Argument Principle and Rouché's Theorem

12.1 The Argument Principle

Theorem 12.1 (Argument Principle). If $f$ is meromorphic inside and on a simple closed contour $\gamma$ with no zeros or poles on $\gamma$Then

$\frac{1}{2\pi i}\int_\gamma \frac{f"(z)}{f(z)}\, dz = N - P$

Where $N$ is the number of zeros and $P$ is the number of poles of $f$ inside $\gamma$ (counting Multiplicities).

12.2 Rouché’s Theorem

Theorem 12.2 (Rouché’s Theorem). If $f$ and $g$ are analytic inside and on a simple closed Contour $\gamma$And $|f(z)| \gt |g(z)|$ on $\gamma$Then $f$ and $f + g$ have the same number of Zeros inside $\gamma$.

Proof. On $\gamma$: $|g(z)/f(z)| \lt 1$. The function $h(z) = 1 + g(z)/f(z)$ satisfies $|h(z) - 1| \lt 1$ on $\gamma$So $h(\gamma)$ does not wind around $0$. By the argument principle Applied to $h$: $0 = N_h - P_h$Meaning $h$ has the same number of zeros and poles inside $\gamma$. But $h = (f + g)/f$So zeros of $h$ are zeros of $f + g$ and poles of $h$ are zeros of $f$. Therefore $f$ and $f + g$ have the same number of zeros. $\blacksquare$

12.3 Worked Example

Problem. Show that $z^4 + 6z + 3$ has exactly one root in $|z| \lt 1$.

Solution. On $|z| = 1$: $|6z| = 6 \gt |z^4 + 3| \leq |z|^4 + 3 = 4$. By Rouché’s theorem with $f(z) = 6z$ and $g(z) = z^4 + 3$: $f + g = z^4 + 6z + 3$ has the same number of zeros in $|z| \lt 1$ as $f(z) = 6z$Which has exactly one zero (at $z = 0$). $\blacksquare$