An application of Rouché’s theorem

Statement of the theorem: Suppose f and g are holomorphic in an open set containing a closed, simple curve \gamma and its interior. If |f(z)| > |g(z)| for all z \in \gamma, then f and f+g have the same number of zeros in the interior of \gamma.

An application: Let f(z) = 2z^5+8z-1. All five zeros of f(z) are inside the disc |z| < 2 and exactly one zero is inside the disc |z| < 1

Proof: Let g(z) = 2z^5 and let h(z) = 8z-1. First, note that g and h are both holomorphic in \mathbb{C}. For |z| = 2, |g(z)| = |2z^5| = |2(2^5)| = 64 > 17 = 8|2|+1 = |8z| + |-1| \ge |8z-1| = |h(z)|. By Rouché’s theorem, the number of zeros of g inside the disc of radius 2 (which is 5, counting multiplicities) is equal to the number of zeros of g+h = f inside the disc. For |z| = 1, |h(z)| = |8z-1| \ge |8z|-1 = 7 > 2 = |2(1^5)| = |2z^5| = |g(z)|. Again, by Rouché’s theorem, the number of zeros of h inside the disc of radius 1 (one) is equal to the number of zeros of g+h = f inside the disc.

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