complex analysis, gamelin, ii.7 problems and solutions

II.7 Fractional Linear Transformations 1

Fractional Linear Transformations

1. Compute explicitly the fractional linear transformations determined by the following correspondences of triples:(a) (1+i , 2, 0)⟼(0,∞ , i−1) (e) (1,2,∞)⟼(0,1,∞)(b) (0,1,∞)⟼(1,1+i , 2) (f) (0,∞ ,i )⟼(0,1,∞)(c) (∞ ,1+i , 2)⟼(0,1,∞) (g) (0,1,∞)⟼(0,∞ ,i)(d) (−2, i , 2)⟼(1−2i , 0,1+2 i ) (h) (1, i ,−1)⟼(1,0,−1) .

(a) Let M 1 map(1+i , 2,0)⟼(1,0,∞). Then, M 1 z= z−2z ( 1+i

1+i−2)=−i z+2 iz

↔(−i i1 0). Let

M 2 map(0,∞ , i−1)⟼(1,0,∞). Then, M 2 z= zi−1

↔(1 00 i−1); M 2

−1↔(i−1 00 1). Then, M 2

−1 M 1

gives the desired map, M z=(1+i)⋅z−1z

.

(b) Following the same process as in (a), M z=−2 z+(1+i)−z+(1+i) gives the desired map.

(c) Following the first part of (a), M z=−1+iz−2 gives the desired map.

(d) We see that the second set of triples is obtained by multiplying the first set by i and translating by 1. Thus, M z=iz+1 gives the desired map.(e) The second set of triples is obtained by translating down by 1. Thus, M z= z−1 gives the desired map.

(f) Since ∞⟼1, we need the limit to be 1 as well as M(0) = 0 and M (i)=∞ . Thus, M z= zz−i

gives the desired map.

(g) Similar to (f), M z= izz−1 gives the desired map.

(h) Following (a), M z= (1+i) z−2i(1−i)z+2 (1+i) gives the desired map.

2. Consider the fractional linear transformation in Exercise 1a above, which maps 1 + i to 0, 2 to ∞ , and 0 to i – 1. Without referring to an explicit formula, determine the image of the circle{∣z−1∣=1} , the image of the disk {∣z−1∣<1} , and the image of the real axis.

The map T z=z+2 maps the circle {∣z−1∣=1} to the circle containing the points i + 1, 2, and 0. Thus, if M is the fractional linear transformation in 1a, M T maps the unit circle to the line through 0 and i – 1, call it ℓ . Thus, M T {∣z−1∣=1}=ℓ , so, M T T −1{∣z−1∣=1}=T −1 ℓ⇔

M {∣z−1∣=1}=T −1 ℓ . Thus, the image of {∣z−1∣=1} under M is the line ℓ shifted left by 2 units. In other words, the image is the line through -2 and i – 3.

The point 2∈{∣z−1∣=1}, so we know M(2) is to the right of M {∣z−1∣=1}. Thus, M {∣z−1∣<1} is the half-plane to the right of the line through -2 and i – 3.


We know the image of the points -2, 0, and 2, so we know the image of the real axis. M (−2)=∞ ,M (0)=i−1, and M (2)=∞ . This doesn't give enough information to determine the image, but we

note that (i+1)e−i π /4∈ℝ , so M [(i+1)e−i π /4] will give us another image point. Thus, sinceM [e−i π /4 e i π /4(i+1)]=0, we know that (composing both sides by the inverse of e−i π /4 z ) thatM [e−i π /4(1+i)]=e i π /4⋅0=0 . Thus, the image of the real axis is the line through 0 and i – 1.

3. Consider the fractional linear transformation that maps 1 to i, 0 to 1 + i, and -1 to 1. Determine the image of the unit circle {∣z∣=1}, the image of the open unit disk {∣z∣<1}, and the image of the imaginary axis. Illustrate with a sketch.

