All chapters Unit 1 · Complex Analysis

Möbius transformations

Chapter 6 of 15
The hook

One fraction that bends a straight line into a circle

Story

Chapter 5 left you with a promise and a frustration. The promise: every analytic function is a conformal map, bending the plane while keeping every angle. The frustration: almost none of them can be written down in a way you can actually compute with. w=z2w = z^2 and w=1/zw = 1/z were about as far as we got before the pictures stopped being drawable.

Here is the exception — a whole family of conformal maps you can write completely, in four numbers, and still it does something that should not be possible for so simple a formula. Take the single fraction

w=1z,w = \frac{1}{z},

feed it the perfectly straight vertical line Re⁡(z)=1\operatorname{Re}(z) = 1, and what comes back is not a line at all. It is a circle — the circle of radius 12\tfrac12 centred at 12\tfrac12, passing through the origin. A straight line, bent into a closed loop, by a one-line formula.

0 Re(z) = 1 z-plane w = 1/z 0 |w − ½| = ½ w-plane
The line never touched the origin; its image is a circle that runs right through it. Straightness is gone — angles, as always, are not.

The culprit is that innocent-looking 1/z1/z. It is the one ingredient that can turn a line into a circle, and the whole of this chapter is, in a sense, the study of what it does and how to control it.

One object, three names

This family has three interchangeable names, and your notes may use any of them. A Möbius transformation, a bilinear transformation, and a linear-fractional transformation are the same thing: the map

w=az+bcz+d,ad−bc≠0.w = \frac{az + b}{cz + d}, \qquad ad - bc \neq 0.

The syllabus and this chapter say Möbius; your class notes and exam paper most likely say bilinear. They are not two topics to learn — they are one topic with two labels. Whenever you read "bilinear transformation" on a question, read "Möbius," and everything here applies unchanged.

Why this matters

Möbius maps are the single most useful family of conformal maps in all of applied complex analysis, for one reason: they are the maps you can actually pin down. Give me where three points should go, and there is exactly one Möbius map that obeys — you will compute it yourself before the chapter is out. That control is why they are the workhorse for moving a hard boundary-value problem (heat, electrostatics, fluid flow) from an awkward region onto a disc or a half-plane, where it can be solved, and then carried back.

And for the exam, this is among the most reliably tested topics in Unit 1: find the fixed points, decompose into elementary maps, find the bilinear map through three points, find an inverse, find the image of a curve. Five shapes, all mechanical once you have seen each one worked. This chapter works each one before it asks you for it.

Intuition before formula

Every Möbius map is just three simple moves, composed

Before the general fraction, meet its three building blocks. Every Möbius transformation, no matter how tangled az+bcz+d\dfrac{az+b}{cz+d} looks, is one of these three moves or a chain of them. Learn what each does to the plane and you have already understood the whole family.

1 · Translation — w=z+bw = z + b

The plainest move there is. Adding the fixed complex number bb slides every point by the same vector. Shapes keep their size, their orientation, everything — the whole plane just shifts. No rotation, no stretch. If b=1+ib = 1 + i, the origin moves to 1+i1 + i and so does everyone else.

2 · Rotation–magnification — w=azw = a z

This is the move you already understood in Chapter 1, where you learned that multiplying by a complex number a=r eiθa = r\,e^{i\theta} does exactly two things at once: it scales by r=∣a∣r = |a| and rotates by θ=arg⁡a\theta = \arg a. So w=azw = az spins the plane about the origin by arg⁡a\arg a and zooms it by ∣a∣|a|. With a=1+i=2 eiπ/4a = 1 + i = \sqrt{2}\,e^{i\pi/4}, the plane turns 45∘45^\circ and grows by a factor of 2\sqrt{2}. Angles and shapes both survive; only orientation and scale change.

3 · Inversion — w=1zw = \dfrac{1}{z}

The strange one — and the only one that does anything surprising. Inversion turns the plane inside out about the unit circle: points close to the origin are flung far away, points far away are pulled in close, and the unit circle stays put. This is the move from the hook, the single ingredient that can take a straight line and bend it into a circle. The other two moves can never do that; inversion is where all the magic of the chapter lives.

