The Schwarz-Christoffel Transformation

Section 10.9 The Schwarz-Christoffel Transformation

To proceed further, we must review the rotational effect of a conformal mapping \(w=f(z)\) at a point \(z_0\text{.}\) If the contour \(C\) has the parameterization \(z(t)=x(t)+iy(t)\text{,}\) then a vector \(\mathbf{\tau }\) tangent to \(C\) at the point \(z_0\) is

\begin{equation*} \mathbf{\tau} = z\,'(t_0) = x\,'(t_0) + iy\,'(t_0)\text{.} \end{equation*}

The image of \(C\) is a contour \(K\) given by \(w=u\big(x(t), \, y(t)\big) + iv\big(x(t), \, y(t)\big)\text{,}\) and a vector \(\mathbf{T}\) tangent to \(K\) at the point \(w_0=f(z_0)\) is

\begin{equation*} \mathbf{T} = w\,'(z_0) = f\,'(z_0)z\,'(t_0)\text{.} \end{equation*}

If the angle of inclination of \(\mathbf{\tau }\) is \(\beta =\mathrm{Arg}(z)\,'(t_0)\text{,}\) then the angle of inclination of \(\mathbf{T}\) is

\begin{equation*} \mathrm{Arg}\big(\mathbf{T}\big) = \mathrm{Arg}[f\,'(z_0) z\,'(t_0)] = \mathrm{Arg}\big(f\,'(z_0)\big) + \beta\text{.} \end{equation*}

Hence the angle of inclination of the tangent \(\mathbf{\tau}\) to \(C\) at \(z_0\) is rotated through the angle \(\mathrm{Arg}\big(f\,'(z_0)\big)\) to obtain the angle of inclination of the tangent \(\mathbf{T}\) to \(K\) at the point \(w_0\text{.}\)

Many applications involving conformal mappings require the construction of a one-to-one conformal mapping from the upper half-plane \(\mathrm{Im}( z) >0\) onto a domain \(G\) in the \(w\) plane where the boundary consists of straight-line segments. Let’s consider the case where \(G\) is the interior of a polygon \(P\) with vertices \(w_1,\,w_2,\,.\,.\,.\,,\,w_n\) specified in the positive sense (counterclockwise). We want to find a function \(w=f(z)\) with the property

\begin{align} w_k \amp = f(x_k) \text{ for } k=1,\,2, \ldots ,n-1, \text{ and }\tag{10.9.1}\\ w_n \amp = f(\infty) \text{ where } x_1 \lt x_2 \lt \cdots \lt x_{n-1}\lt \infty\text{.}\notag \end{align}

Two German mathematicians Herman Amandus Schwarz (1843–1921) and Elwin Bruno Christoffel (1929–1900) independently discovered a method for finding \(f\text{,}\) which we present as Theorem 10.9.1.

Theorem 10.9.1. Schwarz-Christoffel.

Let \(P\) be a polygon in the \(w\) plane with vertices \(w_1,\,w_2, \ldots ,w_n\) and exterior angles \(\alpha_k\text{,}\) where \(-\pi \lt \alpha_k \lt \pi\text{.}\) There exists a one-to-one conformal mapping \(w=f(z)\) from the upper half-plane \(\mathrm{Im}(z)>0\) onto \(G\) that satisfies the boundary conditions in Equations (10.9.1). The derivative \(f\,'(z)\) is

\begin{equation} f\,'(z) = A(z-x_1)^{-\frac{\alpha_1}{\pi}}(z-x_2)^{-\frac{\alpha_2}{\pi}} \cdots (z-x_{n-1})^{-\frac{\alpha_{n-1}}{\pi}}\text{,}\tag{10.9.2} \end{equation}

and the function \(f\) can be expressed as an indefinite integral

\begin{equation} f(z) = B + A\int(z-x_1)^{-\frac{\alpha_1}{\pi}}(z-x_2)^{-\frac{\alpha_2}{\pi}} \cdots (z-x_{n-1})^{-\frac{\alpha_{n-1}}{\pi}}\,dz\text{,}\tag{10.9.3} \end{equation}

where \(A\) and \(B\) are suitably chosen constants. Two of the points \(\{x_k\}\) may be chosen arbitrarily, and the constants \(A\) and \(B\) determine the size and position of \(P\text{.}\)

Proof.