We already know the image of two points on the unit circle, so we just need a third to entirely determine its image. Define T z=z−i. Then, M T (i)=M (0)=1+i . Thus, as we have seen before,

M (i)=T −1(1+i)=1+i+i=1+2 i. Since 1, -1, and 1 + 2i are not collinear, they determine a circle, call it C. To find the equation of C, we need the point of intersection of the perpendicular bisectors of two line segments of the image points from the unit circle considered above. The perpendicular bisector of the line segment from -1 to 1 is clearly the line x = 0. Consider the line segment from 1 to 1 + 2i. The midpoint is 1 + i. Thus, the perpendicular bisector is y = i. Thus, the perpendicular bisectors intersect at z = i. Thus, the image of the unit circle is the circle centered at i of radius √2 . In other words, M {∣z∣=1}={∣z−i∣=√2}.

Since M (0)=1+i and ∣1+i−i∣=1<√2 , we see that 0 maps inside C. Thus, the interior of the unit circle maps to the interior of C.

Now, we have the image points of 3 points on the real line, so we know the image of the real line. Going through a similar process as in the first paragraph, we find that M ℝ={∣z−1/2 i∣<1/2}. Thus,

M Im z=i M ℝ . In other words, the image of ℝ rotated 90 degrees.

4. Consider the fractional linear transformation that maps -1 to -i, 1 to 2i, and i to 0. Determine the image of the unit circle {∣z∣=1}, the image of the open unit disk {∣z∣<1}, and the image of the interval [-1, +1] on the real axis. Illustrate with a sketch.

Since -1, 1, and i are all on the unit circle and -i, 2i, and 0 are all on the imaginary axis, we know that the unit circle gets mapped to the imaginary axis.

Now, let T z=z+1. Then, M T (0)=2 i. Thus, M (0)=−1+2 i . Thus, the inside of the unit circle gets mapped to the left half-plane.

The image of the real axis is the circle passing through the points -i, 2i, and -1 + 2i. By a similar calculation to Q4, the equation of the circle is {∣z−1/2(−1+i)∣=√10 /2}. The portion of the circle we are interested is between z = -i and z = 2i. (Basically, the portion of the circle lying on the left half-plane.


5. What is the image of the horizontal line through i under the fractional linear transformation that interchanges 0 and 1 and maps -1 to 1 + i? Illustrate with a sketch.

Let T 1 z= z−i . Then, M T 1(i)=1. Thus, M (i)=1+i . Also, M T 1(i+1)=M (1)=0. Thus,M (1+i)=i. Finally, M T 1(−1+i)=M (−1)=1+i. Thus, M (−1+i)=1+2 i . Thus, the line maps to

the circle {∣z−(1/ 2+3/ 2i)∣=1}.

6. Show that the image of a straight line under the inversion z⟼1 / z is a straight line or circle, depending on whether the line passes through the origin.

Consider a line passing through the origin with angle θ . Then, the parametric equation for the line isf (t)=t e i θ . Thus, the inversion is 1/ t⋅e−i θ . Clearly, the line has simply been reflected about the real

axis.

Suppose that a line has intercepts at a and ib. We know that a⟼1/a and i b⟼−i /b . Furthermore, if we write the equation of the line as f (t)=t+i [(−b /a) t+b] , we see that ∞⟼0. Thus, we have three non-collinear points, so by the theorem which states that circles in the extended complex plane map to circles in the extended complex plane, the inversion maps the line to a circle. (To be exact, it

maps to the circle {∣z−(1/2 a+i /2 b)∣=∣ 1ab∣√a2+b2

2 } which touches the origin.

7. Show that the fractional linear transformation f (z )=(a z+b)/ (cz+d ) is the identity mapping z if and only if b = c = 0 and a=d≠0 .

(⇐) Trivial.

(⇒) Leta z+bc z+d

= z . Then, if z = 0, we have b /d=0 implying that b = 0. If z=∞, we have that

a /c=∞ . This can only happen if c = 0. Finally, if z = 1, then we have a / d=1 which implies thata=d≠0.

8. Show that any fractional linear transformation can be represented in the formf (z )=(a z+b)/ (c z+d ) , where ad – bc = 1. Is this representation unique?

We are examining the determinant of the matrix associated with any fractional linear transformation. So, without loss of generality, it suffices to look at the possible Jordan canonical forms of 2-by-2

matrices. The possibilities are (λ1 *0 λ2) where * = 1 if the two lambdas are the same and * = 0

otherwise. Then, by the fact that scalar multiples of fractional linear transformations are equivalent, the

last matrix is equivalent (in the sense of fractional linear transformations) to (λ1/ λ2 *0 1). Multiplying

through by √ λ2/ λ1 gives a matrix with determinant 1.