The punchline

Here is the structural fact that makes the whole family tractable: every Möbius transformation is a composition of these three moves — some translations, a rotation–magnification, and (when c≠0c \neq 0) exactly one inversion. Beat 3 proves it with an explicit recipe. The consequence for you: understanding 1/z1/z is most of the battle, because it is the only piece that bends anything. Translations and rotation–magnifications are just bookkeeping around that one interesting step.

Formal statement & worked models

The definition — and every method, worked once

Now the precise statements. Each computational fact is followed immediately by a fully worked example — read these carefully, because every quiz and practice problem later is a variation on one of them.

The definition, and why ad−bc≠0ad - bc \neq 0

Definition · Möbius transformation

For complex constants a,b,c,da, b, c, d with ad−bc≠0ad - bc \neq 0, the map

w=az+bcz+dw = \frac{az + b}{cz + d}

is a Möbius (bilinear) transformation. The quantity ad−bcad - bc is its determinant.

The condition ad−bc≠0ad - bc \neq 0 is not decoration. Suppose it failed — ad=bcad = bc. Then the numerator az+baz + b and denominator cz+dcz + d are proportional: one is a constant multiple of the other. Their ratio is therefore a constant, the same value a/ca/c (when c≠0c \neq 0) for every zz. A map that sends the entire plane to a single point is no transformation at all — it cannot be undone, so there is nothing left to study. The determinant being non-zero is exactly what keeps the map a genuine, invertible transformation.

Conformal wherever it is defined

Differentiate with the quotient rule:

w′=a(cz+d)−c(az+b)(cz+d)2=ad−bc(cz+d)2.w' = \frac{a(cz+d) - c(az+b)}{(cz+d)^2} = \frac{ad - bc}{(cz + d)^2}.

The numerator is the determinant, which we insisted is non-zero. So w′≠0w' \neq 0 everywhere the map is defined (that is, everywhere except the pole z=−d/cz = -d/c (when c≠0c \neq 0)). By Chapter 5, a non-zero derivative means the map is conformal there. Möbius maps have no critical points: they preserve every angle, everywhere they act. That is the Chapter 5 thread paid off — and the reason these maps are so prized.

Decomposition into the three moves

Theorem · decomposition

For c≠0c \neq 0, polynomial long division rewrites the fraction as

w=az+bcz+d=ac+bc−adc (cz+d).w = \frac{az + b}{cz + d} = \frac{a}{c} + \frac{bc - ad}{c\,(cz + d)}.

Read right to left, this is a chain of elementary moves: form cz+dcz + d (a rotation–magnification then a translation), invert it, multiply by the constant bc−adc\dfrac{bc - ad}{c} (another rotation–magnification), and finally translate by ac\dfrac{a}{c}. Exactly the three moves of Beat 2 — with the one inversion in the middle.

Worked example · decompose w=2z−12zw = \dfrac{2z - 1}{2z}

Split the fraction directly:

w=2z−12z=2z2z−12z=1−12z.w = \frac{2z - 1}{2z} = \frac{2z}{2z} - \frac{1}{2z} = 1 - \frac{1}{2z}.

Read the right-hand side as a chain, working from the inside out:

  1. Magnify by 22: z↦2zz \mapsto 2z. (Here is the rotation–magnification move firing — c=2c = 2, not 11.)
  2. Invert: 2z↦12z2z \mapsto \dfrac{1}{2z}.
  3. Multiply by −1-1: 12z↦−12z\dfrac{1}{2z} \mapsto -\dfrac{1}{2z}. Multiplying by −1-1 is a rotation by π\pi (a half-turn).
  4. Translate by +1+1: −12z↦1−12z-\dfrac{1}{2z} \mapsto 1 - \dfrac{1}{2z}.

Four moves; the only interesting one is the inversion in the middle.