The proof relies on finding how much the tangent

\begin{equation*} \mathbf{\tau }_j=1+0i \end{equation*}

(which always points to the right) at the point \((x,0)\) must be rotated by the mapping \(w=f(z)\) so that the line segment \(x_{j-1}\lt x\lt x_j\) is mapped onto the edge of \(P\) that lies between the points \(w_{j-1}=f(x_{j-1})\) and \(w_j=f(x_j)\text{.}\) The amount of rotation is determined by \(\mathrm{Arg}\,f\,'(x)\text{,}\) so Equation (10.9.2) specifies \(f\,'(z)\) in terms of the values \(x_j\) and the amount of rotation \(\alpha_j\) that is required at the vertex \(f(x_j)\text{.}\)

If we let \(x_0=-\infty\) and \(x_n=\infty\text{,}\) then, for values of \(x\) that lie in the interval \(x_{j-1} \lt x \lt x_j\text{,}\) the amount of rotation is

\begin{align*} \mathrm{Arg}\,f\,'(x) \amp = \mathrm{Arg}(A)-\frac{1}{\pi}\Big[\alpha_1\mathrm{Arg}(x-x_1) +\alpha_2\mathrm{Arg}(x-x_2)\\ \amp + \cdots +\alpha_{n-1}\mathrm{Arg}(x-x_{n-1})\Big] \end{align*}

Because \(\mathrm{Arg}(x-x_k)=0\) for \(1\le k\lt j\text{,}\) and \(\mathrm{Arg}(x-x_k) =\pi\) for \(j\le k\le n-1\text{,}\) we can write this equation as

\begin{equation*} \mathrm{Arg}\big(f\,'(x)\big) = \mathrm{Arg}(A) - \alpha_j - \alpha_{j+1} - \cdots - \alpha_{n-1} \end{equation*}

The angle of inclination of the tangent vector \(\mathbf{T}_j\) to the polygon \(P\) at the point \(w=f(x)\) for \(x_{j-1} \lt x \lt x_j\) is

\begin{equation*} \gamma_j = \mathrm{Arg}(A) - \alpha_j - \alpha_{j+1} - \cdots - \alpha_{n-1} \end{equation*}

The angle of inclination of the tangent vector \(\mathbf{T}_{j+1}\) to the polygon \(P\) at the point \(w=f(x)\text{,}\) for \(x_j \lt x \lt x_{j+1}\text{,}\) is

\begin{equation*} \gamma_{j+1} = \mathrm{Arg}(A) - \alpha_{j+1} - \alpha_{j+2} - \cdots - \alpha_{n-1} \end{equation*}

The angle of inclination of the vector tangent to the polygon \(P\) jumps abruptly by the amount \(\alpha_j\) as the point \(w=f(x)\) moves along the side \(\widehat{w_{j-1}w_j}\) through the vertex \(w_j\) to the side \(\widehat{w_jw_{j+1}}\text{.}\) Therefore the exterior angle to the polygon \(P\) at the vertex \(w_j\) is given by the angle \(\alpha_j\) and satisfies the inequality \(-\pi\lt \alpha_j\lt \pi\text{,}\) for \(j=1,\,2,\ldots,n-1\text{.}\) Since the sum of the exterior angles of a polygon equals \(2\pi\text{,}\) we have \(\alpha_n = 2\pi - \alpha_1 - \alpha_2 - \cdots - \alpha_{n-1}\) and only \(n-1\) angles need to be specified. The case \(n=5\) is illustrated in Figure 10.9.2.

Figure 10.9.2. A Schwarz-Christoffel mapping with \(n=5\) and \(\alpha_1+\alpha_2+\cdots +\alpha_4>\pi\)

If the case \(\alpha_1 + \alpha_2 + \cdots + \alpha_{n-1} \le \pi\) occurs, then \(\alpha_n>\pi\text{,}\) and the vertices \(w_1,\,w_2,\ldots,w_n\) cannot form a closed polygon. For this case, Equations (10.9.2) and (10.9.3) will determine a mapping from the upper half plane \(\mathrm{Im}(z)>0\) onto an infinite region in the \(w\) plane, where the vertex \(w_n\) is at infinity. The case \(n=5\) is illustrated in Figure 10.9.3.

