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The rational factors give four simple poles at $z=\pm i,\pm 2i$; none is canceled by the numerator. The square root has branch points at $z=\pm 1$, with local behaviour proportional to $(z\mp 1{)}^{-1/2}$ in the integrand.

For this part, choose $h(z)=\sqrt{z^2-1}$ analytic off $[-1,1]$ and satisfying $h(z)\sim z$ at infinity. Its values are $h(x)=\sqrt{x^2-1}$ for $x>1$, $h(x)=-\sqrt{x^2-1}$ for $x<-1$, andThus the chosen branch cut is $[-1,1]$, and the function is analytic elsewhere apart from the four poles. Infinity is not an intrinsic branch point: $g(z)=O(z^{-2})$ on this branch.

Another legitimate convention applies the positive-axis square root of part (a) to $w=z^2-1$; it places the cuts on $(-\infty ,-1]$ and $[1,\infty )$. Its upper-half-plane values agree with the branch used here, so the contour calculation below and the positive real square root in the required integral are unchanged. The branch choice must be stated because a square root alone does not specify a single-valued complex function.

Factor the polynomial denominator as $(z-i{)}^2(z+i{)}^2$. The numerator has values $i$ and $-i$ at the two roots, so neither cancels. ThereforeThe factor $z^{-1/2}$ gives a branch point at zero, with local behaviour $f(z)\sim z^{-1/2}$. Infinity is also a branch point, since in the coordinate $w=1/z$ the leading behaviour is $f(1/w)\sim w^{5/2}$. A single-valued branch requires a branch cut joining these points. Away from that cut and the two poles, the selected branch is holomorphic. For the integral below, use $0<\arg z<2\pi$, putting the cut along the positive real axis.

Use the Fourier convention $\hat{u}(k,y)={\int }_{\mathbb{R}}u(x,y)e^{ikx}dx$. Transforming the differential equation givesDefine $k_{+}(k)=\sqrt{k}$ analytic above the downward cut from zero, and $k_{-}(k)=\sqrt{k-i}$ analytic below the upward cut from $i$. More explicitly take $\arg k\in (-\pi /2,3\pi /2)$ and $\arg (k-i)\in (-3\pi /2,\pi /2)$. In the strip $0<\mathrm{Im}k<1$, put $q=k_{+}{k}_{-}$. Then $q^2=k^2-ik$ and $\mathrm{Re}q>0$; at $k=i/2$, $q=1/2$, fixing the choice throughout the strip. The decaying $y$ mode is $e^{-qy}$.

The positive-half-line boundary value has transform ${\int }_0^{\infty }{e}^{ikx}dx=i/k$ for $\mathrm{Im}k>0$. Thus $\hat{u}(k,0)=\overline{f}+i/k$ and ${\hat{u}}_y(k,0)=\overline{g}$. The transformed boundary relation isLet $c=k_{-}(0)=e^{-i\pi /4}$. Splitting off the pole at zero givesThe left side is analytic in the lower domain up to $\mathrm{Im}k<1$: its apparent pole at zero is removable. The right side is analytic for $\mathrm{Im}k>0$. By the supplied boundedness of $k\overline{f}$ and $\overline{g}$, both sides tend to zero at infinity in their half-planes. The gluing and Liouville theorem argument therefore makes their common value zero. ConsequentlyChoose the inversion contour $\Gamma =\{s+i\eta :s\in \mathbb{R}\}$ oriented left to right, with any fixed $0<\eta <1$. It passes above the pole at zero and below the branch point at $i$. Fourier inversion gives exactlyThus the appropriate value is $\sqrt{-i}=e^{-i\pi /4}$. The other proposed value reverses the residue of the boundary step and violates its prescribed positive value.

There is also a far-field qualification in the source. The formula is the intended bounded mixed-boundary solution, but it does not tend to zero along every path to infinity. This can be checked explicitly: in parabolic coordinates $x=a^2-b^2$, $y=2ab$, with $a,b>0$, the mixed-boundary convection-diffusion solution in parabolic coordinates isIndeed $\Delta u=u_{bb}/[4(a^2+b^2)]$ and $u_x=-b{u}_b/[2(a^2+b^2)]$; the identity $u_{bb}=-2b{u}_b$ proves the differential equation. At $b=0$ the boundary value is one, and at $a=0$ its normal derivative is zero. Its traces are $f(x)=\mathrm{erfc}\sqrt{-x}$ for $x<0$ and $g(x)=-1/\sqrt{\pi x}$ for $x>0$, whose one-sided transforms are the expressions derived above.

