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Let $f$ be bounded and holomorphic on the right half-plane, continuous on its closure, and satisfy $|f(iy)|\le M$ for every real $y$. Then $|f(z)|\le M$ throughout the half-plane. Rotations and translations give the analogous statement for any half-plane.

For $\epsilon >0$, set $g_{\epsilon }(z)=f(z)/(1+\epsilon z)$. Its denominator has no zero in the closed right half-plane and has modulus at least one on the imaginary boundary. If $|f|\le K$, then on the radius-$R$ semicircle, $|g_{\epsilon }|\le K/(\epsilon R-1)$. The maximum modulus principle on a bounded domain therefore gives $|g_{\epsilon }|\le M$ after letting $R\to \infty$; equivalently first use $M+\eta$ and then $\eta \downarrow 0$ when $M=0$. Finally let $\epsilon \downarrow 0$.

Boundedness in the half-plane cannot be replaced by a boundary bound alone: $e^z$ has modulus one on the imaginary axis and is unbounded to its right. Growth-controlled extensions require a Phragmén–Lindelöf principle.

The Cayley transform gives a conformal bijection between an unbounded half-plane and the unit disc. Two conventions are useful. For the right half-plane, useFor the upper half-plane, useRotating the upper half-plane onto the right half-plane by $z\mapsto -iz$ explains the relationship $C(z)=R(-iz)$. The factors of $i$ distinguish which half-plane is mapped onto the unit disc. ^[ref]

In particular, $C(i)=0$, $C(2i)=1/3$ and $C(1+i)=(1-2i)/5$. The whole positive imaginary axis becomes the real diameter: $C(iy)=(y-1)/(y+1)$ for $y>0$. Horizontal grid lines become circles tangent to the unit circle at $1$ when completed by their endpoint at infinity; they are not hyperbolic geodesics.

On the Riemann sphere, $C$ maps the extended real line onto the unit circle:Every finite real $x$ has $|C(x)|=1$; the remaining point $w=1$ corresponds to infinity. The pole $z=-i$ is outside the upper half-plane, and the pole of $C^{-1}$ at $w=1$ is on the boundary of the unit disc. Neither pole obstructs the interior conformal bijection.

For a short verification, subtract the squared moduli:These identities put each interior image in the required domain; solving $w(z+i)=z-i$ gives the stated inverse function. Both maps are holomorphic, with $C^{\prime }(z)=2i/(z+i{)}^2\ne 0$, so the conformal bijection preserves Euclidean angles between tangent directions.

Euclidean distance is not preserved: $i$ and $2i$ are one unit apart, whereas $0$ and $1/3$ are only one third of a unit apart. The striking feature is that $C$ does preserve length for the hyperbolic metrics of the Poincaré half-plane model and Poincaré disk model:Consequently it is an isometry for these hyperbolic metrics, including their induced hyperbolic distances. For the same example, the corresponding hyperbolic geodesic segments have equal length:Thus the Cayley transform lets us choose the most convenient picture: a straight boundary for calculations, or a bounded unit disc for visualization. It also transports automorphisms of the unit disk into upper half-plane automorphisms by function composition $C^{-1}\circ g\circ C$.

For a real symmetric matrix, or more generally a Hermitian matrix, the finite-dimensional spectral theorem makes all eigenvalues real. Intersecting its Gershgorin discs with the real axis givesThe union preserves gaps that the single outer interval loses. If every $a_{ii}-r_i>0$, all eigenvalues are positive, so the matrix is a positive-definite matrix. This is the positive-diagonal strictly diagonally dominant matrix criterion. For the three-by-three example, $1.7\le \lambda \le 9.4$ gives a positive-definite matrix certificate and bounds its spectral condition number of a positive-definite matrix by $9.4/1.7\approx 5.529412$; the computed value is about $4.544364$.

Strictness matters. The symmetric matrix $\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$ has positive diagonal and is a diagonally dominant matrix, but its eigenvalues are $0$ and $2$: it is singular. For a general complex matrix, the condition $\mathrm{Re}{a}_{ii}>r_i$ only puts the eigenvalues in the open right half-plane. The Hermitian matrix assumption is what converts that statement into positive definiteness.

