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The principal square root maps the upper $Z$ half-plane bijectively onto $\Re z>0$ with the ray $[1,\infty )$ removed. The prevertex zero is the slit tip, and $\pm 1$ represent the two ends at infinity. The inverse is the unique root $Z^2=1-z^{-2}$ with positive imaginary part. Pulling back ${\pi }^{-1}\arg ((Z-1)/(Z+1))$ gives the bounded harmonic function with value one on the slit banks and zero on the imaginary axis. The harmonic function $\Im [Z/(1-Z^2)]$ has zero finite boundary values but grows at infinity, explaining the role of boundedness.

For a Dirichlet character $\chi$ modulo $q$, its Dirichlet L-function is initially defined on the half-plane $\mathrm{Re}s>1$ bySince $|\chi (n)|\le 1$, this series converges absolutely by comparison with ${\sum }_n{n}^{-\mathrm{Re}s}$.

A Dirichlet character is a completely multiplicative function. The fundamental theorem of arithmetic therefore reorganizes the absolutely convergent series into an Euler product:Expanding each factor as a geometric series assigns exactly one term to every positive integer. For the principal Dirichlet character ${\chi }_0$, the factors at primes dividing $q$ disappear, giving $L(s,{\chi }_0)=\zeta (s){\prod }_{p\mid q}(1-p^{-s})$, where $\zeta$ is the Riemann zeta function.

A semi-infinite horizontal plate at unit height and a floor at zero give a velocity potential strip of height $a$ and an upper unit semicircle in the complex velocity $q$. Their matching half-plane maps give $(1+e^{\pi w/a}{)}^{-1}=((q-1)/(q+1){)}^2$. Along the fluid free boundary put $q=e^{-i\theta }$, $-\pi <\theta <0$. Then $dz/d\theta =-(a/\pi )e^{i\theta }\cot (\theta /2)$ and $y=1-(a/\pi )(\theta +\pi +\sin \theta )$. Its downstream limit is $1-a$, which must equal the downstream depth $a$. Thus the contraction is exactly one half.

In the upper half-plane, a line source of outward volume flux $Q$ at height $a$ above an impermeable wall has an equal image source below it. On the wall $u=Qx/[\pi (x^2+a^2)]$ and the normal velocity vanishes. The steady Bernoulli equation gives $p-p_{\infty }=-\rho Q^2{x}^2/[2{\pi }^2(x^2+a^2{)}^2]$. Sincethe ambient-relative force per unit length pulls the wall toward the source with the displayed magnitude. The torque about the source's projection onto the wall is zero in the symmetric Cauchy principal value sense. Individual one-sided moments diverge logarithmically. This planar result differs from the three-dimensional hydrodynamic attraction of a plane wall to a point source.

A useful way to map a circular lens is to send its two corner points to zero and infinity with a Möbius transformation. The two bounding circles then become straight rays, and a suitable power opens their sector into a half-plane.

For the lens $|z+1|<\sqrt{2}$, $|z-1|<\sqrt{2}$, the corners are $\pm i$. Put $w=(z-i)/(z+i)$, with inverse $z=i(1+w)/(1-w)$. For $w=u+iv$, substitution givesThus the image is precisely $u<-|v|$, or $3\pi /4<\arg w<5\pi /4$. Rotation through $-3\pi /4$ followed by squaring gives the displayed conformal map onto the upper half-plane. The open sector contains no opposite pair, so squaring is injective; it has nonzero derivative away from its excluded vertex. These facts also prove bijectivity and conformality, rather than merely identifying boundary images.

With two distinct centroids, K-means clustering assigns points on opposite sides of their nearest-centroid decision hyperplane, which in two dimensions is a straight line. This constraint explains the typical outcomes in the three source plots; exact assignments depend on initialization and the realized points.

Dataset A: one cluster in cluster analysis contains most of the horizontal cloud and the bottom of the vertical cloud; the other contains the upper part of the vertical cloud. The two visually elongated groups are poorly matched to squared Euclidean distance from their centroids. Points at the lower end of the vertical group can be closer to the horizontal group's centroid, so the two original shapes need not be recovered intact.

Dataset B: a roughly straight cut divides the ring and generally the central cloud between the two clusters in cluster analysis. A Voronoi cell for either centroid is a half-plane; it cannot select a surrounding ring while excluding its center. The cut's direction can vary because the geometry is approximately rotationally symmetric.

Dataset C: the two dense lower clouds typically receive different centroids. The much smaller upper cloud is absorbed into one or divided between both clusters in cluster analysis, according to the nearly vertical separating line. With $K=2$, the objective usually benefits more from separating the two large clouds than from spending one centroid on the sparse third group. These are typical qualitative outcomes, rather than guarantees about every initialization.

