Chapter 20

since \(f^{\prime}(z)=3\left(z^{2}-1\right), f\) is conformal at all points except \(z=\pm 1.\)

For \(w=\frac{1}{z}, u=\frac{x}{x^{2}+y^{2}}\) and \(v=\frac{-y}{x^{2}+y^{2}} .\) If \(y=x, u=\frac{1}{2} \frac{1}{x}, v=-\frac{1}{2} \frac{1}{x},\) and so \(v=-u .\) The image is the line \(v=-u\) (with the origin (0,0) excluded.)

Using (3) with \(x_{0}=-2, x_{1}=0, x_{2}=1\) and \(u_{1}=5\) and \(u_{2}=1\) \\[ u=\frac{5}{\pi} \operatorname{Arg}\left(\frac{z}{z+2}\right)+\frac{1}{\pi} \operatorname{Arg}\left(\frac{z-1}{z}\right) \\]

(a) For \(T(z)=\frac{z+1}{z-1}, T(0)=-1, T(1)=\infty,\) and \(T(\infty)=1\). (b) The circle \(|z|=1\) passes through the pole at \(z=1\) and so the image is a line. since \(T(-1)=0\) and \(T(i)=-i,\) the image is the line \(u=0 .\) If \(|z-1|=1\), $$|w-1|=\left|\frac{z+1}{z-1}-1\right|=\frac{2}{|z-1|}=2$$ and so the image is the circle \(|w-1|=2\) in the \(w\) -plane. (c) since \(T(0)=-1,\) the image of the disk \(|z| \leq 1\) is the half-plane \(u \leq 0\).

For \(w=z^{2}, u=x^{2}-y^{2}\) and \(v=2 x y .\) If \(x y=1, v=2\) and so the hyperbola \(x y=1\) is mapped onto the line \(v=2.\)

\(f(z)=z+\operatorname{Ln} z+1\) is analytic for all \(z\) except \(z=y \leq 0 .\) But \(f^{\prime}(z)=1+1 / z\) and \(1+1 / z=0\) for \(z=-1\) Therefore \(f(z)\) is conformal at all points except those on the branch cut \(y \leq 0.\)

The mapping \(f(z)=z^{4}\) maps the wedge \(0 \leq \operatorname{Arg} z \leq \pi / 4\) to the upper half-plane \(R^{\prime}\) and \(U=\frac{1}{\pi}\) Arg \(w\) is the solution to the corresponding Dirichlet problem in \(R^{\prime} .\) Therefore, \\[ \phi(x, y)=U\left(z^{4}\right)=\frac{1}{\pi} \operatorname{Arg} z^{4}=\frac{4}{\pi} \operatorname{Arg} z \\] is the potential in the wedge. A complex potential is \(G(z)=\frac{4}{\pi} \mathrm{Ln} z\) and, since \(\phi(r, \theta)=\frac{4}{\pi} \theta,\) the equipotential lines are the rays \(\theta=\frac{\pi}{4} c .\) Finally \\[ \mathbf{F}=\overline{G^{\prime}(z)}=\frac{4}{\pi} \frac{1}{z}=\frac{4}{\pi}\left(\frac{x}{x^{2}+y^{2}}, \frac{y}{x^{2}+y^{2}}\right) \\]

\(S^{-1}(T(z))=\frac{a z+b}{c z+d}\) where $$\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\operatorname{adj}\left(\left[\begin{array}{rr} i & 1 \\ 1 & -1 \end{array}\right]\right)\left[\begin{array}{rr} 1 & 0 \\ i & -1 \end{array}\right]=\left[\begin{array}{rr} -1-i & 1 \\ -2 & -i \end{array}\right].$$ Therefore, $$S^{-1}(T(z))=\frac{(-1-i) z+1}{-2 z-i}=\frac{(1+i) z-1}{2 z+i} \text { and } S^{-1}(w)=\frac{-w-1}{-w+i}=\frac{w+1}{w-i}.$$