Fixed points

A fixed point is a zz the map leaves where it is: w=zw = z. Setting az+bcz+d=z\dfrac{az+b}{cz+d} = z and clearing the denominator gives a quadratic:

cz2+(d−a)z−b=0.cz^2 + (d - a)z - b = 0.

A quadratic has at most two roots, so a Möbius map has at most two fixed points (unless it is the identity, which fixes everything).

Worked example · fixed points of w=z+3z−1w = \dfrac{z + 3}{z - 1}

Set w=zw = z and clear the denominator:

z=z+3z−1  ⟹  z(z−1)=z+3  ⟹  z2−2z−3=0.z = \frac{z + 3}{z - 1} \;\Longrightarrow\; z(z - 1) = z + 3 \;\Longrightarrow\; z^2 - 2z - 3 = 0.

Factor: (z−3)(z+1)=0(z - 3)(z + 1) = 0. The fixed points are

z=3andz=−1.z = 3 \quad\text{and}\quad z = -1.

Always the same three steps: set w=zw = z, clear the denominator, solve the quadratic.

The cross-ratio and three-points-to-three-points

This is the tool that lets you build a Möbius map to order. The cross-ratio identity says: the map sending z1,z2,z3z_1, z_2, z_3 to w1,w2,w3w_1, w_2, w_3 is the one satisfying

(w−w1)(w2−w3)(w−w3)(w2−w1)=(z−z1)(z2−z3)(z−z3)(z2−z1).\frac{(w - w_1)(w_2 - w_3)}{(w - w_3)(w_2 - w_1)} = \frac{(z - z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)}.

Write it in exactly this order, the same way every time, so the worked numbers always line up. Plug in your six points, then solve for ww in terms of zz.

The ∞\infty rule (both sides)

If one of the six points is ∞\infty, delete the two factors that contain it — they cancel to 11. Concretely:

  • Source z3=∞z_3 = \infty ⇒\Rightarrow the right side becomes z−z1z2−z1\dfrac{z - z_1}{z_2 - z_1}.
  • Target w3=∞w_3 = \infty ⇒\Rightarrow the left side becomes w−w1w2−w1\dfrac{w - w_1}{w_2 - w_1}.

That is the whole rule — strike the two factors carrying the ∞\infty point and continue as normal.

Worked example · the map sending 0,1,2→2,3,60, 1, 2 \to 2, 3, 6

Here z1,z2,z3=0,1,2z_1, z_2, z_3 = 0, 1, 2 and w1,w2,w3=2,3,6w_1, w_2, w_3 = 2, 3, 6; all six are finite, so nothing cancels. The right side:

(z−0)(1−2)(z−2)(1−0)=−zz−2=z2−z.\frac{(z - 0)(1 - 2)}{(z - 2)(1 - 0)} = \frac{-z}{z - 2} = \frac{z}{2 - z}.

The left side:

(w−2)(3−6)(w−6)(3−2)=−3(w−2)w−6=3(w−2)6−w.\frac{(w - 2)(3 - 6)}{(w - 6)(3 - 2)} = \frac{-3(w - 2)}{w - 6} = \frac{3(w - 2)}{6 - w}.

Set them equal and solve for ww. Cross-multiplying 3(w−2)6−w=z2−z\dfrac{3(w-2)}{6-w} = \dfrac{z}{2-z} and collecting the ww terms gives 2w(3−z)=122w(3 - z) = 12, so

w=63−z.w = \frac{6}{3 - z}.

Check. z=0↦63=2z = 0 \mapsto \tfrac63 = 2; z=1↦62=3z = 1 \mapsto \tfrac62 = 3; z=2↦61=6z = 2 \mapsto \tfrac61 = 6. All three land where they should. ✓

The inverse is itself Möbius

Solving w=az+bcz+dw = \dfrac{az + b}{cz + d} for zz gives the inverse map

z=dw−b−cw+a,z = \frac{dw - b}{-cw + a},

which is again a Möbius transformation — same determinant ad−bcad - bc. Notice the pattern: swap a↔da \leftrightarrow d, negate bb and cc. (It is exactly the matrix-inverse pattern, if you have met 2×22 \times 2 matrices.)

