Theorem Proof

Theorem. Let n ≥ 1 be an integer and z1 , z2 , . . . , zn be arbitrary complex numbers. Prove that there is a complex number z0 with |z0 | = 1 such that n Y |zk − z0 | = |z1 − z0 | |z2 − z0 | · · · |zn − z0 | ≥ 1 k=1 Proof. My proof will use Cauchy’s Inequalities from Chapter 2. Theorem A. Let f be analytic on a region A and let γ be a circle with radius R and center z0 that lies in A. Assume that the disk {z such that |z −z0 | < R} also lies in A. Suppose that |f (z)| ≤ M for all z on γ. Then, for any k = 0, 1, 2, . . . , |f (k) (z0 )| ≤ k! M. Rk Now, consider the polynomial P (z) such that P (z) = (z1 − z)(z2 − z) · · · (zn − z). By contradiction, assume that |P (z0 )| = n Y |zk − z0 | < 1 k=1 for all z0 such that |z0 | = 1. In other words, |P (z0 )| ≤ 1 −  for all |z0 | = 1 and some 0 <  ≤ 1. Let us explain why such an  exists. It’s easy to see that |P (z)| attains a maximum on the closed unit disk since we have a continuous function over a compact set. But since P (z) is entire, we further conclude by the Maximum Modulus Principle, that the maximum actually occurs on the unit circle (when restricting |P (z)| to the closed unit disk). This guarantees us the existence of such an  with the properties mentioned above. So if we let γ be the unit circle then by Cauchy’s Inequalities, we get that n! |P (n) (0)| ≤ n (1 − ) = n! − n! < n!. 1 But it is easy to see that P (z) has degree n where the leading the coefficient is 1 or -1. Hence, |P (n) (0)| = n!. Obviously, this contradicts our upper bound for |P (n) (0)| given by Cauchy’s Inequalities. As a result, we conclude that there must exist a z0 on the unit circle such that n Y |zk − z0 | = |z1 − z0 | |z2 − z0 | · · · |zn − z0 | ≥ 1. k=1 1