Figure 10.9.3. A Schwarz-Christoffel mapping with \(n=5\) and \(\alpha_1 + \alpha_2 + \cdots + \alpha_4 \le \pi\)

Equation (10.9.3) gives a representation for \(f\) in terms of an indefinite integral. Note that these integrals do not represent elementary functions unless the image is an infinite region. Also, the integral will involve a multivalued function, and we must select a specific branch to fit the boundary values specified in the problem. Table 10.11.23, which appears at the end of this chapter, is useful for our purposes.

Example 10.9.4.

Use the Schwarz-Christoffel formula to verify that the function \(w=f(z)=\mathrm{Arcsin}z\) maps the upper half-plane \(\mathrm{Im}(z)>0\) onto the semi-infinite strip \(-\frac{\pi}{2}\lt u\lt \frac{\pi}{2}, \; v>0\) shown in Figure 10.9.5.

Solution.

If we choose \(x_1=-1, \, x_2=1, \, w_1 = -\frac{\pi}{2}\text{,}\) and \(w_2=\frac{\pi}{2}\text{,}\) then \(\alpha_1=\frac{\pi}{2}\) and \(\alpha_2=\frac{\pi}{2}\text{,}\) and Equation (10.9.2) for \(f\,'(z)\) becomes

\begin{equation*} f\,'(z) = A(z+1)^{-(\pi /2)/\pi}(z-1)^{-(\pi /2) /\pi}=\frac{A}{(z^2-1)^\frac{1}{2}} \end{equation*}

Then, using Table 10.11.23, the indefinite integral becomes

\begin{equation*} f(z) = i(A)\mathrm{Arcsin}(z) + B\text{.} \end{equation*}

Using the image values \(f(-1) = -\frac{\pi}{2}\) and \(f(1) = \frac{\pi}{2}\text{,}\) we obtain the system

\begin{equation*} -\frac{\pi}{2} = A\left(\!-\frac{\pi}{2}\right)+B \text{ and } \frac{\pi}{2} = A\left(i\frac{\pi}{2}\right) + B\text{.} \end{equation*}

Solving gives \(B=0\) and \(A=-i\text{.}\) Hence, the required function is

\begin{equation*} f(z) = \mathrm{Arcsin}(z) \end{equation*}

Example 10.9.6.

Verify that \(w = f(z) = (z^2-1)^\frac{1}{2}\) maps the upper half-plane \(\mathrm{Im}(z)>0\) onto the upper half-plane \(\mathrm{Im}(w)>0\) slit along the segment from \(0\) to \(i\text{.}\) (Use the principal square root throughout.)

Solution.

If we choose \(x_1=-1 \,x_3=1 \,w_2=i\text{,}\) and \(w_3=d\text{,}\) then the formula

\begin{equation*} g\,'(z) =A(z+1)^{-\frac{\alpha_1}{\pi}}\left(z^{-\frac{\alpha_2}{\pi}}\right)(z-1)^{-\frac{\alpha_3}{\pi}} \end{equation*}

will determine a mapping \(w=g(z)\) from the upper half-plane \(\mathrm{Im}(z)>0\) onto the portion of the upper half-plane \(\mathrm{Im}(w)>0\) that lies outside the triangle with vertices \(\pm d,\,i\) as indicated in Figure 10.9.7(a). If we let \(d \to 0\text{,}\) then \(w_1 \to 0, \; w_3 \to 0, \; \alpha_1 \to \frac{\pi}{2}, \;\alpha_2 \to -\pi\text{,}\) and \(\alpha_3 \to \frac{\pi}{2}\text{.}\) The limiting formula for the derivative \(g\,'(z)\) becomes

\begin{equation*} f\,'(z) = A(z+1)^{-\frac{1}{2}}(z)(z-1)^{-\frac{1}{2}}\text{,} \end{equation*}

which will determine a mapping \(w=f(z)\) from the upper half-plane \(\mathrm{Im}(z)>0\) onto the upper half-plane \(\mathrm{Im}(w)>0\) slit from \(0\) to \(i\) as indicated in Figure 10.9.7(b).

An easy computation reveals that \(f(z)\) is given by

\begin{equation*} f(z) = A\int \frac{z}{(z^2-1)^\frac{1}{2}}\,dz = A(z^2-1)^\frac{1}{2} + B\text{,} \end{equation*}

and the boundary values \(f(\pm 1)=0\) and \(f(0) = i\) lead to the solution

\begin{equation*} f(z) = (z^2-1)^\frac{1}{2} \end{equation*}

Example 10.9.8.