Worked example · inverse of w=z−1z+1w = \dfrac{z - 1}{z + 1}

Here a=1,b=−1,c=1,d=1a = 1, b = -1, c = 1, d = 1. Apply the formula:

z=dw−b−cw+a=(1)w−(−1)−(1)w+1=w+11−w.z = \frac{dw - b}{-cw + a} = \frac{(1)w - (-1)}{-(1)w + 1} = \frac{w + 1}{1 - w}.

So the inverse of w=z−1z+1w = \dfrac{z - 1}{z + 1} is z=w+11−wz = \dfrac{w + 1}{1 - w}. (You will meet z−1z+1\dfrac{z-1}{z+1} again in the quizzes — it is a friendly map to keep nearby.)

Interactive playground

Try it: watch curves cross from one plane to the other

Pick a map, pick a test curve, and see its image. Then drag the black probe point around the left (zz) plane and watch its image move in the right (ww) plane. This is the visual model for everything in Beat 6's "find the image of a curve" problems — see the line become a circle here, before you compute one with algebra.

The Möbius mapper
 · 
z-plane  — drag the probe
w-plane  — the image

What to notice

Three combinations are worth setting up deliberately, because they are exactly the results you will derive or use later:

  • 1/z1/z on the line x=1x = 1 — the straight line becomes the circle ∣w−12∣=12|w - \tfrac12| = \tfrac12 through the origin. This is the hook, Quiz 3, and Reveal 2, all the same picture.
  • Cayley z−iz+i\dfrac{z - i}{z + i} on the real axis — the real axis becomes the unit circle ∣w∣=1|w| = 1. This is why Cayley carries the upper half-plane onto the unit disc (problem P5).
  • 1/z1/z on the unit circle — it stays the unit circle. Inversion fixes ∣z∣=1|z| = 1.

And drag the probe toward the map's pole z=−d/cz = -d/c: the image runs off to infinity, and the readout warns you. Every Möbius map has exactly one point it sends to ∞\infty — its pole.

Retrieval gauntlet

Test yourself before moving on

Each of these follows a method you have now seen worked. Commit to an answer before you read the feedback — recognising the right option once you see it feels like knowing it, and it is not the same thing.

Quiz 1 · why the determinant must be non-zero

The definition always insists ad−bc≠0ad - bc \neq 0. What goes wrong if ad−bc=0ad - bc = 0?

Quiz 2 · fixed points of a familiar map

A fixed point satisfies w=zw = z. What are the fixed points of w=z−1z+1w = \dfrac{z - 1}{z + 1}?

Quiz 3 · image of a line under inversion

Under w=1zw = \dfrac{1}{z}, what is the image of the vertical line Re⁡(z)=1\operatorname{Re}(z) = 1?

Find the Möbius map sending z=1,i,−1z = 1, i, -1 to w=0,1,∞w = 0, 1, \infty.

Predict the next step before you reveal each one: which side of the cross-ratio does the ∞\infty target simplify, and to what?

Step 1 — use the ∞\infty rule. The target w3=∞w_3 = \infty. Which two factors cancel, and what does the left side become?

Target w3=∞w_3 = \infty means the two factors carrying w3w_3 cancel, so the left side collapses to

w−w1w2−w1=w−01−0=w.\frac{w - w_1}{w_2 - w_1} = \frac{w - 0}{1 - 0} = w.

Convenient: the left side is just ww itself.

Step 2 — write the right side. Substitute z1,z2,z3=1,i,−1z_1, z_2, z_3 = 1, i, -1 into (z−z1)(z2−z3)(z−z3)(z2−z1)\dfrac{(z - z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)}.
(z−1) (i−(−1))(z−(−1)) (i−1)=(z−1)(i+1)(z+1)(i−1).\frac{(z - 1)\,(i - (-1))}{(z - (-1))\,(i - 1)} = \frac{(z - 1)(i + 1)}{(z + 1)(i - 1)}.

