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A branch cut is a removed curve or collection of curves used to choose a single-valued branch of a multivalued function of a multivalued complex expression. On $\mathbb{C}\setminus (-\infty ,0]$, choosing $-\pi <\arg z<\pi$ gives the principal logarithm $\mathrm{Log}z=\log |z|+i\arg z$ and hence $\sqrt{z}=\exp (\mathrm{Log}z/2)$.

The limiting values on opposite sides of the negative real axis differ: for $r>0$, the two square-root values approach $i\sqrt{r}$ and $-i\sqrt{r}$. The original algebraic relation $w^2=z$ has no discontinuity; the discontinuity belongs to the selected branch of a multivalued function. Tracking that distinction prevents sign errors in contour integration and motivates a Riemann surface.

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.

In the impermeable strip $x>0$, $0<y<a$, a line source of strength $Q$ at an interior point $c$ has the displayed complex potential of a two-dimensional potential flow, with locally consistent complex logarithm and $f(z)=\cosh (\pi z/a)$. The hyperbolic-cosine map of a half-strip sends the walls to the real axis. An equal image line source at $\overline{f(c)}$ makes that axis a streamline. Since $f^{\prime }(c)\ne 0$, the local singular part remains $Q\log (z-c)/(2\pi )$, proving preservation of strength. With no extra imposed flow, the complex velocity tends to $Q/a$ as $x\to \infty$, so the outgoing volume flux is $Q$. The source position $c$ has positive imaginary part; assuming it real would place it on a wall.

Set $f(z)=\cosh (\pi z/a)$. The hyperbolic cosine addition formula givesFor $x>0$ and $0<y<a$, both factors in the imaginary part are positive. Thus the image lies in the upper half-plane. The three walls map onto the entire real axis:Also $f^{\prime }(z)=(\pi /a)\sinh (\pi z/a)\ne 0$ in the interior.

To verify that this is a one-to-one conformal map onto the whole upper half-plane, factor it through $s=e^{\pi z/a}$. This first map is one-to-one onto $|s|>1$, $\mathrm{Im}s>0$, and $f(z)=(s+s^{-1})/2$. Given any $\mathrm{Im}\zeta >0$, the roots of $s^2-2\zeta s+1=0$ have product one. Neither lies on $|s|=1$, since that would make $\zeta$ real. Exactly one root therefore has $|s|>1$, andforces that root into the upper half of the exterior disk. It has a unique preimage $z=(a/\pi )(\log |s|+i\arg s)$ with $0<\arg s<\pi$. This proves both surjectivity and injectivity. The construction is the hyperbolic-cosine map of a half-strip.

The source location $c$ is complex, with $\mathrm{Re}c>0$ and $0<\mathrm{Im}c<a$. Put $C=f(c)$, so $\mathrm{Im}C>0$. The method of images places an equal line source at $\bar{C}$ below the real axis:For real $\zeta$, the product of the two logarithm arguments is $|\zeta -C{|}^2>0$. Hence the streamfunction is constant along the wall, on locally consistent branches of the complex logarithm, and there is no normal velocity. Pulling back gives the requested complex potential of a two-dimensional potential flow:This is the line source in a closed-end semi-infinite strip, with no extra imposed flow. Since $f^{\prime }(c)\ne 0$, near the physical source $f(z)-f(c)=f^{\prime }(c)(z-c)+O((z-c{)}^2)$. Thus $w=(Q/(2\pi ))\log (z-c)+$ a regular term, verifying that the source strength remains $Q$ under the conformal map. The image source has no preimage inside the physical strip. Finally, as $x\to \infty$, $w(z)\sim Qz/a+$ a constant, so the far-field velocity is $(Q/a,0)$ and the channel carries exactly the injected volume flux $Q$.