Show that the function

\begin{equation*} w = f(z) =\frac{1}{\pi}\mathrm{Arcsin}(z) + \frac{i}{\pi}\mathrm{Arcsin}\left(\frac{1}{z}+\frac{1+i}{2}\right) \end{equation*}

maps the upper half-plane \(\mathrm{Im}(z)>0\) onto the right angle channel in the first quadrant, which is bounded by the coordinate axes and the rays \(x \ge 1, \, y=1\) and \(y \ge 1, \, x=1\text{,}\) as depicted in Figure 10.9.9(b).

Solution.

If we choose \(x_1=-1 \,x_3=1 \,w_2=d\text{,}\) and \(w_3=1+i\text{,}\) then the formula

\begin{equation*} g\,'(z) = A_1(z+1)^{-\frac{\alpha_1}{\pi}}\left(z^{-\frac{\alpha_2}{\pi}}\right)(z-1)^{-\frac{\alpha_3}{\pi}} \end{equation*}

will determine a mapping \(w=g(z)\) of the upper half-plane onto the domain that is indicated in Figure 10.9.9(a). With \(\alpha_1 = \frac{\pi}{2}\text{,}\) we let \(d \to \infty\text{,}\) then \(\alpha_2 \to \pi\) and \(\alpha_3 \to -\frac{\pi}{2}\text{,}\) and the limiting formula for the derivative \(g\,'(z)\) becomes

\begin{align*} f\,'(z) \amp = A_1(z+1)^{-(\pi/2)/\pi}(z)^{-(\pi ) /\pi}(z-1)^{-(-\pi /2) /\pi}\\ \amp = A_1\frac{1}{z}\frac{(z-1)^\frac{1}{2}}{(z+1)^\frac{1}{2}}\\ \amp = A_1\frac{z-1}{z(z^2-1)^\frac{1}{2}}\\ \amp = A\frac{z-1}{z(1-z^2)^\frac{1}{2}}\text{,} \end{align*}

where \(A=-iA_1\text{,}\) which will determine a mapping \(w=f(z)\) from the upper half plane onto the channel as indicated in Figure 10.9.9b. Using Table 11.5.10, we obtain

\begin{align*} f(z) \amp = A\left[ \int \frac{1}{(1-z^2)^\frac{1}{2}}\,dz - i\int \frac{1}{z(z^2-1)^\frac{1}{2}}\,dz\right]\\ \amp = A\left[ \mathrm{Arcsin}(z)+i\mathrm{Arcsin}\left(\frac{1}{z}\right)\right] + B\text{.} \end{align*}

If we use the principal branch of the inverse sine function, then the boundary values \(f(-1)=0\) and \(f(1)=1+i\) lead to the system

\begin{equation*} A\left(-\frac{\pi}{2} - i\frac{\pi}{2}\right) + B = 0, \text{ and } A\left(\frac{\pi}{2} + i\frac{\pi}{2}\right) + B = 1 + i\text{,} \end{equation*}

which we can solve to obtain \(A=\frac{1}{\pi}\) and \(B=\frac{1+i}{2}\text{.}\) Hence the required solution is

\begin{equation*} w=f(z) =\frac{1}{\pi}\mathrm{Arcsin}(z) + \frac{i}{\pi}\mathrm{Arcsin}\left(\frac{1}{z}\right) + \frac{1+i}{2}\text{.} \end{equation*}

Exercises Exercises

1.

Let \(a\) and \(K\) be real constants with \(0\lt K\lt 2\text{.}\) Use the Schwarz-Christoffel formula to show that the function \(w=f(z)=(z-a)^K\) maps the upper half-plane \(\mathrm{Im}(z)>0\) onto the sector \(0\lt \arg_0 w\lt K\pi\) shown in Figure 10.9.10.

Solution.

\(f\,'(z)=A(z-a)^{-\frac{\pi -k\pi}{\pi}}=A(z-a)^{k-1}\text{.}\) Integrate and get \(f(z)=\frac{A}{k}(z-a)^{k}\text{,}\) then choose \(A=k\text{.}\)

2.

Let \(a\) be a real constant. Use the Schwarz-Christoffel formula to show that the function \(w=f(z)=\mathrm{Log}(z-a)\) maps the upper half-plane \(\mathrm{Im}(z)>0\) onto the infinite strip \(0\lt v\lt \pi\) shown in Figure 10.9.11.