The only awkward piece is the constant i+1i−1\dfrac{i + 1}{i - 1} — simplify it next. (This is the spot a sign error usually creeps in.)

Step 3 — simplify and read off ww. Reduce i+1i−1\dfrac{i + 1}{i - 1}, then equate to the left side.

Multiply top and bottom by the conjugate i−1‾=−1−i\overline{i - 1} = -1 - i (or just compute): i+1i−1=−i\dfrac{i + 1}{i - 1} = -i. So the right side is −i z−1z+1-i\,\dfrac{z - 1}{z + 1}, and since the left side is ww,

w=−i z−1z+1=i 1−zz+1.w = -i\,\frac{z - 1}{z + 1} = i\,\frac{1 - z}{z + 1}.

Check. z=1↦0z = 1 \mapsto 0; z=i↦1z = i \mapsto 1; z=−1z = -1 makes the denominator zero, so z=−1↦∞z = -1 \mapsto \infty. All three correct. ✓

Prove it: why does w=1zw = \dfrac{1}{z} turn the line Re⁡(z)=1\operatorname{Re}(z) = 1 into a circle? This is the hook, made rigorous — and the one place this chapter touches CO-2.

Predict the next step before you reveal each one: after you write Re⁡(z)\operatorname{Re}(z) in terms of u,vu, v, what equation does the line Re⁡(z)=1\operatorname{Re}(z) = 1 impose?

Step 1 — set up. Write w=u+ivw = u + iv. Since w=1/zw = 1/z, express Re⁡(z)\operatorname{Re}(z) in terms of uu and vv.

Invert the map: z=1wz = \dfrac{1}{w}. For any complex ww,

z=1w=wˉ∣w∣2=u−ivu2+v2,soRe⁡(z)=uu2+v2.z = \frac{1}{w} = \frac{\bar w}{|w|^2} = \frac{u - iv}{u^2 + v^2}, \qquad\text{so}\qquad \operatorname{Re}(z) = \frac{u}{u^2 + v^2}.
Step 2 — impose the line. The condition on the line is Re⁡(z)=1\operatorname{Re}(z) = 1. Substitute.
uu2+v2=1  ⟹  u=u2+v2.\frac{u}{u^2 + v^2} = 1 \;\Longrightarrow\; u = u^2 + v^2.

One clean equation in uu and vv — the equation of the image curve.

Step 3 — recognise the curve. Rearrange u=u2+v2u = u^2 + v^2 and complete the square in uu.
u2−u+v2=0  ⟹  (u−12)2+v2=14.u^2 - u + v^2 = 0 \;\Longrightarrow\; \left(u - \tfrac12\right)^2 + v^2 = \tfrac14.

A circle: centre (12,0)\left(\tfrac12, 0\right), radius 12\tfrac12, passing through the origin. Exactly the circle in the hook, and the right answer to Quiz 3.

Step 4 — why this always happens. What is special about lines and circles together that makes inversion send one family into the other?

Lines and circles share a single equation:

A(x2+y2)+Bx+Cy+D=0.A(x^2 + y^2) + Bx + Cy + D = 0.

It is a line when A=0A = 0 and a circle when A≠0A \neq 0. (For A≠0A \neq 0 a rare choice of the other constants can collapse it to a single point or to nothing — neither happens for the curves here.) Substituting z=1/wz = 1/w into this equation swaps the roles of the constants AA and DD. So inversion sends this combined family into itself, and a line not through the origin (A=0, D≠0A = 0,\ D \neq 0) becomes a circle through the origin (A≠0, D=0A \neq 0,\ D = 0) — which is exactly what just happened to Re⁡(z)=1\operatorname{Re}(z) = 1.

Takeaway. Möbius maps preserve the combined "lines-and-circles" family. Translations and rotation–magnifications keep lines as lines and circles as circles; only inversion can move a curve between the two sub-types. That single fact is the whole "circles and lines go to circles and lines" headline.