At a fixed angle away from the positive $x$ axis, $b\to \infty$ and $u\to 0$. But at fixed $y>0$ and $x\to +\infty$, $b\sim y/(2\sqrt{x})$ and $u\to 1$. Hence the radial decay statement cannot hold uniformly as printed; the intended decay is away from the downstream boundary layer. The exact expression also identifies its thickness as $O(\sqrt{x})$.

The square root produces branch points at $z=\pm 1$, with local behaviour proportional to $(z\mp 1{)}^{-1/2}$. One branch cut can join them along $[-1,1]$. Another choice uses the two rays $(-\infty ,-1]\cup [1,\infty )$. Each choice prevents a loop from enclosing just one branch point within a single-valued domain. For the finite-segment choice fix $\sqrt{z^2-1}\sim z$ at infinity. Infinity is then a pole of order three, not a branch point, since

Use the principal complex square root, $\sqrt{z}=\sqrt{r}{e}^{i\theta /2}$ with $-\pi <\theta <\pi$. For $F(z)=e^{z-\sqrt{z}}/z$, the local expansion on this holomorphic branch isThe fractional power prevents a single-valued Laurent series in a punctured disc. In particular, zero is a branch point, not an isolated singularity classified as a pole. Merely observing that $zF(z)\to 1$ on the slit plane would not establish a simple pole: $zF(z)$ does not extend as a holomorphic function through a whole disc around zero.

The negative real half-line is a chosen branch cut. Its upper and lower boundary values differ because $\sqrt{-x+i0}=i\sqrt{x}$ and $\sqrt{-x-i0}=-i\sqrt{x}$ for $x>0$. Continuation once around zero changes the sign of the complex square root, and hence changes $F$ in general. Other branch cuts are possible; this one defines the principal holomorphic branch.

Singularities and cuts. There are four simple poles $z_j=e^{i(\pi /4+j\pi /2)}$, $j=0,1,2,3$, because the roots of $1+z^4$ are simple and do not coincide with $\pm 1$. At $z=\pm 1$, the local behavior is a nonzero constant times $(z\mp 1{)}^{-1/2}$: these are square-root branch points. Infinity is not a branch point; on either branch $f(z)=O(z^{-5})$, so it extends there with a zero of order five. Three valid choices of branch cuts are the segment $[-1,1]$, the two rays $(-\infty ,-1]\cup [1,\infty )$ joined through infinity, or a semicircular arc joining $-1$ to $1$ and avoiding the poles. A circuit about either endpoint changes the sign, while a circuit about both endpoints does not.

Branch and residues. On $\mathbb{C}\setminus [-1,1]$, let $s(z)=\sqrt{z^2-1}$ be the branch with $s(z)/z\to 1$ at infinity, and define $\sqrt{1-z^2}=-is(z)$. Its upper boundary value on $(-1,1)$ is positive. The residue at a simple pole $z_j$ isFor fully explicit values, putIn counterclockwise order $z_0,z_1,z_2,z_3$, the values of $s$ are $A+iB,-A+iB,-A-iB,A-iB$. Since $A^2+B^2=\sqrt{2}$, substitution yieldsA branch point has no ordinary residue, since it has no single-valued Laurent series on a punctured neighborhood. At infinity the residue is zero.