The three contacts in the strip. Work with the external-contact arrangement drawn in the original diagram: both disks lie between the two parallel lines and touch opposite boundaries. A rigid motion puts those boundaries at $x=0$ and $x=d$, with $d>0$, and makes $P=(0,0)$. Write $R=(d,h)$, and let the positive radii be $r_1,r_2$. Because a tangent to a circle is perpendicular to the corresponding radius, the centres areAt the external tangency, the two radial segments form the segment $O_1{O}_2$, with $|O_{1}Q|=r_1$ and $|Q{O}_2|=r_2$. Hence the internal division formula givesThusThe coefficient is strictly between zero and one, so $Q$ lies on the actual line segment $PR$. This proves collinearity for tangent circles between parallel lines. We will also use the fact that $PQ$ and $QR$ are chords of the respective circles; the entire line segment $PR$ therefore lies in the union of their closed disks.

The four contacts. Label the successive contact points $T_{12},T_{23},T_{34},T_{41}$, with $T_{jk}$ belonging to $C_j$ and $C_k$. The original diagram has four positive-radius circles with disjoint interiors, successive external tangencies, and distinct contact points. Set $T=T_{41}$ and apply the reciprocal Möbius transformationBy part (b), the two circles $C_1,C_4$ through $T$ become straight lines ${\ell }_1,{\ell }_4$. These straight lines cannot meet at a finite point, since their inverse images would give a second intersection of $C_1,C_4$. They are therefore distinct parallel lines. The other two circles avoid $T$, so their images $D_2,D_3$ are ordinary circles missing zero.

The three remaining contacts retain their tangencies. Indeed, away from $T$ the map and its inverse are holomorphic functions with nonzero derivatives, hence are local conformal maps. In particular they preserve a shared tangent line. Equivalently, bijectivity preserves the unique intersection of two generalized circles, which characterizes their tangency here.

The finite closed disks bounded by $D_2,D_3$ lie between ${\ell }_1,{\ell }_4$, have external tangency to each other, and touch opposite boundaries at the images of $T_{12}$ and $T_{34}$. To track the sides as well as the boundaries, a closed disk avoiding $T$ maps to the bounded closed disk of its image circle, while each disk with $T$ on its boundary maps to a closed half-plane. The two latter half-planes have disjoint interiors, so their complement is the strip containing zero. Disjointness of the original interiors puts $D_2,D_3$ inside that strip. Neither image closed disk contains zero, since the reciprocal Möbius transformation sends only infinity to zero.

The first part now makes the three finite transformed contact points collinear, on a straight line $\ell$. It is essential that $\ell$ misses zero. If it passed through zero, its segment between ${\ell }_1$ and ${\ell }_4$ would contain zero, since zero lies strictly within the strip. But the first part shows that this whole line segment is covered by the two image closed disks, neither of which contains zero. This contradiction proves $0\notin \ell$.

The inverse image of a straight line missing zero is a genuine circle through $T$, by the same coefficient transformation in part (b), or by applying the involution $u\mapsto 1/u$ in reverse and translating back. This circle contains $T_{12},T_{23},T_{34}$, while $T$ is the inverse image of the point at infinity on $\ell$. ThereforeThis is the four contact points of a circle chain result; the four contacts form a cyclic quadrilateral.

The pictured arrangement matters. Arbitrary mixtures of internal tangency and external tangency can instead give four distinct collinear contacts: take four circles whose diameter endpoints on the real axis are respectively $[0,3]$, $[0,1]$, $[1,2]$, and $[2,3]$. Their successive contacts are $0,1,2,3$, so no ordinary circle contains all four. The proof above establishes the requested genuine-circle conclusion for the external-contact arrangement actually drawn, rather than silently treating every generalized circle as an ordinary circle.

Choose each factor with its argument between zero and $\pi$ in the upper half-plane. On an interval between consecutive real prevertices, each factor has a constant boundary argument, so $f^{\prime }$ has constant argument and the image is a straight side. Crossing $X_j$ from left to right decreases $\arg (Z-X_j)$ by $\pi$, hence increases $\arg f^{\prime }$ by ${\beta }_j\pi$, giving the prescribed exterior turn. Locally,so an upper half-disc maps to angle $(1-{\beta }_j)\pi$, the required interior angle. The overall constant rotates and scales the polygon; integration adds translation. For a general prescribed polygon, the prevertex positions must also satisfy the side-length conditions: the derivative formula fixes angles but arbitrary choices do not fix arbitrary side lengths.

For the slit domain, the slit tip has interior angle $2\pi$, so its exterior parameter is $-1$. The two vertices at infinity give the factors at $\pm 1$, leading to $f^{\prime }(Z)$ proportional to $Z(1-Z^2{)}^{-3/2}$. Integrating and imposing the tip value givesThe square root is the principal one: for $\Im Z>0$, $1-Z^2$ never lies on the nonpositive real axis. Thus $\Re f>0$. Boundary points $-1<X<1$ map to the two banks of the slit, with $X=0$ at the tip; the outside intervals map to the imaginary axis. The derivative is nonzero in the open half-plane. Conversely each point of the slit domain determines exactly one root $Z$ of $Z^2=1-z^{-2}$ with positive imaginary part, proving bijectivity.