Exam translator & practice

How this appears on exams — and a worked practice set

Exam shapes

Möbius (bilinear) questions come in five recognisable shapes. Each is mechanical once you have seen it worked, and you now have:

  • "Find the fixed points" — set w=zw = z, solve the quadratic cz2+(d−a)z−b=0cz^2 + (d - a)z - b = 0.
  • "Express as a composition / decompose into elementary maps" — long-divide to ac+bc−adc(cz+d)\dfrac{a}{c} + \dfrac{bc - ad}{c(cz + d)} and read off the chain.
  • "Find the bilinear transformation mapping three points" — the cross-ratio identity, with the ∞\infty rule when a point is at infinity.
  • "Find the inverse" — apply z=dw−b−cw+az = \dfrac{dw - b}{-cw + a}.
  • "Find the image of [a curve] under [a map]" — substitute z=(inverse in w)z = (\text{inverse in } w) and simplify, or test the modulus directly.

Work the set below. The two hardest types each open with a faded bridge — a half-finished solution with the key steps blank — before the cold problems. Fill the blanks yourself, then reveal. Every solution is complete; nothing points outside this page.

Fixed points

Shape · find the fixed points

P1. Find the fixed points of w=3z−2zw = \dfrac{3z - 2}{z}.

Set w=zw = z and clear the denominator:

z⋅z=3z−2  ⟹  z2−3z+2=0  ⟹  (z−1)(z−2)=0.z \cdot z = 3z - 2 \;\Longrightarrow\; z^2 - 3z + 2 = 0 \;\Longrightarrow\; (z - 1)(z - 2) = 0.

Fixed points: z=1z = 1 and z=2z = 2.

Shape · find the fixed points (complex pair)

P6. Find the fixed points of w=z−2z+1w = \dfrac{z - 2}{z + 1}.

Set w=zw = z and clear the denominator:

z(z+1)=z−2  ⟹  z2+2=0  ⟹  z2=−2.z(z + 1) = z - 2 \;\Longrightarrow\; z^2 + 2 = 0 \;\Longrightarrow\; z^2 = -2.

Fixed points: z=±2 iz = \pm\sqrt{2}\,i. A perfectly good answer — fixed points are allowed to be complex.

The quadratic can do three things

The fixed-point equation cz2+(d−a)z−b=0cz^2 + (d - a)z - b = 0 is an ordinary quadratic, so it can come out three ways: two distinct roots (P1), a complex-conjugate pair (P6), or a single repeated root. The repeated case is the one worth seeing once — for example w=3z−4z−1w = \dfrac{3z - 4}{z - 1}:

z(z−1)=3z−4  ⟹  z2−4z+4=0  ⟹  (z−2)2=0,z(z - 1) = 3z - 4 \;\Longrightarrow\; z^2 - 4z + 4 = 0 \;\Longrightarrow\; (z - 2)^2 = 0,

so there is the one double fixed point z=2z = 2. A map with a single fixed point like this is called parabolic. That "one fixed point vs two" split is the only structural classification you need here.

Decomposition into elementary maps

Faded bridge · complete the chain

Decompose w=2z+12zw = \dfrac{2z + 1}{2z}. Fill the blanks, then reveal:

Using w=ac+bc−adc(cz+d)w = \dfrac{a}{c} + \dfrac{bc - ad}{c(cz + d)} with a=2, b=1, c=2, d=0a = 2,\ b = 1,\ c = 2,\ d = 0: the leading term ac=\dfrac{a}{c} = ___; the multiplier after inversion bc−adc=\dfrac{bc - ad}{c} = ___. Then write the chain (form cz+dcz + d → invert → multiply by the outer factor → translate by ac\dfrac{a}{c}).

ac=22=1\dfrac{a}{c} = \dfrac{2}{2} = 1;   outer multiplier bc−adc=(1)(2)−(2)(0)2=1\dfrac{bc - ad}{c} = \dfrac{(1)(2) - (2)(0)}{2} = 1. So

w=2z+12z=1+12z.w = \frac{2z + 1}{2z} = 1 + \frac{1}{2z}.

