Questions: Schwarz's Lemma and Automorphisms of the Disk

Question 1

Let f be holomorphic on the unit disk with |f(z)| < 1 and f(0) = 0. What is the largest possible value of |f(1/2)|?

• A. 1/4
• B. 1/2
• C. 1
• D. 2

Answer: (B)

Schwarz's lemma gives |f(z)| ≤ |z|, so |f(1/2)| ≤ 1/2. The bound is reached by the identity f(z) = z, and in the equality case f must be a rotation.

Question 2

If f maps the unit disk into itself holomorphically with f(0) = 0 and f'(0) = i, then f(z) = iz.

• T. True
• F. False

Answer: True

|f'(0)| = |i| = 1 is the equality case of Schwarz's lemma, so f(z) = e^(iθ)z for some θ. Then f'(0) = e^(iθ) = i, which gives f(z) = iz.

Question 3

Explain why g(z) = f(z)/z, with g(0) = f'(0), is holomorphic on the disk when f(0) = 0, and how the maximum modulus principle gives |g(z)| ≤ 1.

Model answer: Since f(0) = 0, f(z) = f'(0)z + a₂z² + ..., so f(z)/z = f'(0) + a₂z + ... is a power series and holomorphic. On |z| = r < 1, |g(z)| ≤ 1/r, so by maximum modulus |g| ≤ 1/r on |z| ≤ r; letting r → 1 gives |g| ≤ 1.

The singularity of f(z)/z at 0 is removable because f vanishes there. The maximum modulus principle moves the boundary estimate to the interior, and the limit r → 1 removes the factor 1/r.

Question 4

Which map is a holomorphic bijection of the unit disk onto itself that sends a = 1/2 to 0?

• A. z - 1/2
• B. (z - 1/2)/(1 - z/2)
• C. 2z - 1
• D. (z - 1/2)/(1 + z/2)

Answer: (B)

The map φₐ(z) = (z - a)/(1 - āz) with a = 1/2 is a disk automorphism sending a to 0. The map z - 1/2 shifts part of the disk outside it, 2z - 1 doubles distances, and (z - 1/2)/(1 + z/2) sends z = -0.9 to about -2.5, outside the disk. The denominator must use the conjugate with a minus sign, 1 - āz.