The complex modulus is a nonnegative real number. Equality therefore requires $z-1$ itself to be real and nonnegative. Its imaginary part gives $y=0$, and its real part then gives $x-1\ge 0$. Conversely every such point satisfies the equality. Hence the complete solution set isGeometrically this is the closed ray on the real axis starting at $z=1$.

Take $r>0$ and select a complex argument $\theta$ for $z$. To specify the complex exponentiation, use $z^{1+i}=\exp ((1+i)(\ln r+i\theta ))$. Expanding its exponent givesIts complex logarithm therefore has valuesThe continuous logarithm obtained by retaining the displayed exponent uses $k=0$. If the outer logarithm is the principal logarithm, choose $k$ to put the imaginary part in $(-\pi ,\pi ]$ instead; the real part is unchanged. In particular one cannot generally assert $\mathrm{Log}(e^w)=w$ without checking its imaginary part.

The condition on the real part gives $\ln r=\theta$, so the required logarithmic spiral has polar coordinatesAs $\theta$ increases, it runs from the negative real axis near radius $e^{-\pi }$ through $-i{e}^{-\pi /2}$, $1$ and $i{e}^{\pi /2}$ to the negative real axis at radius $e^{\pi }$. The angle increases by one complete turn. Using the conventional principal argument $-\pi <\theta \le \pi$, the inner negative-axis endpoint is excluded and the outer endpoint is included. The convention $-\pi \le \theta <\pi$ reverses those endpoint choices. If both limiting arguments are admitted by the paper's wording, both limiting endpoints are drawn; the interior curve is the same. The origin is never included because the complex logarithm is undefined there.

Put $s=t^2-t^{-2}$. Part (b)(ii) gives $u=s+2i$, soThe parameter $s$ increases strictly from $-15/4$ to $15/4$. Therefore the imaginary part $Y=4s$ increases from $-15$ to $15$, while the real part is $X=s^2-4$. The endpoints areThe real axis crossing is at $s=0$, giving $\boxed{(-4,0)}$ at $t=1$. The two imaginary axis crossings satisfy $s^2=4$, givingIn order, the trajectory passes from its lower endpoint through $(0,-8)$, then the leftmost point $(-4,0)$, then $(0,8)$, to its upper endpoint. It is a right-opening parabola segment, traversed continuously upward as shown in the right panel.

Squaring givesThus $\mathrm{Im}u=2$ is constant, while $d(\mathrm{Re}u)/dt=2t+2t^{-3}>0$. The trajectory is the horizontal line segmentIt crosses the imaginary axis at $\boxed{(0,2)}$, when $t=1$, and never crosses the real axis. The middle panel shows the rightward direction.

Write $z=X+iY$. Its coordinates are $X=t$, $Y=1/t$, so $XY=1$. For the given positive range of $t$, this traces a segment of the first-quadrant hyperbola, moving rightward and downward:The complex conjugate is $z^{\ast }=t-i/t$. It is the reflection in the real axis, tracing $XY=-1$ in the fourth quadrant, rightward and upward:Neither trajectory crosses either axis, because $t$ and $1/t$ remain nonzero. The left panel of the figure marks the endpoints and directions.

For $z\in \mathbb{C}\setminus [1,\infty )$, the number $w=1-z$ avoids the nonpositive real axis. Define the branch of the complex logarithm byHere the principal argument is selected as a continuous map on that slit domain, so this is a holomorphic function, with derivative $-1/(1-z)$ and $L(1)=0$. For example, locally writing $w=\rho e^{i\varphi }$ gives $d(\log \rho +i\varphi )=dw/w$, verifying the complex derivative. The normalization also makes this branch of the complex logarithm unique: any two continuous choices differ by a constant integer multiple of $2\pi i$ on the connected slit domain, and their values at $w=1$ agree.