\(f(z)=\left(z^{2}-1\right)^{1 / 2}=e^{\frac{1}{2} \operatorname{Ln}\left(z^{2}-1\right)}\) is analytic for all \(z\) outside the interval [-1,1] on the real axis. This follows from the fact that \(z^{2}-1=\left(x^{2}-y^{2}-1\right)+2 x y i\) and so we must exclude values of \(z\) for which \(v=2 x y=0\) and \(u=x^{2}-y^{2}-1 \leq 0 .\) Therefore \(y=0\) and \(x^{2} \leq 1 . f^{\prime}(z)=z /\left(z^{2}-1\right)^{1 / 2}\) is non-zero outside this interval. Therefore \(f\) is conformal except for \(z=x,-1 \leq x \leq 1.\)

For \(w=\operatorname{Ln} z, u=\log _{e}|z|\) and \(v=\operatorname{Arg} z .\) The semi-circle \(|z|=1, y>0\) may also be described by \(r=1\) \(0<\theta<\pi .\) Therefore \(u=0\) and \(0

From Theorem 20.5 \\[ u(x, y)=\frac{y}{\pi} \int_{-\infty}^{\infty} \frac{\cos t}{(x-t)^{2}+y^{2}} d t=\frac{y}{\pi} \int_{-\infty}^{\infty} \frac{\cos (x-s)}{s^{2}+y^{2}} d s \\] letting \(s=x-t .\) But \(\cos (x-s)=\cos x \cos x+\sin x \sin s .\) It follows that \\[ u(x, y)=\frac{y \cos x}{\pi} \int_{-\infty}^{\infty} \frac{\cos s}{s^{2}+y^{2}} d s+\frac{y \sin x}{\pi} \int_{-\infty}^{\infty} \frac{\sin s}{s^{2}+y^{2}} d s=\frac{y \cos x}{\pi}\left(\frac{\pi e^{-y}}{y}\right)=e^{-y} \cos x, \quad y>0 \\]

\(S^{-1}(T(z))=\frac{a z+b}{c z+d}\) where $$\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\operatorname{adj}\left(\left[\begin{array}{ll} 2 & 1 \\ 1 & 1 \end{array}\right]\right)\left[\begin{array}{rr} i & 0 \\ 1 & -2 i \end{array}\right]=\left[\begin{array}{cr} -1+i & 2 i \\ 2-i & -4 i \end{array}\right].$$ Therefore, $$S^{-1}(T(z))=\frac{(-1+i) z+2 i}{(2-i) z-4 i} \quad \text { and } \quad S^{-1}(w)=\frac{w-1}{-w+2}.$$

The function \(f(z)=\pi i-\frac{1}{2}[\operatorname{Ln}(z+1)+\operatorname{Ln}(z-1)]\) is analytic except on the branch cut \(x-1 \leq 0\) or \(x \leq 1,\) and $$f^{\prime}(z)=-\frac{1}{2}\left(\frac{1}{z+1}+\frac{1}{z-1}\right)=-\frac{z}{z^{2}-1}$$ is non-zero for \(z \neq 0, \pm 1 .\) Therefore \(f\) is conformal except for \(z=x, x \leq 1.\)

\(S^{-1}(T(z))=\frac{a z+b}{c z+d}\) where $$\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\operatorname{adj}\left(\left[\begin{array}{ll} 1 & -2 \\ 1 & -1 \end{array}\right]\right)\left[\begin{array}{ll} 2 & -3 \\ 1 & -3 \end{array}\right]=\left[\begin{array}{rr} 0 & -3 \\ -1 & 0 \end{array}\right].$$ Therefore, $$S^{-1}(T(z))=\frac{-3}{-z}=\frac{3}{z} \quad \text { and } \quad S^{-1}(w)=\frac{-w+2}{-w+1}=\frac{w-2}{w-1}.$$

If \(r=2\) and \(0 \leq \theta \leq \frac{\pi}{2}, w=\sqrt{2} e^{i \theta / 2} .\) Therefore \(|w|=\sqrt{2}\) and \(0 \leq \operatorname{Arg} w \leq \pi / 4 .\) This image is a circular arc.

