# 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.