\hint{Set \(x_1=a-1 \, x_2=a, \, w_1=i\pi\text{,}\) and\(w_2=-d\text{.}\) Then let \(d \to \infty\text{.}\)}

3.

For the remaining exercises, construct the derivative \(f\,'(z)\) and use the Schwarz-Christoffel formula, Equation (10.9.3), and techniques of integration to determine the required conformal mapping \(w=f(z)\text{.}\)

\end{task}

(a)

Show that \(w=f(z)=\frac{1}{\pi}(z^2-1)^\frac{1}{2} + \frac{1}{\pi}\mathrm{Log}\left[ z+(z^2-1)^\frac{1}{2}\right] - i\) maps the upper half-plane onto the domain indicated in Figure 10.9.12.

\hint{Set \(x_1=-1 \,w_1=0\text{,}\) and \(w_2=-i\text{.}\)}

Solution.

\(f\,'(z) = A(z+1)^\frac{1}{2}(z-1)^\frac{1}{2} = A\left[\frac{z}{(z^2-1)^\frac{1}{2}} + \frac{1}{(z^2-1)^\frac{1}{2}}\right]\text{.}\) Integrate with the boundary conditions \(f(-1)=0\) and \(f(1)=-1\) to get \(w=f(z)=\frac{1}{\pi }\left[(z^2-1)^\frac{1}{2} + \mathrm{Log}(z+(z^2-1)^\frac{1}{2})\right]-i\text{.}\)

(b)

Show that \(w=f(z)=\frac{2}{\pi}(z^2-1)^\frac{1}{2}+\frac{2}{\pi}\mathrm{Arcsin}\frac{1}{z}\) maps the upper half-plane onto the domain indicated in Figure 10.9.13.

\hint{Set \(x_1=w_1=-1 ,\, x_2=0, \, x_3=w_3=1\text{,}\) and \(w_2=-id\text{.}\) Then let \(d \to \infty\text{.}\)}

(c)

Show that \(w=f(z)=\frac{1}{2}\mathrm{Log}(z^2-1)=\mathrm{Log}\left[(z^2-1)^\frac{1}{2}\right]\) maps the upper half-plane \(\mathrm{Im}(z) >0\) onto the infinite strip \(0\lt v\lt \pi\) slit along the ray \(u\le 0\)\(,\,v=\frac{\pi}{2}\text{,}\) per Figure 10.9.14.

\hint{Set \(x_1=-1, \, x_2=0, \, x_3=1, \, w_1=i\pi -d, \ ,w_2=\frac{i\pi}{2}\text{,}\) and \(w_3=-d\text{.}\) Then let \(d \to \infty\text{.}\)}

Solution.

\(f\,'(z) = A(z+1)^{-1}z(z-1)^{-1}\text{,}\) and \(w=f(z)=\mathrm{Log}(z^2-1)^\frac{1}{2}\text{.}\)

(d)

Show that \(w=f(z) = -\frac{2}{\pi}z(1-z^2)^\frac{1}{2}-\frac{2}{\pi}\mathrm{Arcsin}(z)\) maps the upper half-plane onto the domain indicated in Figure 10.9.15.

\hint{Set \(x_1=-1, \, x_2=1, \, w_1=1\text{,}\) and \(w_2=-1\text{.}\)}

(e)

Show that \(w=f(z)=z+\mathrm{Log}(z)\) maps the upper half-plane \(\mathrm{Im}(z)>0\) onto the upper half-plane \(\mathrm{Im}(w)>0\) slit along the ray \(u \le -1, \, v=\pi\text{,}\) as shown in Figure 10.9.16.

\hint{Set \(x_1=-1, \, x_2=0, \, w_1=-1+i\pi\text{,}\) and \(w_2=-d\text{.}\) Then Let \(d \to \infty\text{.}\)}

Solution.

\(f\,'(z) = A(z+1)^{1}z^{-1} = A(1+\frac{1}{z})\text{.}\) Integrate to get \(f(z)=z+\mathrm{Log}(z)\text{.}\)

(f)

Show that \(w=f(z)=i\pi+2(z+1)^\frac{1}{2}+\mathrm{Log}\left[\frac{1-(z+1)^\frac{1}{2}}{1+(z+1)^\frac{1}{2}}\right]\) maps the upper half-plane onto the domain indicated in Figure 10.9.17.