(a) \(\psi(x, y)=\operatorname{Im}\left(z^{4}\right)=4 x y\left(x^{2}-y^{2}\right)\) and so \(\psi(x, y)=0\) when \(y=x\) and \(y=0\) (b) \(\mathbf{V}=\overline{G^{\prime}(z)}=\overline{4 z^{3}}=4\left(x^{3}-3 x y^{2}, y^{3}-3 x^{2} y\right)\) (c) In polar coordinates \(r^{4} \sin 4 \theta=c\) or \(r=(c \csc 4 \theta)^{1 / 4},\) for \(0 < \theta<\pi / 4\), are the streamlines. See the figure.

Since \(\alpha_{1}=3 \pi / 2=\alpha_{2}, f^{\prime}(z)=A(z+1)^{1 / 2}(z-1)^{1 / 2}=A\left(z^{2}-1\right)^{1 / 2} .\) Therefore, $$f(z)=A\left[\frac{z\left(z^{2}-1\right)^{1 / 2}}{2}-\frac{1}{2} \operatorname{Ln}(z+\left(z^{2}-1\right)^{1 / 2}\right]+B.$$ but \(f(-1)=-a i\) and \(f(1)=a i .\) It follows that $$a i=f(1)=B, \quad-a i=f(-1)=A\left(-\frac{\pi i}{2}\right)+B$$ and so \(B=a i\) and \(A=4 a / \pi .\) Therefore, $$f(z)=\frac{4 a}{\pi}\left[\frac{z\left(z^{2}-1\right)^{1 / 2}}{2}-\frac{1}{2} \operatorname{Ln}\left(z+\left(z^{2}-1\right)^{1 / 2}\right)\right]+a i=\frac{4 a}{\pi}\left[\frac{z\left(z^{2}-1\right)^{1 / 2}}{2}-\frac{1}{2} \cosh ^{-1} z\right]+a i.$$

(a) \(\operatorname{since} G\left(r e^{i \theta}\right)=r^{2 / 3} e^{i 2 \theta / 3}, \psi(r, \theta)=\operatorname{Im}\left(G\left(r e^{i \theta}\right)\right)=r^{2 / 3} \sin \frac{2 \theta}{3} .\) Note that \(\psi=0\) on the boundary where \(\theta=0\) and \(\theta=3 \pi / 2\) (b) \(\mathbf{V}=\overline{G^{\prime}(z)}=\overline{\frac{2}{3} z^{-1 / 3}} .\) Therefore, letting \(z=r e^{i \theta}, \mathbf{V}=\frac{2}{3} r^{-1 / 3}(\cos (\theta / 3), \sin (\theta / 3))\) for \(0<\theta<3 \pi / 2\) (c) \(r^{2 / 3} \sin (2 \theta / 3)=c\) implies that \(r=[c \csc (2 \theta / 3)]^{2 / 3}\) for \(0<\theta<3 \pi / 2\). The streamlines are shown in the figure.

Using \(\mathrm{H}-5, f(z)=\left(\frac{1+z}{1-z}\right)^{2}\) maps \(R\) onto the upper half-plane \(R^{\prime} .\) The corre sponding Dirichlet problem in \(R^{\prime}\) is shown in the figure. From (3) \\[ U=\frac{1}{\pi} \operatorname{Arg}\left(\frac{w}{w+1}\right)+\frac{1}{\pi}[\pi-\operatorname{Arg}(w-1)]=1-\frac{1}{\pi} \operatorname{Arg}(w-1)+\frac{1}{w} \operatorname{Arg}\left(\frac{w}{w+1}\right) \\] The harmonic function \(u=U(f(z))\) may be simplified to \\[ u=1-\frac{1}{\pi} \operatorname{Arg}\left(\frac{4 z}{(1-z)^{2}}\right)+\frac{1}{\pi} \operatorname{Arg}\left(\frac{(1+z)^{2}}{2\left(1+z^{2}\right)}\right) \\]

If \(w=z+\frac{1}{z}\) and \(z=e^{i t}, w=e^{i t}+e^{-i t}=2 \cos t .\) Therefore \(u=2 \cos t\) and \(v=0\) and so the image is the closed interval [-2,2] on the \(u\) -axis.

(a) \(\psi(x, y)=\operatorname{Im}(\sin z)=\cos x \sinh y\) and \(\psi(x, y)=0\) when \(x=\pm \pi / 2\) or when \(y=0\) (b) \(\mathbf{V}=\overline{G^{\prime}(z)}=\overline{\cos z}=(\cos x \cosh y, \sin x \sinh y)\) (c) \(\cos x \sinh y=c \Rightarrow y=\sinh ^{-1}(c \sec x)\) and the streamlines are shown in the figure.

\(S(w)=\frac{\left(w-w_{1}\right)\left(w_{2}-w_{3}\right)}{\left(w-w_{3}\right)\left(w_{2}-w_{1}\right)}\) maps \(w_{1}, w_{2}, w_{3}\) to \(0,1, \infty\) and so \(S^{\prime}\) maps \(0,1, \infty\) to \(w_{1}, w_{2}, w_{3} .\) Therefore, \(z=\frac{(w-0)(i-2)}{(w-2)(i-0)}\) and so \(w=\frac{2 z}{z-1-2 i}\) maps \(0,1, \infty\) to \(0, i, 2\).

The first quadrant may be described by \(r>0,0<\theta<\pi / 2 .\) If \(w=1 / z\) and \(z=r e^{i \theta}, w=\frac{1}{r} e^{-i \theta} .\) Therefore \(\operatorname{Arg} w=-\theta\) and \(\operatorname{so}-\pi / 2<\operatorname{Arg} w<0 .\) The image is therefore the fourth quadrant.

Using the cross-ratio formula (7), $$\frac{(1)(-i-1)}{(w-1)(1)}=\frac{(z+1)(-1)}{(z-1)(1)}$$ and so \(w=\frac{(2+i) z-i}{z+1}\) maps -1,0,1 to \(\infty,-i, 1\).

For \(u\left(e^{i \theta}\right)=\cos 2 \theta,\) the Fourier series solution (6) reduces to \\[ u(r, \theta)=r^{2} \cos 2 \theta=\operatorname{Re}\left(z^{2}\right) \quad \text { or } \quad u(x, y)=x^{2}-y^{2} \\] The corresponding system of level curves is shown in the figure.

The mapping \(w=z+4 i\) is a translation which maps the circle \(|z|=1\) to a circle of radius \(r=1\) and with center \(w=4 i .\) This circle may be described by \(|w-4 i|=1.\)

(a) For \(f(z)=\left(z^{2}-1\right)^{1 / 2}\) \\[ f(t)=\left|t^{2}-1\right|^{1 / 2} \cos \left(\frac{1}{2} \operatorname{Arg}\left(t^{2}-1\right)\right)+i\left|t^{2}-1\right|^{1 / 2} \sin \left(\frac{1}{2} \operatorname{Arg}\left(t^{2}-1\right)\right) \\] $$\text { and so } f(t)=\left\\{\begin{array}{ll} \left|t^{2}-1\right|^{1 / 2}, & |t|>1 \\ i\left|t^{2}-1\right|^{1 / 2}, & |t|<1 \end{array} \text { Hence } \operatorname{Im}(G(z))=0 \text { on the boundary of } R\right.$$ $$\text { (b) } x=\left|(t+i c)^{2}-1\right|^{1 / 2} \cos \left(\frac{1}{2} \operatorname{Arg}\left((t+i c)^{2}-1\right)\right), \quad y=\left|(t+i c)^{2}-1\right|^{1 / 2} \sin \left(\frac{1}{2} \operatorname{Arg}\left((t+i c)^{2}-1\right)\right) \text { for } c>0$$ c.

If \(w=2 z-1\) and \(|z|=1,\) then, since \(z=\frac{w+1}{2},\left|\frac{w+1}{2}\right|=1\) or \(|w+1|=2 .\) The image is a circle with center at \(w=-1\) and with radius \(r=2.\)

The mapping \(w=i z\) is a rotation through \(90^{\circ}\) since \(i=e^{i \pi / 2} .\) Therefore the strip \(0 \leq y \leq 1\) is rotated through \(90^{\circ}\) and so the strip \(-1 \leq u \leq 0\) is the image in the \(w\) -plane.

Since \(w=(1+i) z=\sqrt{2} e^{i \pi / 4} z,\) the mapping is the composite of a rotation through \(45^{\circ}\) and a magnification by \(\alpha=\sqrt{2} .\) The image of the first quadrant is therefore the angular wedge \(\pi / 4 \leq \operatorname{Arg} w \leq 3 \pi / 4.\)

\(\left|w-w_{1}\right|=\lambda\left|w-w_{2}\right| \Longrightarrow\left(u-u_{1}\right)^{2}+\left(v-v_{1}\right)^{2}=\lambda^{2}\left[\left(u-u_{2}\right)^{2}+\left(v-v_{2}\right)^{2}\right]\) The latter equation may be put in the form $$A u^{2}+B v^{2}+C u+D v+F=0$$ where \(A=B=1-\lambda^{2}\). If \(\lambda=1\), the line \(C u+D v+F=0\) results. If \(\lambda > 0\) and \(\lambda \neq 1\), then the equation defines a circle.

The mapping \(w=z-i\) lowers the strip \(1 \leq y \leq 4\) one unit so that the image is \(0 \leq v \leq 3\) in the \(w\) -plane.

(a) \(\mathbf{V}=\frac{a+i b}{\bar{z}}=\left(\frac{a x-b y}{x^{2}+y^{2}}, \frac{b x+a y}{x^{2}+y^{2}}\right)\) and since \(\left(x^{\prime}(t), y^{\prime}(t)\right)=\mathbf{V},\) the path of the particle satisfies $$\frac{d x}{d t}=\frac{a x-b y}{x^{2}+y^{2}}, \quad \frac{d y}{d t}=\frac{b x+a y}{x^{2}+y^{2}}$$ (b) Switching to polar coordinates, \\[ \begin{array}{l} \frac{d r}{d t}=\frac{1}{r}\left(x \frac{d x}{d t}+y \frac{d y}{d t}\right)=\frac{1}{r}\left(\frac{a x^{2}-b x y}{r^{2}}+\frac{b x y+a y^{2}}{r^{2}}\right)=\frac{a}{r} \\ \frac{d \theta}{d t}=\frac{1}{r^{2}}\left(-y \frac{d x}{d t}+x \frac{d y}{d t}\right)=\frac{1}{r^{2}}\left(\frac{-a x y+b y^{2}}{r^{2}}+\frac{b x^{2}+a x y}{r^{2}}\right)=\frac{b}{r^{2}} \end{array} \\] Therefore \(\frac{d r}{d \theta}=\frac{a}{b} r\) and so \(r=c e^{a \theta / b}\) (c) \(\frac{d r}{d t}=\frac{a}{r}<0\) if and only if \(a<0,\) and in this case \(r\) is decreasing and the curve spirals inward. \(\frac{d \theta}{d t}=\frac{b}{r^{2}}<0\) if and only if \(b<0,\) and in this case \(\theta\) is decreasing and the curve is traversed clockwise.

We first use \(z_{1}=e^{-i \pi / 4} z\) to rotate the wedge \(\pi / 4 \leq \operatorname{Arg} z \leq \pi / 2\) to the wedge \(0 \leq \operatorname{Arg} z_{1} \leq \pi / 4 .\) The power function \(w=z_{1}^{4}\) then changes the opening of this wedge by a factor of 4 so that the strip \(0 \leq \operatorname{Arg} w \leq \pi\) results. The composite of these two mappings is \(w=\left(e^{-\pi / 4 i} z\right)^{4}=e^{-\pi i} z^{4}=-z^{4}.\)