To evaluate the corresponding analytic branch of a square root on the unit circle, factorFor $0<\theta <2\pi$, we have $\sin (\theta /2)>0$ and $-\pi /2<\theta /2-\pi /2<\pi /2$. Thus this factorization already uses the selected principal argument, with no hidden change by $2\pi$. It follows thatHalving this expression and applying the exponential function givesSince $\sqrt{2}{e}^{-i\pi /4}=1-i$, the desired branch value isFor instance, $\theta =\pi$ gives $\sqrt{2}>0$, a quick check that the sign agrees with the normalization.

Put $D(z)={\sinh }^{2}z+{\cosh }^{2}a$ and $S=\sinh (2a)>0$. Traverse the rectangle $-R\to R\to R+\pi i\to -R+\pi i\to -R$ with positive orientation of a contour, taking $R>a$. Part (i) shows that the only enclosed poles of $f(z)=z/D(z)$ areThey are simple poles becauseThe residue formula for a quotient at a simple denominator zero givesTheir imaginary terms cancel and their residues sum to $-2a/S$. Thus Cauchy's residue theorem gives

The choice of numerator $z$ makes the horizontal sides useful. The hyperbolic sine satisfies $\sinh (z+\pi i)=-\sinh z$, so $D(x+\pi i)=D(x)$. Consequently the lower and upper contour integrals, with their actual orientations, add toIt remains to justify discarding both vertical sides. On either side $z=\pm R+iy$, $0\le y\le \pi$, the reverse triangle inequality and part (b)(ii) implyThis lower bound is positive for sufficiently large $R$. Also $|z|\le R+\pi$ and each vertical side has length $\pi$. The estimation lemma bounds the absolute value of the sum $V_R$ of the vertical contributions byThe real improper integral converges absolutely, since its integrand is continuous on the real axis and decays like $4e^{-2|x|}$ at infinity. Combining all four sides and taking $R\to \infty$ givesThereforeAn independent real-variable check uses the hyperbolic tangent substitution $u=\tanh x$: the integral becomes ${\int }_{-1}^{1}du/({\cosh }^{2}a-u^2{\sinh }^{2}a)$. Its antiderivative uses the inverse hyperbolic tangent, giving $2\mathrm{arctanh}(\tanh a)/(\sinh a\cosh a)=4a/\sinh (2a)$. As $a\downarrow 0$, this tends to $2$, agreeing with ${\int }_{-\infty }^{\infty }{\mathrm{sech}}^{2}xdx=2$ and checking the factor of two.

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.

For $a\ne b$, integratealong the contour consisting of $[-R,R]$ and the upper semicircle, with positive orientation of a contour and $R>\max \{a,b\}$. Its enclosed simple poles are $ia$ and $ib$, with residuesOn the upper semicircle, $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$. The reverse triangle inequality gives $|z^2+a^2|\ge R^2-a^2$ and $|z^2+b^2|\ge R^2-b^2$, so the estimation lemma givesThe integral along the real axis is absolutely convergent: the denominator stays nonzero on bounded real intervals, and the integrand is $O(x^{-4})$ at infinity. The residue theorem therefore givesTake real parts. The real integrand is an even function, so

To include $a=b$, fix $a>0$ and let $b\to a$. For $a/2\le b\le 3a/2$, the absolute value of the integrand is bounded byan integrable function on $[0,\infty )$. The dominated convergence theorem justifies passing the limit under the integral. If $h(t)=e^{-t}/t$, then $h^{\prime }(a)=-e^{-a}(a+1)/a^2$, andHence

Use the principal logarithm $\mathrm{Log}z=\log |z|+i\mathrm{Arg}z$ with $-\pi <\mathrm{Arg}z<\pi$, whose branch cut is the negative real axis. For this choice,has three simple poles away from the cut: $z=1$, $z=ia$, and $z=-ia$. The pole at $1$, caused by $\mathrm{Log}1=0$, must be included.

