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For a parameter $k$, one form of the incomplete elliptic integral of the first kind isIts value depends on the choices of square-root branches and, under analytic continuation around the branch points, on the integration path.

Let $w$ be a local value of the complex inverse sine. Since the limiting values at the two branch points are $\pi /2$ at $z=1$ and $-\pi /2$ at $z=-1$, monodromy reflection at a square-root branch point givesFollowing a counterclockwise loop that encloses both branch points in the convention where $\sqrt{1-z^2}$ is positive just above $(-1,1)$ applies the two reflections in the order that givesThe reverse orientation gives $w+2\pi$. Thus the translational part of the monodromy is generated by $2\pi$.

On the stated branch of the principal complex logarithm, the only denominator zero is $z=1$, and $\log z=(z-1)+O((z-1{)}^2)$. ThereforeZero is a logarithmic branch point, not a removable isolated singularity, even though $1/\log z$ tends to zero as $z\to 0$ within the cut plane. Continuation around zero changes $\log z$ by $2\pi i$. The nonpositive real axis is the chosen branch cut; its nonzero points are not intrinsically branch points and admit local analytic continuation. There are no other finite singularities inside the specified branch domain.

At $z=1$, numerator and denominator both have simple zeroes. Their derivatives give the removable valueAt each positive integer $n\ge 2$, the numerator is nonzero and the denominator has a simple zero, so these are simple poles. At zero the expansion on the principal sheet iswhich contains a fractional power. Thus zero is a branch point with pole-like leading growth, not an ordinary isolated pole. The negative real axis is the chosen branch cut. On local continuations across its nonzero points, negative integers also give simple poles because the continued square root is $\pm i\sqrt{|n|}$, never one. They lie outside the open principal branch domain; noninteger negative points are merely cut points, not intrinsic singularities.

Linearity permits subtraction of the incident acoustic membrane wave. The scattered pressure solves the homogeneous Helmholtz equation, and the kinematic relation defines its line displacement ${\eta }^{\prime }$ on both halves. On $x<0$, subtracting the incident elastic membrane equation gives the scattered dynamic equation. On $x>0$, there is no elastic membrane, so the total pressure jump must vanish, giving $[p^{\prime }]=-[p_I]$. The pinned endpoint has $\eta (0)=0$, hence ${\eta }^{\prime }(0)=-1$, rather than zero.

Use full and Half-range Fourier transforms with the same $e^{ikx}$ convention as part (a), and putWrite $P^{-}$ for the left transform of the upper scattered pressure, $P^{+}$ for its right transform, and ${\eta }^{\pm }$ for the corresponding transforms of ${\eta }^{\prime }$. Outgoing antisymmetry in $y$ impliesThe second relation comes from the prescribed pressure cancellation on $x>0$.

The endpoint terms in the left Fourier transform of ${\eta }_{xx}^{\prime }$ are crucial:Thus $2P^{-}=a{\eta }^{-}+TA+iT(k+k_I)$. Eliminating ${\eta }^{-}$ gives the Wiener-Hopf equationThere is a defect in the printed right side: with the stated incident exponential, pressure jump, and pinned total displacement, it omits the incident endpoint term and has the opposite sign on the incident-pressure term. The displayed corrected equation follows directly from both boundary traces. The given right side cannot be derived with these definitions. Using $a(k_I)=-2\beta /{\gamma }_I$, an especially useful equivalent form isThe correction is necessary for the reconstructed pressure to satisfy the original physical boundary conditions.

For Wiener-Hopf factorization, take $K=K^{+}{K}^{-}$ with the plus factor analytic and nonzero in the upper half-plane and the minus factor analytic and nonzero in the lower half-plane. Singularities listed next refer to continuation out of each factor's own analytic half-plane. Let $b=\omega \sqrt{m/T}$, so $a=-T(k-b)(k+b)$ and $b$ lies below the real axis. Generically $K^{+}$ inherits the pole at $b$ and the square-root branch point at $k_0$, while $K^{-}$ inherits the pole at $-b$ and the branch point at $-k_0$. The kernel is finite but nonanalytic at the acoustic branch points: its local behavior is a constant plus a square-root term, not an inverse-square-root divergence. Coincident branch points and poles require a limiting treatment.

