Argument Principle and Rouché's Theorem

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).

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. $\blacksquare$

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 $z = 0$). $\blacksquare$

Application 1: Roots of $z^n + a$. The polynomial $z^n + a$ with $|a| \lt 1$ has all $n$ roots inside $|z| = 1$. On $|z| = 1$, $|z^n| = 1 \gt |a|$, so by Rouché with $f(z) = z^n$ and $g(z) = a$, $z^n + a$ has $n$ zeros in $|z| \lt 1$.

Application 2: Location of zeros. Show that all roots of $z^4 + z + 1 = 0$ satisfy $|z| \lt 2$.

On $|z| = 2$: $|z^4| = 16 \gt |z + 1| \leq 3$. By Rouché with $f(z) = z^4$ and $g(z) = z + 1$: $z^4 + z + 1$ has $4$ zeros in $|z| \lt 2$ (same as $z^4$).

Application 3: Two roots in the unit disk. Show that $z^5 + 3z^2 + 1$ has exactly two roots in $|z| \lt 1$.

On $|z| = 1$: $|3z^2 + 1| \geq |3z^2| - |1| = 2 \gt |z^5| = 1$. By Rouché with $f(z) = 3z^2 + 1$ and $g(z) = z^5$: $z^5 + 3z^2 + 1$ has the same number of zeros as $3z^2 + 1$ in $|z| \lt 1$. $3z^2 + 1 = 0 \Rightarrow z = \pm i/\sqrt{3}$, both in $|z| \lt 1$. So $2$ zeros.

Rouché’s theorem is often used to bound the location of polynomial roots.

Theorem 12.3 (Eneström-Kakeya). If $a_0 \geq a_1 \geq \cdots \geq a_n > 0$, then every root of $P(z) = a_0 z^n + a_1 z^{n-1} + \cdots + a_n$ satisfies $|z| \leq 1$.

Proof. On $|z| = 1$, $|a_0 z^n| = a_0$ and $|a_1 z^{n-1} + \cdots + a_n| \leq a_1 + \cdots + a_n \leq a_0$ (strict unless all coefficients are equal). By Rouché, $P(z)$ has $n$ zeros inside $|z| = 1$. $\blacksquare$

The integral $\frac{1}{2\pi i}\int_\gamma f'(z)/f(z)\, dz$ equals the winding number of the curve $f(\gamma)$ around the origin. This geometric interpretation is useful in proving the argument principle:

$N - P = \frac{1}{2\pi} \Delta_\gamma \arg f(z)$

where $\Delta_\gamma \arg f(z)$ is the net change in the argument of $f(z)$ as $z$ traverses $\gamma$.

As a corollary of the argument principle, we have:

Theorem 12.4 (Open Mapping Theorem). A non-constant analytic function maps open sets to open sets.

Proof. For any $z_0$ in the domain, apply the argument principle to a small circle around $z_0$ on which $f(z) \neq f(z_0)$. The winding number of $f(\gamma)$ around $f(z_0)$ is positive, so points near $f(z_0)$ are in the image. $\blacksquare$

Problem 1. Determine the number of zeros of $z^7 - 2z^5 + 6z^3 - z + 1$ in $|z| \lt 1$.

Solution. On $|z| = 1$, $|6z^3| = 6 \gt |z^7 - 2z^5 - z + 1| \leq 1 + 2 + 1 + 1 = 5$. By Rouché, the function has $3$ zeros in $|z| \lt 1$ (same as $6z^3$). $\blacksquare$

Problem 2. Show that $z^4 + 5z + 2$ has exactly one root in $|z| \lt 1$.

Problem 3. Prove that $e^z - 3z^2 = 0$ has two roots in $|z| \lt 1$.

Solution. On $|z| = 1$, $|e^z| = e^{\mathrm{Re}\, z} \leq e \lt 3 = |3z^2|$. By Rouché with $f(z) = -3z^2$ and $g(z) = e^z$, $e^z - 3z^2$ has $2$ zeros in $|z| \lt 1$. $\blacksquare$

Problem 4. Show that all roots of $z^6 + 6z^2 + 1 = 0$ lie in the annulus $1/2 \lt |z| \lt 2$.

A practical application of the argument principle is counting zeros in a region without solving the equation. For $f(z) = z^5 + z + 1$ in $|z| < 1$:

On $|z| = 1$, $|z^5| = 1 \lt |z + 1| \geq |1 - 1| = 0$. We cannot directly apply Rouché here. Instead, check $|z^5 + 1| \geq |1 - |z|^5| = 0$ and $|z| = 1$ so $|z^5 + 1| \geq 0$ and $|z| = 1$ gives $|z| = 1 > |z^5 + 1|$? Let us check with $f(z) = z$ and $g(z) = z^5 + 1$. On $|z| = 1$, $|z| = 1$ and $|z^5 + 1| \leq |z|^5 + 1 = 2$. This does not give a strict inequality. Let us try $f(z) = z^5$ and $g(z) = z + 1$. On $|z| = 1$, $|z^5| = 1$ but $|z + 1|$ can be 0 at $z = -1$. So Rouché fails. Numerical computation shows there is 1 root in $|z| < 1$ and 4 roots outside.

If $f$ is meromorphic with poles, the argument principle counts $N - P$. This can be used to determine the number of roots of equations of the form $f(z) = a$ by applying the argument principle to $f(z) - a$.

Problem 5. Use the argument principle to show that $z^4 - 4z^3 + 6z^2 - 4z + 1 = (z-1)^4$ has all four roots inside $|z| = 2$.

Problem 6. Prove that the equation $z^3 + 2z + 1 = 0$ has two roots in $|z| < 1$ and one in $|z| > 1$.

Solution. On $|z| = 1$, $|2z| = 2 \gt |z^3 + 1| \leq 2$, so equality is possible. Instead use $f(z) = 2z + 1$, $g(z) = z^3$. On $|z| = 1$, $|2z + 1| \geq |2z| - 1 = 1$ and $|z^3| = 1$. The inequality $|2z + 1| \geq 1$ does not guarantee $> |z^3|$. So a more refined contour or splitting is needed.

Problem 7. Determine the number of zeros of $z^9 - 2z^5 + 3z^2 - 1$ in $|z| < 1$ and $1 < |z| < 2$.

Problem 8. Show that $e^z = 3z$ has exactly two solutions in $|z| < 2$.