Choose $0<\epsilon <\min \{1,a\}$ and $R>\max \{1,a\}$, and use a positively oriented keyhole contour around the negative real axis: the large circle is traversed counterclockwise, the upper bank from $-R$ to $-\epsilon$, the small circle clockwise, and the lower bank from $-\epsilon$ to $-R$. Formally, use radial segments at arguments $\pm (\pi -\eta )$, then let $\eta \downarrow 0$ with $R,\epsilon$ fixed. The function has continuous limiting values on each compact bank, which justifies this limit.

Put $L=\log a$. The residues areThe first follows from $\mathrm{Log}z=(z-1)+O((z-1{)}^2)$; the other two follow from the simple zeros of $z^2+a^2$. Their sum isThese expressions are valid also for $a=1$, since $L\pm i\pi /2$ never vanish for real $L$.

On the outer circle, $|\mathrm{Log}z|\ge \log R$ for $R>1$ and $|z^2+a^2|\ge R^2-a^2$. On the inner circle, $|\mathrm{Log}z|\ge |\log \epsilon |$ and $|z^2+a^2|\ge a^2-{\epsilon }^2$. Thus the estimation lemma bounds the two arc integrals byrespectively, as $R\to \infty$ and $\epsilon \downarrow 0$.

On the upper bank, $z=-x+i0$ gives $\mathrm{Log}z=\log x+i\pi$; on the lower bank it gives $\log x-i\pi$. Accounting for the opposite orientations, their combined integral isThe real improper integral converges: its integrand is bounded near zero by $1/(a^2{\pi }^2)$ and at infinity by $1/({\pi }^2{x}^2)$. Taking the arc limits in the residue theorem gives $-2\pi iI=2\pi i\Sigma$, henceThe final subtraction is exactly the contribution of the pole at $1$.

Use the same principal square root $S(z)=\sqrt{(z-a)/(z-b)}$, which is holomorphic on $A$ because its argument avoids the nonpositive real axis. Thenis a holomorphic function satisfyingThus it is a holomorphic branch of the requested square root. This construction does not require a global holomorphic logarithm of $z-a$ or $z-b$ separately. Its sign is fixed by $S(z)\to 1$ at infinity, so $g(z)/z\to 1$. The negative of this function is the other holomorphic branch; any two branches have a continuous quotient taking values in $\{1,-1\}$, which is constant because $A$ is connected.

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.

The lower wall has streamfunction zero. The streamline through the lip and along the upper fluid free boundary has streamfunction equal to the downstream volumetric flow rate, namely $a$ because the downstream speed is one and its depth is $a$. The complex potential of a two-dimensional potential flow therefore fillsThe chosen velocity potential origin gives $w(D)=ia$. The remote reservoir endpoints $A,E$ have velocity potential tending to $-\infty$; the downstream endpoints $B,C$ have velocity potential tending to $+\infty$.

Put $q=dw/dz=u-iv$. On the floor its values run from zero to one along the positive real axis. Above the plate the flow goes left toward the lip, so the plate velocities run along the negative real axis from zero at $E$ to $-1$ at $D$. On the fluid free boundary Bernoulli's equation and its constant pressure give $|q|=1$. Its downward velocity has $v<0$, hence $\mathrm{Im}q>0$. Consequently the hodograph plane domain isthe upper unit semicircle. Its marked boundary values are $q(A)=q(E)=0$, $q(D)=-1$, and $q(B)=q(C)=1$.

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 $(-\pi /2,3\pi /2)$.

Use the Fourier transform convention $\overline{U}(k,y)={\int }_{\mathbb{R}}u(x,y)e^{ikx}dx$, initially in a class where the integral transforms and subsequent limits are defined. Transforming Laplace's equation gives ${\overline{U}}_{yy}-k^2\overline{U}=0$ on real $k$. Decay for $y\to \infty$ selectsThe boundary trace is the sum of its negative-half-line unknown and the given positive-half-line datum. The normal derivative is zero on the negative side. ThereforeFor the $e^{ikx}$ convention, negative support gives a integral transform analytic for $\mathrm{Im}k<0$, and positive support gives one analytic for $\mathrm{Im}k>0$, subject to the appropriate endpoint interpretation. The normal trace in this particular problem requires a finite-part interpretation, detailed below.