The two bounding circles meet at zero and one, which the Möbius transformation sends to zero and infinity. They therefore become straight lines through zero. More explicitly write $w=f(z)$, so $z=w/(1+w)$. The two disk inequalities becomeSince $\zeta =1/2+i\sqrt{3}/2$, these say $\mathrm{Re}w>\sqrt{3}|\mathrm{Im}w|$. ThusThe positive real axis belongs to the image because $f(1/2)=1$, fixing which sector is intended. Cubing maps this sector bijectively onto the right half-plane: its argument interval of width $\pi /3$ becomes $(-\pi /2,\pi /2)$. The Cayley transform between the half-plane and disk $v\mapsto (v-1)/(v+1)$ then maps that half-plane bijectively onto the unit disk. Hence a holomorphic bijection isEvery denominator is nonzero on the corresponding domain, and the sector width ensures the power map is injective. This is the conformal map of a symmetric circular lens.

Put $c=2\alpha /\pi$. In the right half-plane,The last two cancel, so $\boxed{{\Phi }_{uu}+{\Phi }_{vv}=0}$. The possible singular point $(0,0)$ lies outside the open domain $u>0$.

For $\alpha \ne 0$, a level curve of value ${\Phi }_0$ has $v/u=\tan (\pi {\Phi }_0/(2\alpha ))$. These are rays from the origin, restricted to the right half-plane. For positive $\alpha$, the values increase with angle from $-\alpha$ on the lower boundary to $+\alpha$ on the upper boundary. If $\alpha =0$, the function is identically zero. The left panel of the figure in (c) shows these ray-shaped level curves.

There is no such map for the domain printed in the paper. The planar domain $A=\mathbb{C}\setminus [a,b]$ is not simply connected. For example, a sufficiently large positively oriented circle $\gamma$ surrounds $[a,b]$ and lies in $A$, whileThe integrand is holomorphic on $A$. If $A$ were simply connected, the Cauchy integral theorem would force this integral to be zero. A conformal bijection is a homeomorphism and preserves simple connectivity, whereas $H$ is simply connected. Thus a conformal bijection $A\to H$ is impossible.

The explicit construction also pinpoints the missing point. DefineThis Möbius transformation sends the removed line segment to the nonpositive real axis, together with infinity. More precisely, its image on the stated planar domain issince $T(z)=1$ has no finite solution when $a\ne b$. Conversely, its inverse $z=(bw-a)/(w-1)$ is defined for every $w$ in this set and cannot lie on $[a,b]$, proving this image description exactly.

Take the principal square root of $T(z)$, characterized by positive real part, and then multiply by $i$. The result is the conformal mapEach stage is injective and onto the indicated image; the square root maps the slit plane to the right half-plane, with the point $1$ omitted here. Hence this is a conformal bijection onto the punctured upper half-plane. Its inverse isIf the source is instead $(\mathbb{C}\cup \{\infty \})\setminus [a,b]$ on the Riemann sphere, the same map extends with $F(\infty )=i$ and becomes a conformal bijection onto all of $H$. Indeed, in the local coordinate $u=1/z$ it has expansion $F=i+\frac{i}{2}(b-a)u+O(u^2)$, whose derivative at $u=0$ is nonzero. This gives the sphere version of the requested construction without silently changing the printed planar domain.

Use the branch of $\sqrt{1-Z^2}$ that is positive for $Z=iY$, $Y>0$. The right-half-plane slit mapis holomorphic in the upper half-plane, with derivative $Z(1-Z^2{)}^{-3/2}$, the required Schwarz-Christoffel formula form. Its derivative has no interior zero.

To verify the image and global injectivity, factor the construction. Squaring maps the upper half-plane bijectively onto $\mathbb{C}\setminus [0,\infty )$. Thus $1-Z^2$ ranges over $\mathbb{C}\setminus (-\infty ,1]$. The positive-real-part square root maps this onto the right half-plane with the interval $(0,1]$ removed; reciprocation maps it onto the right half-plane slit along $[1,\infty )$. This also identifies the slit tip as a boundary point; the strict inequality in the printed slit description must be understood with that boundary convention.

Approaching the real axis from above givesIn particular $Z=-1$ maps to the lower infinite prime end $A$, $Z=0$ to the tip $B=1$, and $Z=1$ to the upper infinite prime end $C$. The real-axis point at infinity maps to zero on the vertical wall.

The right half-plane consists of complex numbers with positive real part. Its boundary is the imaginary axis. For example, $1+i$ is inside, $i$ is on the boundary, and $-1+i$ is outside. Rotation $z\mapsto iz$ takes the right half-plane onto the upper half-plane; the Cayley transform between the half-plane and disk uses $R(z)=(z-1)/(z+1)$ for this convention.