\hint{Set \(x_1=-1, \, x_2=0, \, w_1=i\pi\text{,}\) and \(w_2=-d\text{.}\) Then let \(d \to \infty\text{.}\)}

(g)

Show that \(w=f(z)=(z-1)^{\alpha}\left[\frac{1+\alpha z}{1-\alpha}\right]^{1-\alpha}\) maps the upper half-plane \(\mathrm{Im}(z) >0\) onto the upper half-plane \(\mathrm{Im}(w)>0\) slit along the segment from \(0\) to \(e^{i\alpha \pi}\text{,}\) per Figure 10.9.18.

\hint{Show that \(f\,'(z) = A\left[\frac{z+(1-\alpha)}{\alpha} \right]^{-\alpha}(z)(z-1)^{\alpha -1}\text{.}\)}

Solution.

Select \(x_1 = -\frac{1-\alpha}{\alpha }\) and \(x_2=x_3=1\text{,}\) then form \(f\,'(z) = A(z+\frac{1-\alpha}{\alpha })^{-\alpha}(z)(z-1)^{\alpha -1}\text{.}\)

Computation then reveals that \(A=(\frac{1-\alpha }{\alpha})^{\alpha -1}\text{,}\) which is used to construct the desired function \(w=f(z) =\int A(z+\frac{1-\alpha}{\alpha })^{-\alpha}(z)(z-1)^{\alpha -1}\,dz = (z-1)^{\alpha}(1+ \frac{\alpha z}{1-\alpha})^{1-\alpha}\text{.}\)

(h)

Show that \(w = f(z) = 4(z+1)^\frac{1}{4} + \log\left[\frac{(z+1)^\frac{1}{4}-1}{(z+1)^\frac{1}{4}+1}\right] + i\log\left[\frac{i-(z+1)^\frac{1}{4}}{i+(z+1)^\frac{1}{4}}\right]\) maps the upper half-plane onto the domain indicated in Figure 10.9.19.

\hint{Set \(z_1=-1, \, z_2=0, \, w_1=i\pi\text{,}\) and \(w_2=-d\text{.}\) Then let \(d \to \infty\text{.}\) Use the change of variable \(z+1=s^4\) in the resulting integral.}

(i)

Show that \(w=f(z) = -\frac{i}{z}z^\frac{1}{2}(z-3)\) maps the upper half-plane onto the domain indicated in Figure 10.9.20.

\hint{Set \(x_1=0\text{,}\) \(x_2=1\text{,}\) \(\,w_1=-d\text{,}\) and \(w_2=i\) and let \(d\to 0\text{.}\)}

Solution.

\(f\,'(z) = Az^{-\frac{1}{2}}(z-1) = A(z^\frac{1}{2}-z^{-\frac{1}{2}})\text{.}\) Integrate and get \(f(z)=-\frac{i}{2}z^\frac{1}{2}(z-3)\text{.}\)

(j)

Show that \(\displaystyle w = f(z) = \int \frac{1}{(1-z^2)^\frac{3}{4}}\,dz\) maps the upper half-plane \(\mathrm{Im}(z)>0\) onto a right triangle with angles \(\frac{\pi}{2}, \, \frac{\pi}{4}\text{,}\) and \(\frac{\pi}{4}\text{.}\)

(k)

Show that \(\displaystyle w=f(z) =\int\frac{1}{(1-z^2)^\frac{2}{3}}\,dz\) maps the upper half-plane onto an equilateral triangle.

(l)

Show that \(\displaystyle w=f(z)=\int\frac{1}{(z-z^3)^\frac{1}{2}}\,dz\) maps the upper half-plane onto a square.

(m)

Show that \(w=f(z)=2(z+1)^\frac{1}{2}-\mathrm{Log}\left[\frac{1-(z+1)^\frac{1}{2}}{1+(z+1)^\frac{1}{2}}\right]\) maps the upper half-plane \(\mathrm{Im}(z) >0\) onto the domain indicated in Figure 10.9.21.

\hint{Set \(x_1=-1, \, x_2=0, \, x_3=1, \, w_1=0, \,w_2=d\text{,}\) and \(w_3 = 2\sqrt{2} - 2\ln(\sqrt{2}-1) + i\pi\text{.}\) Then let \(d \to \infty\text{.}\)}

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