Chain: magnify by 22 (z↦2zz \mapsto 2z) → invert → (multiply by 11, i.e. do nothing) → translate by +1+1.

Shape · express as a composition

P2. Express w=z+12zw = \dfrac{z + 1}{2z} as a composition of elementary maps.

Split the fraction:

z+12z=z2z+12z=12+12z.\frac{z + 1}{2z} = \frac{z}{2z} + \frac{1}{2z} = \frac12 + \frac{1}{2z}.

Chain: magnify by 22 (z↦2zz \mapsto 2z) → invert → translate by +12+\tfrac12. (Because c=2≠1c = 2 \neq 1, the magnification step genuinely fires; in a map with c=1c = 1 it would be the identity, which is why some decompositions look like only two moves.)

Bilinear map through three points

Faded bridge · the source-∞\infty rule

Find the map sending z=0,1,∞z = 0, 1, \infty to w=1,2,3w = 1, 2, 3. Fill the blanks, then reveal:

The source z3=∞z_3 = \infty, so the right side becomes z−z1z2−z1=\dfrac{z - z_1}{z_2 - z_1} = ___. The left side (all targets finite) is (w−1)(2−3)(w−3)(2−1)=\dfrac{(w - 1)(2 - 3)}{(w - 3)(2 - 1)} = ___. Equate and solve for ww.

Right side: z−01−0=z\dfrac{z - 0}{1 - 0} = z. Left side: (w−1)(−1)(w−3)=−w−1w−3\dfrac{(w - 1)(-1)}{(w - 3)} = -\dfrac{w - 1}{w - 3}. Setting −w−1w−3=z-\dfrac{w - 1}{w - 3} = z and solving,

w=3z+1z+1.w = \frac{3z + 1}{z + 1}.

Check. z=0↦1z = 0 \mapsto 1; z=1↦42=2z = 1 \mapsto \tfrac42 = 2; z→∞↦3z \to \infty \mapsto 3. ✓

Shape · bilinear map + its inverse

P3. (a) Find the bilinear transformation mapping z=0,1,2z = 0, 1, 2 to w=0,1,3w = 0, 1, 3. (b) Find its inverse.

(a) All six points finite. Right side: (z−0)(1−2)(z−2)(1−0)=−zz−2\dfrac{(z - 0)(1 - 2)}{(z - 2)(1 - 0)} = \dfrac{-z}{z - 2}. Left side: (w−0)(1−3)(w−3)(1−0)=−2ww−3\dfrac{(w - 0)(1 - 3)}{(w - 3)(1 - 0)} = \dfrac{-2w}{w - 3}. Equate −2ww−3=−zz−2\dfrac{-2w}{w - 3} = \dfrac{-z}{z - 2}, cross-multiply and solve:

w=3z4−z.w = \frac{3z}{4 - z}.

Check. 0↦00 \mapsto 0; 1↦34−1=11 \mapsto \dfrac{3}{4 - 1} = 1; 2↦64−2=32 \mapsto \dfrac{6}{4 - 2} = 3. ✓

(b) Write w=3z4−z=3z+0−z+4w = \dfrac{3z}{4 - z} = \dfrac{3z + 0}{-z + 4}, so a=3, b=0, c=−1, d=4a = 3,\ b = 0,\ c = -1,\ d = 4. The inverse formula gives

z=dw−b−cw+a=4w−0w+3=4ww+3.z = \frac{dw - b}{-cw + a} = \frac{4w - 0}{w + 3} = \frac{4w}{w + 3}.

Answers: w=3z4−zw = \dfrac{3z}{4 - z}; inverse z=4ww+3z = \dfrac{4w}{w + 3}.