The Möbius map $(Z-1)/(Z+1)$ takes the upper half-plane to itself, so its principal argument lies in $(0,\pi )$. The imaginary part of its analytic logarithm is harmonic, equals $\pi$ on $(-1,1)$ and zero on the other real intervals. ThusIt lies between zero and one and has the required boundary limits away from the two discontinuities. Pulling this harmonic function back by the inverse conformal map yieldsThis specifies all branches and remains bounded even as $z$ tends to either end at infinity.

To exhibit nonuniqueness without the growth restriction, putThis is harmonic in the domain. On its boundary the corresponding $Z$ is real and the rational expression is real, except at $\pm 1$, which represent infinity. At $Z=\infty$, representing $z=0$, the expression tends to zero. Thus $H$ has zero finite boundary data. It is nonzero, since for $Z=iY$ its value is $Y/(1+Y^2)>0$, and it is unbounded as $Z$ approaches either pole $\pm 1$ non-tangentially. Therefore $u+cH$ gives distinct solutions with the same finite boundary data when boundedness at infinity is removed.

Letbe the root with positive real and positive imaginary parts. More explicitly, with $r_j=\sqrt{{\kappa }_j^4+{ϵ}^2}$,Decay at each infinity and continuity at zero force $y=A{e}^{-i{q}_{0}x}$ for $x<0$ and $y=A{e}^{i{q}_{1}x}$ for $x>0$. The derivative jump is $i(q_1+q_0)A=1$, soThe two roots lie in the first quadrant, so their sum cannot vanish.

Integrating the differential equation twice by parts on each half-line givesTheir sum is the stated identity with right side one. Equivalently the global distributional equation is $y^{\prime \prime }+{\kappa }^{2}y={\delta }_0$, since the derivative has jump one.

The direct half-transforms areHere ${\hat{y}}_{+}$ is analytic above its lower-half-plane pole $-q_1$, while ${\hat{y}}_{-}$ is analytic below its upper-half-plane pole $q_0$.

To recover the same result from the transform identity alone, use Wiener-Hopf factorization:Multiply by $(q_1+k)/(q_0+k)$. The coefficient of ${\hat{y}}_{-}$ becomes $(q_0-k)/(q_1-k)$, analytic and zero-free in the lower half-plane, while the coefficient of ${\hat{y}}_{+}$ is analytic and zero-free in the upper half-plane. Split the right side asThe first term is analytic above and the second below. Equating the separated functions and using their decay at infinity fixes the entire remainder to zero, yielding the boxed ${\hat{y}}_{+}$, with the branches fixed before factorization.

For the velocity potential strip, $e^{\pi w/a}$ maps to the upper half-plane. The real Möbius transformationpreserves that half-plane and sends its marked endpoints to $-1,\infty ,0$ in the order $A/E,D,B/C$.

For the hodograph plane, first put $r=(q-1)/(q+1)$. Its image is the second quadrant, bounded by the negative real ray and positive imaginary ray. Squaring takes that quadrant to the lower half-plane, and negating restores the upper half-plane. Thushas exactly the same three marked values. Uniqueness of a conformal map to a half-plane with three assigned boundary values makes these maps equal. Cancel the minus signs to obtainThe sign in the intermediate map is important for identifying the upper, rather than lower, half-plane.

For $\mathrm{Im}k>0$, put $s=-ik$, so $\mathrm{Re}s>0$. The Laplace transform for the Gamma function, analytically continued from positive real $s$, givesChoose $\arg k\in (0,\pi )$ in the upper half-plane. Since $\arg (-ik)=\arg k-\pi /2$,This integral is absolutely convergent throughout the upper half-plane. On the nonzero real axis it converges conditionally by Dirichlet test for improper integrals, and its value is the upper boundary limit of the same branch. In particular negative real $k$ has argument $\pi$, not $-\pi$ in this formula.

At $k=0$ the integral diverges at infinity; in the lower half-plane its exponential function factor grows and the ordinary improper integral diverges. The origin itself causes no local-integrability problem for $x^{-1/4}$. A useful continuation of the chosen power uses a branch cut descending from $k=0$ with argument in $(-\pi /2,3\pi /2)$.

The poles are the four roots of $z^4=-1$. The upper half-plane contains $a=e^{i\pi /4}$ and $b=e^{3i\pi /4}$, both enclosed when $R>1$. At a simple pole $q$,The sum of the two residues isThe residue theorem, with the real segment traversed from $-R$ to $R$ and the upper arc from $R$ to $-R$, yields

Along a level curve, $V_{x}dx+V_{y}dy=0$, with $V_x=V_0\theta /\pi$ and $V_y=V_0(1+\log r)/\pi$. ThereforeIn the intended upper half-plane $0<\theta <\pi$, the slope is positive for $0<r<e^{-1}$, vertical at $r=e^{-1}$, and negative for $r>e^{-1}$. At the positive real boundary $\theta =0$ the slope is zero except at the stationary point $(e^{-1},0)$, where both derivatives vanish and the quotient is indeterminate. More generally, outside the chosen half-plane its sign is the sign of $-\theta (1+\log r)$, with the argument branch held fixed. A convenient parametric description of the curves to sketch is $V\pi /V_0=r(\sin \theta \log r+\theta \cos \theta )$.