Contour evaluation. The original contour goes counterclockwise around a large circle and clockwise round the slit, with its upper lip directed to the right. The large-circle contribution is $O(R^{-4})$. Small endpoint arcs contribute $O({ϵ}^{1/2})$. The artificial slit extension from $1$ to $R$ has equal boundary values on both sides, so its contributions cancel. On $(-1,1)$ the two boundary values of $f$ are opposite. Thus, with $I$ denoting the requested real integral, the residue theorem givesTherefore

The square root produces branch points at $z=\pm 1$, with local behaviour proportional to $(z\mp 1{)}^{-1/2}$. One branch cut can join them along $[-1,1]$. Another choice uses the two rays $(-\infty ,-1]\cup [1,\infty )$. Each choice prevents a loop from enclosing just one branch point within a single-valued domain. For the finite-segment choice fix $\sqrt{z^2-1}\sim z$ at infinity. Infinity is then a pole of order three, not a branch point, since

The square-root branch points are zero and infinity. At zero the local behavior is $z^{-1/2}$; in the coordinate $u=1/z$ near infinity it is $u^{1/2}$. In either case one circuit reverses the sign and two restore it. Thus the ramification index is $2$: these are two-sheeted algebraic branch points, sometimes described as branch order one when order means ramification index minus one. The decay to zero at infinity does not remove the branching.

Use the complex logarithm with $0\le \arg w\le \pi /2$ on the nonzero points of the first-quadrant square. Put $z=x+iy=\log |w|+i\arg w$. The square bounds $\mathrm{Re}w,\mathrm{Im}w\le a$ giveThe two edges meeting at $w=0$ map to the horizontal half-lines $y=0$ and $y=\pi /2$, extending indefinitely to the left. The other edges map to $x=\log a-\log \cos y$ for $0\le y\le \pi /4$ and $x=\log a-\log \sin y$ for $\pi /4\le y\le \pi /2$. They meet at $\log (a\sqrt{2})+i\pi /4$.

The angles are preserved at $a$, $a(1+i)$ and $ia$. There $d\log w/dw=1/w$ is finite and nonzero. The origin is a branch point, has no finite logarithmic image, and supplies no fourth finite vertex at which angle preservation could be asserted.

The finite singularities are square-root branch points at $z=\pm 1$. Choose the cut between them and the branch $f(z)\sim z$ at infinity. The two banks then have values $+i\sqrt{1-x^2}$ above and $-i\sqrt{1-x^2}$ below. A counterclockwise loop around the cut traverses its upper bank right-to-left, so its integral is $-2iI$, where $I$ is the required real integral. At large radius,Therefore $\boxed{I=\pi /2}$. Infinity is not another square-root branch point: this branch has an ordinary pole there on its sheet. Small detours about the finite endpoints vanish as their radius to power $3/2$.

A branch point obstructs a single-valued analytic continuation around a closed circuit. Continuing $\log z=\log |z|+i\theta$ once around zero increases its value by $2\pi i$. A branch cut removes a curve joining branch points or extending to infinity so that a continuous argument can be chosen on the remainder. Zero and infinity are logarithmic branch points; the positive-axis cut used here selects $0<\arg z<2\pi$. The cut is a choice of representation, not a line of poles of the multivalued logarithm.

A branch point is a point around which analytic continuation changes the value of a multivalued function. For $z^{1/2}=r^{1/2}{e}^{i\theta /2}$, one circuit around zero changes the sign. A branch cut removes paths producing this ambiguity; for example $-\pi <\arg z<\pi$ gives a single holomorphic square root off the nonpositive real axis. The position of a cut is a convention, while the branch point is intrinsic.

Set $q(z)=(z-1)/(z+1)$. For $z$ outside the closed segment joining $-1$ to $1$, $q(z)$ avoids the nonpositive real axis. Therefore the principal square root of $q$ defines a holomorphic function there, with the normalization $f(\infty )=1$. The endpoints are branch points. Expanding at infinity,For a positively oriented contour surrounding the cut, its integral equals the integral on a sufficiently large positively oriented circle, since the intervening annulus is holomorphic. ThusOn the upper and lower banks, $f_{+}(x)=+i\sqrt{(1-x)/(1+x)}$ and $f_{-}(x)=-i\sqrt{(1-x)/(1+x)}$. The upper bank is traversed from $1$ to $-1$, and the lower bank from $-1$ to $1$. Their sum is $-2i$ times the required real integral. Small endpoint arcs vanish because their singularities have exponent less than one. ConsequentlyFor $\alpha =1/n$, use $f_{\alpha }=q^{\alpha }$ with the same normalization. Its expansion is $1-2\alpha /z+O(z^{-2})$, and its bank values are $e^{\pm i\pi \alpha }[(1-x)/(1+x){]}^{\alpha }$. Equating the contour integral $-4\pi i\alpha$ with $-2i\sin (\pi \alpha )$ times the real integral givesThe condition $n\ge 2$ ensures endpoint integrability. Choosing the other square-root branch reverses $I_{\Gamma }$, but also reverses both bank values, leaving the real integral unchanged.