Zeros of $K$ are the fluid-loaded dispersion relation roots. Lower-half-plane roots, including the incoming guided root $k_I$, belong to the continued $K^{+}$; upper-half-plane roots, including the reflected root $-k_I$ for the even kernel, belong to the continued $K^{-}$. The roots represent membrane-guided modes; the branch cuts represent radiated acoustic waves. Only roots on the selected outgoing sheet are included, not spurious roots created by squaring the dispersion relation. No explicit factorization is required.

Here is an explicit solution in terms of those factors. DefineThe pole of $B$ at $-b$ has residue $C$; the pole at $b$ is canceled by the pole of $K^{+}$. Thus $B^{\pm }$ have the required respective analyticity. Divide the corrected Wiener-Hopf equation by $K^{+}$ and split its right side as $F^{+}+F^{-}$, whereThe difference quotient in $F^{-}$ is removable at $k_I$ and is analytic below. Moving the plus and minus terms to opposite sides yields the common entire function. With the assumed $E(k)=0$, the solution isCombining $P^{-}$ with the known $P^{+}$ gives the full scattered pressure transformThe requested pressure integral, for $y\ne 0$, is consequentlyThe real contour $\mathcal{C}$ uses the causal continuation $\mathrm{Im}\omega <0$; its undamped limit retains the induced pole and branch-cut prescriptions. The residue theorem shows that upper-half-plane zeros of $K^{-}$ yield left-going scattered elastic membrane modes. On the right, the lower incident-pole residue of the scattered integral cancels $p_I$ at the open line, as required.

Finally reconstruct the left scattered displacement:For a generic unspecified $A$, this has a lower-half-plane pole at the bare elastic membrane wavenumber $b$. Such a pole is incompatible with analyticity of an outgoing left-supported scattered displacement: it would represent an additional right-going incoming elastic membrane contribution. It must be removed. Its residue is $-[2P^{-}(b)-TA-iT(b+k_I)]/(2Tb)$, so the incoming-pole cancellation at a pinned membrane edge condition isSince $P^{-}$ is linear in $A$, this fixes the endpoint slope generically. Explicitly it isThe physical lower incident pole of the total displacement is already prescribed by ${\eta }_I$; it must not be confused with this removable spurious pole of the scattered field.

Set $F(z,w)=w^2-z^n+1$. Its complex gradient is $(-n{z}^{n-1},2w)$, never zero on $F=0$: simultaneous zero would give $z=w=0$, which is not on the curve. The complex implicit function theorem therefore gives a one-dimensional complex manifold, hence a Riemann surface. Where $w\ne 0$ use $z$ as coordinate, and where $z\ne 0$ use $w$; the transition maps and both projections are holomorphic.

For $\pi$, a ramification point of a holomorphic map has $w=0$ and $z=\zeta$ with ${\zeta }^n=1$. In the coordinate $w$, $z-\zeta =w^2/(n{\zeta }^{n-1})+O(w^4)$, giving ramification index $2$. Thus there are $n$ such points, and their branch points are the $n$th roots of unity.

For $\tau$, a ramification point of a holomorphic map has $z=0,w=\pm i$. In the coordinate $z$, $w-w_0=z^n/(2w_0)+O(z^{2n})$, giving ramification index $n$ at both points and branch points $\pm i$.

Write $n=2m$. At infinity set $u=1/z$, $v=w{u}^m$; then $v^2=1-u^n$. At $u=0$ there are two smooth points, $v=\pm 1$, and $u$ is a local coordinate. The extended $\pi$ has a simple pole at each, so neither is ramified. The degree-two Riemann-Hurwitz formula for the map to the Riemann sphere gives $2g-2=2(-2)+n$, henceThe list for $\tau$ above concerns the original affine Riemann surface; on the compactification its poles at the two added points have order $m$ and are additionally ramified if $m>1$.