9. Show that the fractional linear transformation that are real on the real axis are precisely those that can be expressed in the form (a z+b)/(c z+d ) where a, b, c, and d are real.

Suppose for some x0∈ℝ ,a x0+bc x0+d

= y∈ℝ . Then, in order for this to happen, a, b, c, and d have to be

real or else a, b, c, d are pure imaginary (because the imaginary unit must be common to the numerator and denominator in order for cancellation to occur). But if they are all pure imaginary, then the i can be

factored out to obtain the equivalent fractional linear transformationA x0+BC x0+d where A, B, C, and D are

real.

10. Suppose the fractional linear transformation (a z+b)/(c z+d ) maps ℝ to ℝ , and ad – bc = 1. Show that a, b, c, and d are all real or they are all pure imaginary.

This follows from Q9, regardless of the value of the determinant.

11. Show the following. (a) If f is conjugate to g, then g is conjugate to f. (b) If f 1 is conjugate tof 2 and f 2 is conjugate to f 3 , then f 1 is conjugate to f 3 . (c) If f is conjugate to g, then f ∘ f is

conjugate to g ∘g , and more generally, the m-fold composition f ∘ f ∘⋯∘ f (m times) is conjugate tog ∘g ∘⋯∘g (m times). (d) If f and g are conjugate, then the conjugating function h maps the fixed

points of f to the fixed points of g. In particular, f and g have the same number of fixed points.

(a) f is conjugate to g ⇔g=h∘ f ∘h−1⇔h−1∘g ∘h= f .(b) f 1 is conjugate to f 2 and f 2 is conjugate to f 3⇒ f 1=h∘ f 2∘h−1 and f 2=k f 3 k−1 . Thus,

f 1=h∘(k ∘ f 3∘k−1)∘h−1 .(c) If f =h∘ g∘h−1 , then f ∘ f =h∘g ∘h−1∘h∘ g∘h−1=h∘g ∘g ∘h−1 . Thus, f ∘ f is conjugate to

g ∘g . This can be continued inductively.(d) f (z 0)=h−1∘g ∘h (z0)=z 0⇒ g∘h(z0)=h(z0)⇒ g (h(z 0))=h (z0)⇒h( z0) is a fixed point of g.

12. Classify the conjugacy classes of fractional linear transformations by establishing the following:(a) A fractional linear transformation that is not the identity has either 1 or 2 fixed points, that is, points satisfying f (z 0)= z0 .(b) If a fractional linear transformation f (z ) has two fixed points, then it is conjugate to the dilation

z⟼a z with a≠0 , a≠1 , that is, there is a fractional linear transformation h(z ) such thath( f (z ))=a h( z) . Is a unique?

(c) If a fractional linear transformation f (z ) has exactly one fixed point, then it is conjugate to the translation ζ ⟼ ζ +1. In other words, there is a fractional linear transformation h(z ) such that

h( f (h−1(ζ )))=ζ +1 , or equivalently, such that h( f (z ))=h( z )+1.

(a) Any continuous function has at least one fixed point on an interval, so since fractional linear transformations are continuous, we have at least one fixed point.


Suppose a fractional linear transformation M has three fixed points, say z1 , z2 , and z 3 . Then, the circle in ℂ containing z1 , z2 , and z 3 must be mapped to itself by a combination of the preservation of circles property and the preservation of orientation property. Then, any diameter of the circle, when extended to a line (circle in ℂ ) must be mapped to itself since it passes through two fixed points (i.e., the points of the circle). But every point in ℂ falls on one of these lines. Thus, M = id.

(b) Let M be any fractional linear transformation. Then, the Jordan canonical form of M is either

(λ1 00 λ2) or (λ 1

0 λ). In the first case, there is a T such that T −1 M T =λ1/ λ2 z . In this case, the

constant a is almost uniquely determined (the constant could be negative). Thus, M ~ a z . As such, since the conjugating function maps fixed points to fixed points, M must have two fixed points since a z does (0 and ∞ are the fixed points).

(c) Now suppose M has the second Jordan canonical form discussed above. Then, clearly M ~ ζ +1.For the same reasons as in (b), M has one fixed point and ζ is “almost” uniquely determined.

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