(a) If \(u=\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{\partial^{2} \phi}{\partial y^{2}},\) $$\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=\frac{\partial^{4} \phi}{\partial x^{4}}+2 \frac{\partial^{4} \phi}{\partial x^{2} \partial y^{2}}+\frac{\partial^{4} \phi}{\partial y^{4}}=0$$ since \(\phi\) is assumed to be biharmonic. (b) If \(g=u+i v,\) then \(\phi=\operatorname{Re}(\bar{z} g(z))=x u+y v.\) $$\begin{aligned}&\frac{\partial^{2} \phi}{\partial x^{2}}=3 \frac{\partial u}{\partial x}+x \frac{\partial^{2} u}{\partial x^{2}}+y \frac{\partial^{2} v}{\partial x^{2}}\\\&\frac{\partial^{2} \phi}{\partial y^{2}}=2 \frac{\partial v}{\partial y}+x \frac{\partial^{2} u}{\partial y^{2}}+y \frac{\partial^{2} v}{\partial y^{2}}\end{aligned}.$$ since \(u\) and \(v\) are harmonic and \(\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},\) $$\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{\partial^{2} \phi}{\partial y^{2}}=2 \frac{\partial u}{\partial x}+2 \frac{\partial v}{\partial y}=4 \frac{\partial u}{\partial x}.$$ Now \(u_{1}=\frac{\partial u}{\partial x}\) is also harmonic and so \(\frac{\partial^{2} u_{1}}{\partial x^{2}}+\frac{\partial^{2} u_{1}}{\partial y^{2}}=0 .\) But $$\frac{\partial^{2} u_{1}}{\partial x^{2}}+\frac{\partial^{2} u_{1}}{\partial y^{2}}=\frac{1}{4}\left[\frac{\partial^{4} \phi}{\partial x^{4}}+2 \frac{\partial^{4} \phi}{\partial x^{2} \partial y^{2}}+\frac{\partial^{4} \phi}{\partial y^{4}}\right]$$ and so \(\phi\) is biharmonic.

The power function \(z_{1}=z^{2 / 3}\) maps the wedge \(0 \leq \operatorname{Arg} z \leq 3 \pi / 2\) to the upper half-plane \(0 \leq \operatorname{Arg} z_{1} \leq \pi .\) We then let \(w=e^{-i \pi / 2} z_{1}+2=-i z_{1}+2\) to rotate the upper half-plane through \(-90^{\circ}\) and then translate 2 units to the right. Therefore the composite function is \(w=-i z^{2 / 3}+2\) and the image region is the half-plane \(u \geq 2.\)

(a) Letting \(z=x+i y\) and noting that in this case \(x^{2}+y^{2}=R^{2},\) the transformation \(w=z+k^{2} / z\) becomes \\[ \begin{aligned} w &=x+i y+\frac{k^{2}}{x+i y}=x+i y+\frac{k^{2}}{x^{2}+y^{2}}(x-i y) \\ &=x+i y+\frac{k^{2}}{R^{2}}(x-i y)=\left(1+\frac{k^{2}}{R^{2}}\right) x+i\left(1-\frac{k^{2}}{R^{2}}\right) y \end{aligned} \\] We identify \(u=\left(1+k^{2} / R^{2}\right) x\) and \(v=\left(1-k^{2} / R^{2}\right) y .\) Then \\[ \frac{u^{2}}{\left(1+\frac{k^{2}}{R^{2}}\right)^{2}}=x^{2} \quad \text { and } \quad \frac{v^{2}}{\left(1-\frac{k^{2}}{R^{2}}\right)^{2}}=y^{2}, \quad k \neq R ,\\] so that \\[ \frac{u^{2}}{\left(1+\frac{k^{2}}{R^{2}}\right)^{2}}+\frac{v^{2}}{\left(1-\frac{k^{2}}{R^{2}}\right)^{2}}=x^{2}+y^{2}=R^{2}, \quad k \neq R \\] (b) When \(R=k\) the circle \(z=k e^{i t}\) is transformed into \(w=z+k^{2} / z=k e^{i t}+k e^{-i t}=2 k \cos t+0 i .\) Thus, the circle \(z=k e^{i t}\) is transformed into the closed interval \([-2 k, 2 k]\) on the \(u\) -axis. (c) Letting \(w=z+k^{2} / z\) we have \\[ \frac{w-2 k}{w+2 k}=\frac{z+\frac{k^{2}}{z}-2 k}{z+\frac{k^{2}}{z}+2 k}=\frac{z^{2}+k^{2}-2 k z}{z^{2}+k^{2}+2 k z}=\frac{(z-k)^{2}}{(z+k)^{2}}=\left(\frac{z-k}{z+k}\right)^{2} .\\]