At a regular point, $f(z_0+\zeta )-f(z_0)=f^{\prime }(z_0)\zeta +O({\zeta }^2)$. Multiplication by a nonzero complex number rotates every tangent by the same angle and multiplies every length by the same positive factor. Oriented angles are preserved. If the first derivative vanishes but the second does not, the leading term is $f^{\prime \prime }(z_0){\zeta }^2/2$: tangent arguments are doubled, so angles are doubled, modulo the identification of rays. This is a local statement; holomorphy alone does not imply global injectivity.

Write $z=x+iy$. ThenFor $-1<x<1$ and $y>0$, the imaginary part is positive. The bottom edge maps monotonically to $[-1,1]$; the left and right vertical edges map to $(-\infty ,-1]$ and $[1,\infty )$ respectively. The inverse below establishes that every point of the open upper half-plane occurs exactly once. At fixed $y=a$, putting $u=\mathrm{Re}w$, $v=\mathrm{Im}w$, one obtainsThis is the upper half of an ellipse, traversed from its left endpoint to its right endpoint.

Choose the branch $s(w)=\sqrt{1-w^2}$ analytic in the upper half-plane and positive on the boundary interval $(-1,1)$. Then the inverse branch into the strip isThe logarithm is chosen continuously so that $-1<\mathrm{Re}z<1$ and $\mathrm{Im}z>0$; for instance $w=iv$, $v>0$, gives $z=(2i/\pi )\mathrm{arsinh}v$. Differentiation gives $dz/dw=2/(\pi \sqrt{1-w^2})$. The only zeros of the original derivative in the closed strip are $z=\pm 1$, with images $w=\pm 1$. The inverse has square-root branch points there, and cannot extend holomorphically across them. At either corner the original map doubles the right angle to a straight angle; a holomorphic inverse with nonzero derivative could not halve it. The open strip interior has no such failure.

Write $\nu =m\gamma /{\hslash }^2$, so $t(k)\simeq k/(k+i\nu )$. Its nontrivial pole isFor negative $\gamma$ this is on the positive imaginary $k$ axis, the physical bound-state sheet. For positive $\gamma$ it lies on the negative imaginary axis and is a virtual-state pole reached by analytic continuation through the continuum cut. These different sheets can have the same negative numerical energy; specifying only $E_{\ast }$ would conceal the distinction. As a function of $E$, the momentum square root also introduces the continuum threshold branch point at zero. If $\gamma =0$, numerator and denominator cancel at this order and no pole is predicted by this leading approximation.

Let $z=E-V_0+i0$. The outgoing analytic branch of Klein-Gordon step momentum iswhere each square root is continued from the upper half-plane. It behaves as $z/(\hslash c)$ at large positive or negative real $z$. The branch points are $E=V_0\pm M$; replacing this product by a principal square root of $z^2-M^2$ can choose the wrong sign below the lower threshold.

I. For $E-V_0>M$, $q>0$ and both transmitted kinetic energy and group velocity are positive. Thus $0\le \mathcal{R}<1$ and $\mathcal{T}=4kq/(k+q{)}^2>0$; the transmission decreases to zero on approaching the upper threshold from above.

II. For $|E-V_0|<M$, write $q=i\kappa$ with $\kappa =\sqrt{M^2-(E-V_0{)}^2}/(\hslash c)>0$. The right wave decays, $r=(k-i\kappa )/(k+i\kappa )$ has modulus one, and $\mathcal{T}=0$. There is an evanescent tail but no flux transmitted to infinity. Its reflection quantum phase changes continuously between the two thresholds.

