2011 AMC 12A Problem 23

Question

Let

f(z) = \dfrac{z + a}{z + b}

and

g(z) = f(f(z)),

where

a

and

b

are complex numbers. Suppose that

|a| = 1

and

g(g(z)) = z

for all

z

for which

g(g(z))

is defined. What is the difference between the largest and smallest possible values of

|b|?

0

\sqrt{2} - 1

\sqrt{3} - 1

1

2

Answer: C

Concepts:complex number matrix roots of unity

Difficulty rating: 2560

Small hint:

Represent

f

by the matrix

\begin{pmatrix} 1 & a \\ 1 & b \end{pmatrix};

then

g(g(z)) = z

means

f^4

is the identity

Big hint:

The order-

4

case forces

(\operatorname{tr})^2 = 2\det,

which gives

b^2 = -(1 + 2a)

Solution:

Direct composition gives

g(z)=\dfrac{Az+B}{Cz+D},

where

A=1+a,

B=a(1+b),

C=1+b,

and

D=a+b^2.

The matrix

\begin{pmatrix}A&B\\C&D\end{pmatrix}

represents

g.

For

g\circ g

to be the identity, its square must be scalar. Comparing the off-diagonal entries and the two diagonal entries gives two possibilities: either

B=C=0

and

A=D,

or

A+D=0.

The first gives

b=-1

(with

a\ne-1

); the second gives

b^2=-(1+2a).

In the second case,

|b|^2=|1+2a|.

As

a

runs around the unit circle,

|1+2a|

ranges from

1

to

3,

so

1\le|b|\le\sqrt3.

Both endpoints occur: take

(a,b)=(-1,1)

and

(a,b)=(1,i\sqrt3).

The separate case

b=-1

also has

|b|=1.

Therefore the requested difference is

\sqrt3-1.

Thus, the correct answer is C.