Shape · bilinear map (target-∞\infty)

P4. Find the bilinear transformation mapping z=−1,0,1z = -1, 0, 1 to w=0,i,∞w = 0, i, \infty.

Target w3=∞w_3 = \infty, so the left side reduces to w−w1w2−w1=w−0i−0=wi\dfrac{w - w_1}{w_2 - w_1} = \dfrac{w - 0}{i - 0} = \dfrac{w}{i}. Right side (all sources finite):

(z−(−1))(0−1)(z−1)(0−(−1))=−(z+1)z−1.\frac{(z - (-1))(0 - 1)}{(z - 1)(0 - (-1))} = \frac{-(z + 1)}{z - 1}.

Equate wi=−z+1z−1\dfrac{w}{i} = -\dfrac{z + 1}{z - 1} and multiply by ii:

w=−i z+1z−1.w = -i\,\frac{z + 1}{z - 1}.

Check. z=−1↦0z = -1 \mapsto 0; z=0↦−i⋅1−1=iz = 0 \mapsto -i\cdot\tfrac{1}{-1} = i; z=1↦∞z = 1 \mapsto \infty. ✓

Image of a curve

Shape · find the image of a curve

P5. Find the image of the real axis under the Cayley transformation w=z−iz+iw = \dfrac{z - i}{z + i}.

A point on the real axis is z=tz = t with tt real. Take the modulus of ww:

∣w∣=∣t−i∣∣t+i∣=t2+1t2+1=1for every real t.|w| = \frac{|t - i|}{|t + i|} = \frac{\sqrt{t^2 + 1}}{\sqrt{t^2 + 1}} = 1 \qquad\text{for every real } t.

Image: the unit circle ∣w∣=1|w| = 1. Every point of the real axis lands on it, which is precisely why the Cayley transformation carries the upper half-plane onto the unit disc — the standard bridge between those two regions.

Six things to carry out of this chapter

  1. A Möbius (bilinear, linear-fractional) map is w=az+bcz+dw = \dfrac{az + b}{cz + d} with ad−bc≠0ad - bc \neq 0. The non-zero determinant is what stops it collapsing to a constant.
  2. It is conformal everywhere it is defined. Because w′=ad−bc(cz+d)2w' = \dfrac{ad - bc}{(cz + d)^2} is never zero — Möbius maps have no critical points.
  3. Every Möbius map is three moves composed: translation, rotation–magnification, and (when c≠0c \neq 0) one inversion 1/z1/z — read off from ac+bc−adc(cz+d)\dfrac{a}{c} + \dfrac{bc - ad}{c(cz + d)}.
  4. Fixed points solve cz2+(d−a)z−b=0cz^2 + (d - a)z - b = 0 — set w=zw = z, clear, solve. At most two: distinct, a complex pair, or one repeated (parabolic).
  5. The cross-ratio builds the map through three points. Use it in one fixed order; if a point is ∞\infty, strike the two factors carrying it. The inverse is z=dw−b−cw+az = \dfrac{dw - b}{-cw + a} — again Möbius.
  6. Möbius maps send lines-and-circles to lines-and-circles. Inversion is the only move that can turn a line into a circle (or back); the line Re⁡(z)=1\operatorname{Re}(z) = 1 under 1/z1/z becomes ∣w−12∣=12\left|w - \tfrac12\right| = \tfrac12.
Before you move on

Come back to the cold problems tomorrow. You worked P1–P6 with the methods fresh; the real test is whether they are still there after a night's sleep. Re-doing two or three of them from a blank page tomorrow — not re-reading them, re-doing them — is worth more than another hour now, and it is exactly the kind of spaced practice that pays off at revision time in week 13.

What comes next. Chapter 7 opens complex integration — integrating along contours, and the remarkable fact that for analytic functions the path often stops mattering. The conformal-and-Möbius geometry you have built across Chapters 5 and 6 is the last of the "what do these maps do to shapes" thread; from here the subject turns to integrals.

Same material, another voice

If a different explanation would help, this one is worth your time — free, from MIT OpenCourseWare: