transform $(z-i)/(z+i)$ sends the starting point to the center of the unit disc. Its exit angle is uniform by rotational symmetry. The inverse on the unit circle gives $X=-\cot (\Theta /2)$, $\Theta$ uniform on $(0,2\pi )$, whose change-of-variable density is exactly the one above.

For a start $z_0=x_0+i{y}_0$, $y_0>0$, translation and positive scaling send it to $i$. Thus the exit coordinate has law $x_0+y_{0}C$, where $C$ is standard Cauchy. The corresponding formulas areThis is the upper-half-plane Poisson kernel. The printed density is correct; it is the preceding asserted tail from which it is meant to follow that needs correction.

Under the product order, every point of the open unit disc has both coordinates below one, so $(1,1)$ is an upper bound. Any upper bound $(u,v)$ must have $u\ge 1$ and $v\ge 1$, since the two coordinate projections of the open unit disc are $(-1,1)$. Therefore $(1,1){\le }_P(u,v)$, proving ${\sup }_{{\le }_P}S=(1,1)$.

Under the lexicographic order, every point $(1,v)$, with arbitrary real $v$, is an upper bound, because every point of $S$ has first coordinate strictly below one. Conversely, an upper bound cannot have first coordinate $u<1$: a point $(x,0)\in S$ with $x>u$ would exceed it. If an upper bound has $u>1$, the upper bound $(1,0)$ is strictly smaller. If it is $(1,v)$, the upper bound $(1,v-1)$ is strictly smaller. Hence there is no least upper bound under the lexicographic order. This illustrates the supremum of an open disc under product and lexicographic orders.

A connected Riemann surface whose universal cover is the unit disc inherits the complete curvature-minus-one metric displayed on its cover. Möbius transformations preserving the disc are isometries, so deck transformations preserve the metric and it descends to the surface. This uniformization notion must be distinguished from the potential-theoretic existence of a Green function. Holomorphic maps between such surfaces contract their metric by lifting to Schwarz-Pick theorem on the covers.

On the unit disc, the pseudohyperbolic distance isIt is invariant under disc Möbius transformations. For the curvature-minus-one hyperbolic metric with length element $2|dz|/(1-|z{|}^2)$, $\delta (z,w)=\tanh (\rho (z,w)/2)$ and $e^{-\rho (z,w)}=(1-\delta )/(1+\delta )$. The Schwarz-Pick theorem says that holomorphic functions from the disc into itself contract this distance. A Blaschke factor has modulus $\delta (z,a)$, so the distance also controls the size of a product with prescribed zeros.

A meromorphic function on the unit disc has bounded characteristic if and only if it has this representation. Its pole Blaschke product $B$ makes $g=Bf$ analytic with bounded positive logarithmic means. The least harmonic majorant by expanding disk lifts gives $H\ge {\log }^{+}|g|$. A harmonic conjugate gives an analytic $A$ with $\mathrm{Re}A=H$, so $u=g{e}^{-A}$ and $v=B{e}^{-A}$ are bounded. Conversely the Nevanlinna first main theorem bounds the characteristic of $u/v$ by that of $u$ and $1/v$, hence by a constant.

Every nonempty simply connected proper open subset of the complex plane is conformally equivalent to the unit disc.

A holomorphic self-map of the unit disc contracts its pseudohyperbolic distance. Conjugating by Möbius transformations that send one chosen source point and its image to zero reduces the statement to the Schwarz lemma. In the complex upper half-plane this givesIf $f(iY)=iY+o(Y)$ at infinity, multiply the equivalent inequality between one minus the squared distances by $Y$ and let $Y\to \infty$. It yields $\mathrm{Im}f(z)\ge \mathrm{Im}z$. Applied to the inverse mapping-out function of a compact H-hull, this proves the nonnegativity of its half-plane capacity.

For a positive harmonic function on the unit disc, choose a global harmonic conjugate and apply the Carathéodory derivative bound for the right half-plane to $u+iv$. This gives $|\nabla u(0)|\le 2u(0)$ by the Cauchy-Riemann equations. A disk automorphism sending zero to $z$ has derivative modulus $1-|z{|}^2$, so the chain rule gives the displayed bound. Composing a boundary-pole Poisson kernel with a disk automorphism attains equality, proving the coefficient is optimal at every point.

A fixed-aperture approach region at a boundary point of the unit disc. Its points approach the boundary point without becoming tangent to the circle. Convergence inside every such fixed region is a nontangential limit; the corresponding supremum defines a non-tangential maximal function.

Let $R$ be a compact Riemann surface of genus $g$, and remove $n$ distinct points. The punctured surface is uniformized by the unit disc exactly when $2g-2+n>0$. It is uniformized by the complex plane exactly when $2g-2+n$ is $-1$ or $0$, corresponding respectively to $\mathbb{C}$, and to ${\mathbb{C}}^{\ast }$ or a complex torus. The sole spherical case is $g=n=0$.

The upper half-plane is the open setIt contains precisely the complex numbers with positive imaginary part: $1+i$ is inside, $1$ lies on its boundary, and $1-i$ belongs to the lower half-plane. In the complex plane its boundary is the real line; in the Riemann sphere the boundary also includes infinity. It is a convex set, hence a simply connected domain, and the Cayley transform between the half-plane and disk makes it biholomorphic to the unit disc.

With the hyperbolic metric $ds=|dz|/\mathrm{Im}z$, the same domain becomes the Poincaré half-plane model. Its hyperbolic geodesics are vertical lines and upper semicircles whose centres lie on the real line. Under the Cayley transform, these become arcs of circles orthogonal to the unit circle, with diameters as the straight-line cases. ^[ref]

Matching colours identify each hyperbolic geodesic and its image. The vertical line through zero maps to a diameter; the vertical line through one maps to an arc ending at $-i$ and $1$. A semicircle from $-1$ to $1$ also maps to a diameter, whereas the semicircle from $0$ to $2$ maps to a curved arc. Their endpoints lie on the boundary, outside the domains: the endpoints are infinitely far from interior points in hyperbolic distance.

The upper half-plane is useful because its straight boundary and explicit upper half-plane automorphisms make conformal maps and hyperbolic geometry calculable. The picture explains why shortest hyperbolic geodesic routes can look curved in Euclidean coordinates; their lengths are measured with the position-dependent hyperbolic metric.

For historical context, in his letter to Klein of 22 September 1882, Poincaré constructs a quadrilateral bounded by four circles orthogonal to a fundamental circle, and studies transformations generating repeated copies. In modern language, the sides inside that circle are hyperbolic geodesics in a disk model. The letter directly exhibits the connection between transformation groups, geometry and complex functions. ^[ref]

For a holomorphic function on the unit disc, this boundary condition characterizes Blaschke products, including unimodular constants. The nondecreasing radial means of the nonnegative subharmonic function ${\log }^{+}|f|$ must vanish, so $|f|\le 1$. In the Blaschke factorization of a bounded holomorphic function $f=Bg$, both logarithmic means tend to zero. The mean value property for harmonic functions applied to $\log |g|$ gives $|g(0)|=1$, and the maximum modulus principle makes $g$ unimodular constant. The converse follows from the radial logarithmic mean of a Blaschke product. Omitting the outer absolute value changes the condition: $e^z$ has zero mean logarithmic modulus, but is not a Blaschke product.

If a degree-$N$ polynomial $f(\zeta )={\sum }_{j=1}^N{a}_j{\zeta }^j$ maps the unit disc conformally onto $D$, Green's theorem expresses $M_n={\iint }_D{z}^{n}dxdy$ as the residue of $f^n{f}^{\prime }{\sum }_j{\bar{a}}_j{\zeta }^{-j}$ times $\pi$. Its lowest possible power is $n-N$, so no residue occurs when $n\ge N$. For a quadratic map, $M_0=\pi (|a_1{|}^2+2|a_2{|}^2)$ and $M_1=\pi a_1^2{\bar{a}}_2$. The first moment can be complex; the zeroth is area.

If a holomorphic function $f$ on the unit disc satisfies $|f|\le M$ and $f(a)=0$, the Schwarz-Pick lemma applied to $f/M$ gives $|f(0)|/M\le |a|$. Equality for a nonzero central value forces a scaled automorphism of the unit disk. If $f$ extends past the unit circle and $|f(0)|$ is smaller than its minimum boundary modulus, the maximum modulus principle applied to $1/f$ proves that a zero occurs inside the circle.

A nonzero holomorphic function on the unit disc can have exactly a prescribed sequence of zeros, counted with multiplicity, if and only if that sequence is locally finite in the disc. Necessity is the identity theorem for holomorphic functions. For sufficiency, remove a finite number $m$ of zeros at zero and enumerate the others as $a_n$, so $|a_n|\to 1$. Use the Weierstrass elementary factors $E_n(t)=(1-t)\exp ({\sum }_{j=1}^n{t}^j/j)$. On every compact subdisc, eventually $|z/a_n|\le q<1$ and $|\log E_n(z/a_n)|\le q^{n+1}/((n+1)(1-q))$. Hence $z^m{\prod }_n{E}_n(z/a_n)$ converges locally uniformly and has exactly the prescribed zeros. Finite sequences simply give finite products.