### 17th Putnam 1957

**Problem A4**

p(z) is a polynomial of degree n with complex coefficients. Its roots (in the complex plane) can be covered by a disk radius r. Show that for any complex k, the roots of n p(z) - k p'(z) can be covered by a disk radius r + |k|.

**Solution**

Let the roots of p(z) be a_{1}, a_{2}, ... , a_{n}. Suppose they all lie in the disk centre c, radius r. Then |c - a_{n}| ≤ r. Suppose that |c - w| > r + |k|. We show that w is not a root of n p(z) - k p'(z). We have |w - a_{i}| ≥ |w - c| - |c - a_{i}| > r + |k| - r = |k|. Now p'(z)/p(z) = ∑ 1/(z - a_{i}) (note that this is still true if we have repeated roots), so |p'(w)/p(w)| < n/|k| and hence |k p'(w)/p(w)| < n. So |n - k p'(w)/p(w)| > 0. But |p(w)| > 0 (since w lies outside the disk containing all the roots of p(z) ), so |n p(w) - k p'(w)| = |p(w)| |n - k p'(w)/p(w)| > 0.

17th Putnam 1957

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002