Section 27

Fractional-Linear (Möbius) Mappings

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Problem 27.1

Let x=Re(z),y=Im(z)x = \mathrm{Re}\left(z\right), y = \mathrm{Im}\left(z\right), and let CC be a positive constant. Find the images of each line of the indicated families under the mapping w=1zw = \frac{1}{z}:

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(1)

the family of circles {(x,y):x2+y2=Cx}\left\{ (x, y): x^{2}+y^{2} = C x\right\};

(2)

the family of lines {(x,y):y=x+C}\left\{ (x, y): y = x+C\right\};

(3)

the family of lines {(x,y):y=Cx}\left\{ (x, y): y = C x\right\}.

Problem 27.2

Let x=Re(z),y=Im(z)x = \mathrm{Re}\left(z\right), \quad y = \mathrm{Im}\left(z\right). Find the images of each domain of the indicated families under the mapping w=1zw = \frac{1}{z}:

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(1)

the family of disks {(x,y):x2+y2<Cx}\left\{ (x, y): x^{2}+y^{2} < C x\right\} (here CC is a positive constant);

(2)

the family of disks {(x,y):x2+y2<Cx}\left\{ (x, y): x^{2}+y^{2} < C x\right\} (here CC is a negative constant);

(3)

the family of disks {(x,y):x2+y2<Cy}\left\{ (x, y): x^{2}+y^{2} < C y\right\} (CC is a positive constant);

(4)

the family of half-planes {(x,y):y>Cx}\left\{ (x, y): y > C x\right\} (CC is a positive constant);

(5)

the family of disks {z:∣z−a∣<R}\left\{ z:\left|z-a\right| < R\right\}, where aa is a fixed point, and the positive constant RR satisfies the condition R<∣a∣R < \left|a\right|;

(6)

the family of disks {z:∣z−a∣<R}\left\{ z:\left|z-a\right| < R\right\}, where aa is a fixed point, and the constant RR satisfies the condition R>∣a∣R > \left|a\right|.

Problem 27.3

Find the image of the disk {z:∣z−1∣<2}\left\{ z:\left|z-1\right| < 2\right\} under the following mappings:

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(1)

w=1−2izw = 1-2 i z;

(2)

w=2izz+3w = \frac{2 i z}{z+3};

(3)

w=z+1z−2w = \frac{z+1}{z-2};

(4)

w=z−12z−6w = \frac{z-1}{2 z-6}.

Problem 27.4

Find the image of the half-plane {z:Re(z)<1}\left\{ z: \mathrm{Re}\left(z\right) < 1\right\} under the following mappings:

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(1)

w=(1+i)z+1w = (1+i) z+1;

(2)

w=zz−1+iw = \frac{z}{z-1+i};

(3)

w=zz−2w = \frac{z}{z-2};

(4)

w=4zz+1w = \frac{4 z}{z+1};

(5)

w=z−3+iz+1+iw = \frac{z-3+i}{z+1+i}.

Problem 27.5

Find the images of the indicated domains DD under the indicated mappings:

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(1)

D={z:∣z∣<1,Im(z)>0},w=1−z1+zD = \left\{ z:\left|z\right| < 1, \mathrm{Im}\left(z\right) > 0\right\} , w = \frac{1-z}{1+z};

(2)

D={z:z∉[−2,1]},w=z+21−zD = \left\{ z: z \notin [-2,1]\right\} , w = \frac{z+2}{1-z};

(3)

D={z:∣z−i∣>1,Im(z)>0},w=1zD = \left\{ z:\left|z-i\right| > 1, \mathrm{Im}\left(z\right) > 0\right\} , w = \frac{1}{z};

(4)

D={z:1<∣z∣<2},w=2z−1D = \left\{ z: 1 < \left|z\right| < 2\right\} , w = \frac{2}{z-1}.

Problem 27.6

Find linear-fractional functions w(z)w(z) satisfying the conditions:

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(1)

w(0)=4,w(1+i)=2+2i,w(2i)=0w(0) = 4, w(1+i) = 2+2 i, w(2 i) = 0;

(2)

w(0)=0,w(1+i)=2+2i,w(2i)=4w(0) = 0, w(1+i) = 2+2 i, w(2 i) = 4;

(3)

w(0)=0,w(1+i)=∞,w(2i)=2iw(0) = 0, w(1+i) = \infty , w(2 i) = 2 i.

Find the image of the disk {z:∣z−i∣<1}\left\{ z:\left|z-i\right| < 1\right\} under the mappings given by these functions.

Problem 27.7

Find linear-fractional functions w(z)w(z) satisfying the conditions:

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(1)

w(i)=2,w(∞)=1+i,w(−i)=0w(i) = 2, w(\infty ) = 1+i, w(-i) = 0;

(2)

w(i)=0,w(∞)=1,w(−i)=∞w(i) = 0, w(\infty ) = 1, w(-i) = \infty;

(3)

w(i)=−2,w(∞)=2i,w(−i)=2w(i) = -2, w(\infty ) = 2 i, w(-i) = 2.

Find the image of the half-plane {z:Re(z)>0}\left\{ z: \mathrm{Re}\left(z\right) > 0\right\} under the mappings given by these functions.

Problem 27.8

Find a function w(z)w(z) that conformally maps the domain DD onto the domain D1D_{1} and satisfies the indicated conditions:

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(1)

D={z:∣z∣<1},D1={w:∣w∣<1},w(z0)=0,arg⁡w′(z0)=α(∣z0∣<1)D = \left\{ z:\left|z\right| < 1\right\} , \quad D_{1} = \left\{ w:\left|w\right| < 1\right\} , \quad w\left(z_{0}\right) = 0, \quad \arg w^{\prime }\left(z_{0}\right) = \alpha \left(\left|z_{0}\right| < 1\right);

(2)

D={z:∣z∣<1},D1={w:∣w∣<1},w(z0)=w0,arg⁡w′(z0)=α(∣z0∣<1,∣w0∣<1)D = \left\{ z:\left|z\right| < 1\right\} , D_{1} = \left\{ w:\left|w\right| < 1\right\} , w\left(z_{0}\right) = w_{0}, \arg w^{\prime }\left(z_{0}\right) = \alpha \left(\left|z_{0}\right| < 1,\left|w_{0}\right| < 1\right);

(3)

D={z:∣Im(z)∣>0},D1={w:∣w∣<1},w(z0)=0,arg⁡w′(z0)=α(Im(z0)>0)D = \left\{ z:\left|\mathrm{Im}\left(z\right)\right| > 0\right\} , D_{1} = \left\{ w:\left|w\right| < 1\right\} , w\left(z_{0}\right) = 0, \arg w^{\prime }\left(z_{0}\right) = \alpha \left(\mathrm{Im}\left(z_{0}\right) > 0\right);

(4)

D={z:∣Im(z)∣>0},D1={w:∣w∣<1},w(z0)=w0,arg⁡w′(z0)=α(Im(z0)>0,∣w0∣<1)D = \left\{ z:\left|\mathrm{Im}\left(z\right)\right| > 0\right\} , D_{1} = \left\{ w:\left|w\right| < 1\right\} , w\left(z_{0}\right) = w_{0}, \arg w^{\prime }\left(z_{0}\right) = \alpha \left(\mathrm{Im}\left(z_{0}\right) > 0,\left|w_{0}\right| < 1\right);

(5)

D={z:∣Im(z)∣>0},D1={w:Im(w)>0},w(z0)=w0D = \left\{ z:\left|\mathrm{Im}\left(z\right)\right| > 0\right\} , D_{1} = \left\{ w: \mathrm{Im}\left(w\right) > 0\right\} , w\left(z_{0}\right) = w_{0}, arg⁡w′(z0)=α(Im(z0)>0,Im(w0)>0);\arg w^{\prime }\left(z_{0}\right) = \alpha \left(\mathrm{Im}\left(z_{0}\right) > 0, \mathrm{Im}\left(w_{0}\right) > 0\right) ;

(6)

D={z:∣z∣<1},D1={w:∣w∣<1},w(1)=1,w(i)=3i−45D = \left\{ z:\left|z\right| < 1\right\} , D_{1} = \left\{ w:\left|w\right| < 1\right\} , w(1) = 1, w(i) = \frac{3 i-4}{5}, w(−1)=−1w(-1) = -1;

(7)

D={z:∣z∣<1},D1={w:∣w∣<1},w(i)=i,w(i2)=4i5D = \left\{ z:\left|z\right| < 1\right\} , D_{1} = \left\{ w:\left|w\right| < 1\right\} , w(i) = i, w\left(\frac{i}{2}\right) = \frac{4 i}{5};

(8)

D={z:∣z∣<1},D1={w:∣w∣<1},w(12)=12D = \left\{ z:\left|z\right| < 1\right\} , D_{1} = \left\{ w:\left|w\right| < 1\right\} , w\left(\frac{1}{2}\right) = \frac{1}{2}, arg⁡w′(12)=π2\arg w^{\prime }\left(\frac{1}{2}\right) = \frac{\pi }{2};

(9)

D={z:Im(z)>0},D1={w:∣w∣<1},w(0)=i,w(−1)=1D = \left\{ z: \mathrm{Im}\left(z\right) > 0\right\} , D_{1} = \left\{ w:\left|w\right| < 1\right\} , w(0) = i, w(-1) = 1, w(∞)=−1w(\infty ) = -1;

(10)

D={z:Im(z)>0},D1={w:∣w∣<1},w(0)=−i,w(2i)=i3D = \left\{ z: \mathrm{Im}\left(z\right) > 0\right\} , D_{1} = \left\{ w:\left|w\right| < 1\right\} , w(0) = -i, w(2 i) = \frac{i}{3};

(11)

D={z:Im(z)>0},D1={w:∣w∣<1},w(1+i)=0D = \left\{ z: \mathrm{Im}\left(z\right) > 0\right\} , \quad D_{1} = \left\{ w:\left|w\right| < 1\right\} , \quad w(1+i) = 0, arg⁡w′(1+i)=π\arg w^{\prime }(1+i) = \pi;

(12)

D={z:Im(z)>0},D1={w:Im(w)>0},w(−1)=0,w(0)=2D = \left\{ z: \mathrm{Im}\left(z\right) > 0\right\} , D_{1} = \left\{ w: \mathrm{Im}\left(w\right) > 0\right\} , w(-1) = 0, w(0) = 2, w(1)=∞w(1) = \infty;

(13)

D={z:Im(z)>0},D1={w:Im(w)>0},w(−1)=−2D = \left\{ z: \mathrm{Im}\left(z\right) > 0\right\} , D_{1} = \left\{ w: \mathrm{Im}\left(w\right) > 0\right\} , w(-1) = -2, w(−2+i)=1+3i;w(-2+i) = 1+3 i ;

(14)

D={z:Im(z)>0},D1={w:Im(w)>0},w(1+i)=iD = \left\{ z: \mathrm{Im}\left(z\right) > 0\right\} , \quad D_{1} = \left\{ w: \mathrm{Im}\left(w\right) > 0\right\} , \quad w(1+i) = i, arg⁡w′(1+i)=π2\arg w^{\prime }(1+i) = \frac{\pi }{2};

(15)

D={z:∣z−1−i∣<2},D1={w:∣w∣<1},w(i)=0,arg⁡w′(i)=π2D = \left\{ z:\left|z-1-i\right| < 2\right\} , D_{1} = \left\{ w:\left|w\right| < 1\right\} , w(i) = 0, \arg w^{\prime }(i) = \frac{\pi }{2};

(16)

D={z:Re(z)>−1},D1={w:∣w∣<1},w(0)=0,arg⁡w′(0)=πD = \left\{ z: \mathrm{Re}\left(z\right) > -1\right\} , D_{1} = \left\{ w:\left|w\right| < 1\right\} , w(0) = 0, \arg w^{\prime }(0) = \pi.