For the Wiener-Hopf method, regularize the real-axis kernel byTake the branch cut of $a_{+}$ vertically downward from $-iϵ$ and that of $a_{-}$ vertically upward from $iϵ$. Their product on the real axis is the positive square root $\sqrt{k^2+{ϵ}^2}$. The branch of $H_{+}$ has its cut downward from zero. Dividing the regularized equation by $a_{+}$ givesThe forcing $P$ is analytic in $0<\mathrm{Im}k<ϵ$, between its two cuts. Decompose $P=P_{+}-P_{-}$ by Cauchy integrals on two horizontal lines within that strip. Separation giveswhen the endpoint growth class makes the common entire remainder bounded. For $\mathrm{Im}k<0$ one can take the minus Cauchy integral on $\mathrm{Im}\hat{k}=\delta$, with $0<\delta <ϵ$, obtainingThe integrand is $O(|\hat{k}{|}^{-5/4})$ at infinity, so the Cauchy integral converges there. In the limit choose $\delta =\delta (ϵ)$, for example $\delta =ϵ/2$; a fixed positive $\delta$ cannot remain between zero and $ϵ$. With $\sqrt{k}$ denoting the limiting lower-analytic branch and with the prescribed limiting interpretation,The numerator square root has a cut going upward from $iϵ$; the power ${\hat{k}}^{-3/4}$ has a cut going downward from zero. The contour passes between them. The pole $\hat{k}=k$ is below this contour when $k$ is in the lower integral transform domain. These branch choices are essential, rather than independent principal powers of the factors.

There is an important mathematical limitation in the source's ordinary integral definition of $\bar{g}$. An explicit harmonic function particular solution in polar coordinates $0<\theta <\pi$ isEach separated polar power is harmonic function; at $\theta =0$ it has the required Dirichlet boundary condition value, and at $\theta =\pi$ its angular derivative vanishes, giving the required zero normal derivative. It also decays at infinity. Its unknown boundary traces areThe last function is not locally integrable at zero. Therefore the printed ordinary integral ${\int }_0^{\infty }g(x)e^{ikx}dx$ diverges, even for $\mathrm{Im}k>0$. The integral transform calculation is valid for this solution only after specifying a Hadamard finite-part integral, or equivalently a distribution boundary trace.

For the canonical homogeneous extension of this power, the two transformed traces areThe first is analytic below and the second above. The second formula follows from $\Gamma (3/4)=(-1/4)\Gamma (-1/4)$ and the finite-part continuation of the Laplace transform. Substituting their real-axis boundary values directly verifies the boxed integral transform relation for both signs of $k$, so the qualification can be checked without relying only on formal factorization.

There is also homogeneous corner freedom:satisfies zero Dirichlet boundary condition data on the positive axis, zero Neumann boundary condition data on the negative axis and decay at infinity. Its negative trace has integral transform $C\Gamma (1/2)(ik{)}^{-1/2}$, corresponding to the term $A/\sqrt{k}$. Higher singular modes $r^{-(n+1/2)}\sin ((n+1/2)\theta )$ would also satisfy the homogeneous boundary conditions for integers $n\ge 1$, but are excluded if the negative trace is required to be locally integrable. Thus the one-constant formula belongs to a specified corner-growth class. The original boundary conditions alone, without that class, do not assert uniqueness. These facts form the finite-part normal trace for mixed quarter-power boundary data.

For real $x\ne -1$, $w=1-2/(x+1)$ is real. Its finite image is the entire real axis except $w=1$, which is approached only as $x\to \pm \infty$. The point $x=-1$ maps to infinity. On the extended complex plane the extended real axis maps onto itself, with $z=\infty$ mapping to one.