III. For $E-V_0<-M$, the analytic continuation gives $q=-Q$, where $Q=\sqrt{(E-V_0{)}^2-M^2}/(\hslash c)>0$. The group velocity $c^2\hslash q/(E-V_0)$ is again rightward, while the charge density and transmitted charge current are negative. ThereforeChoosing $q=+Q$ would describe a wavepacket arriving from the right, inconsistent with the chosen incoming channel. As the lower threshold is approached, $q\to 0^{-}$ and $\mathcal{R}\to 1$ from above. This Klein-Gordon step amplification has a real denominator zero at $E=V_0/2$ when $V_0>2M$: there $Q=k$, and no finite stationary solution with unit incoming amplitude satisfies both matching equations. The resulting formal pole must not be described as an infinite single-particle probability. Finite switching, wavepackets and the treatment of the external source require a multiparticle analysis.^[ref]

A square-root branch point has ramification index $2$: one circuit switches local branches and two restore the original branch. Both zero and infinity are square-root branch points of $z^{-1/2}$, since the coordinate $u=1/z$ changes the latter local expression to $u^{1/2}$. A finite limiting value at a branch point does not imply a removable singularity.

Let $a>0$ and let $R$ be a rational function with $R(z^2)$ finite for real $z$ and $R(u)=O(u^{-1})$ at infinity. Choose the branch $h(z)=\sqrt{z^2-a^2}$ analytic off $[-a,a]$, with $h(z)\sim z$ at infinity. Put $g(z)=zR(z^2)/h(z)$. On the real exterior, $g(-x)=g(x)$; on the upper bank of the cut, $h(x+i0)=i\sqrt{a^2-x^2}$, so $g(x+i0)$ is odd in $x$.

Close a real-axis contour with a large upper semicircle, making clockwise upper indentations at the two branch points. The large arc vanishes because $g=O(z^{-2})$. Each small indentation is $O(\sqrt{\epsilon })$ and vanishes. The integral along the upper bank of $[-a,a]$ is zero by oddness; the two exterior pieces sum to twice the required integral. The residue theorem therefore givesResidues must use the chosen branch, notably $h(iy)=i\sqrt{y^2+a^2}$ for $y>0$. The substitution $u=\sqrt{x^2-a^2}$ provides an independent real-variable check, converting the integral to ${\int }_0^{\infty }R(u^2+a^2)du$.

On $\mathbb{C}\setminus ((-\infty ,-1]\cup [1,\infty ))$, choose the holomorphic branch $f(z)=\sqrt{1-z}\sqrt{1+z}$, taking each factor's principal square root. This gives $f(x)>0$ for $-1<x<1$. Across the branch cuts, its limits are $f(x+i0)=-i\sqrt{x^2-1}$ and $f(x-i0)=+i\sqrt{x^2-1}$ for $x>1$, and $f(x+i0)=+i\sqrt{x^2-1}$ and $f(x-i0)=-i\sqrt{x^2-1}$ for $x<-1$. The endpoints are branch points; analytic continuation around a single endpoint changes the square-root sign.

The branch points are $\pm 1$; infinity has two unbranched meromorphic choices. On the complex plane slit along $[-1,1]$, normalization $f(z)\sim z$ at infinity selects a holomorphic square root with boundary values $f(x+i0)=i\sqrt{1-x^2}$ and $f(x-i0)=-i\sqrt{1-x^2}$ for $-1<x<1$. On the exterior real intervals, $f(x)=\mathrm{sgn}(x)\sqrt{x^2-1}$. A counterclockwise contour integral around the finite branch cut satisfies $\oint d\zeta /f(\zeta )=2\pi i$: the upper lip runs right to left, the lower lip left to right, and the endpoint arcs contribute $O(\sqrt{\epsilon })$. The opposite global branch reverses these signs.

Count ramification points upstairs in $X$, weighted by $e_p-1$, rather than counting distinct branch values of a holomorphic map downstairs in $Y$. Grouping by fibres givesThe cardinality counts distinct points; the degree of a holomorphic map counts them with their analytic ramification indices. Thus each summand is the number of missing distinct preimages in that fibre.^[ref]

The phrase branch point varies across conventions: to keep the two surfaces distinct, use ramification point for $p\in X$ and branch value of a holomorphic map for $f(p)\in Y$. The existing name branch point of a holomorphic map here denotes the latter. Several ramification points can share one branch value of a holomorphic map.