Problem 27.9

Find the general form of the conformal mapping of the following domains onto the annulus {w:1<∣w∣<R}\left\{ w: 1 < \left|w\right| < R\right\}:

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(1)

{z:∣z−3∣>9,∣z−8∣<16}\left\{ z:\left|z-3\right| > 9,\left|z-8\right| < 16\right\};

(2)

{z:∣z−5∣>4, Re z>0}\left\{ z:\left|z-5\right| > 4\text{, Re }z > 0\right\}.

Problem 27.10

Let w(z)w(z) be an arbitrary linear-fractional function, and let z1,z2,z3,z4z_{1}, z_{2}, z_{3}, z_{4} be four pairwise distinct points of the extended complex plane. Denote

wk=w(zk),k=1,2,3,4. w_{k} = w\left(z_{k}\right), \quad k = 1,2,3,4.

Prove that

w4−w1w4−w2:w3−w1w3−w2=z4−z1z4−z2:z3−z1z3−z2. \frac{w_{4}-w_{1}}{w_{4}-w_{2}}: \frac{w_{3}-w_{1}}{w_{3}-w_{2}} = \frac{z_{4}-z_{1}}{z_{4}-z_{2}}: \frac{z_{3}-z_{1}}{z_{3}-z_{2}}.
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Problem 27.11

Prove that the level lines of the modulus of a linear-fractional function are circles or straight lines.

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Problem 27.12

Prove that the level lines of the real part of a linear-fractional function are circles or straight lines.

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Problem 27.13

Find the condition that the points z1,z1∗,z2,z2∗z_{1}, z_{1}^{*}, z_{2}, z_{2}^{*} must satisfy for there to exist a circle (or line) with respect to which the points zkz_{k} are symmetric to the points zk∗(k=1,2)z_{k}^{*}(k = 1,2).

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Problem 27.14

Let the function w(z)w(z) be meromorphic in the domain DD. Prove the following statements:

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(1)

if under the mapping w=w(z)w = w(z) the image of any line segment (lying in the domain DD) is a line segment, then w(z)w(z) is a linear function;

(2)

if under the mapping w=w(z)w = w(z) the image of any line segment (lying in the domain DD) is an arc of a circle or a line segment, then w=w(z)w = w(z) is a linear-fractional function.

Remark.
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A point a∈C‾a \in \overline{\mathbb {C}} is called a fixed point of the linear-fractional transformation ff if f(a)=af(a) = a. This term is used in problems 27.15-27.17 below.

Problem 27.15

Prove the statements:

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(1)

every linear-fractional transformation has at least one fixed point (finite or infinite);

(2)

every linear-fractional transformation other than the identity has at most two fixed points (finite and infinite).

Problem 27.16

Prove that a linear-fractional transformation ff with a unique fixed point a∈C‾a \in \overline{\mathbb {C}} satisfies the equation

1f(z)−a=1z−a+A,A∈C \frac{1}{f(z)-a} = \frac{1}{z-a}+A, \quad A \in \mathbb {C}

for a≠∞a \neq \infty, while for a=∞a = \infty it has the form

f(z)=z+A,A∈C. f(z) = z+A, \quad A \in \mathbb {C}.
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Problem 27.17

Prove that a linear-fractional transformation ff with two distinct finite fixed points aa and bb satisfies the equation

f(z)−af(z)−b=A⋅z−az−b,A∈C \frac{f(z)-a}{f(z)-b} = A \cdot \frac{z-a}{z-b}, \quad A \in \mathbb {C}
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