The argument principle counts zeros by winding numbers. Rouché's theorem turns that into a quick comparison: if one part of a function dominates the rest on a circle, the whole function has as many zeros inside as that part alone.
Idea clave
Rouché's theorem. If $f$ and $g$ are analytic on and inside a simple closed contour $C$, and
$$|g(z)| < |f(z)| \quad\text{at every point of } C,$$
then $f$ and $f + g$ have the same number of zeros inside $C$ (counted with multiplicity).
Por qué funciona
Picture a person walking along the curve $f(C)$ around a post at $0$, with a dog on a leash: the dog's position is $f + g$, and the leash has length $|g|$. Since the leash is always shorter than the person's distance $|f|$ to the post, the dog can never reach the post, so both go around it the same number of times. By the argument principle, those winding numbers are the zero counts of $f$ and of $f + g$.
Worked examples
Ejemplo resuelto: Inside the unit circle
How many zeros does $p(z) = z^5 + 6z^2 - 1$ have in $|z| < 1$?
- On $|z| = 1$ take $f = 6z^2$: $|f| = 6$. The rest, $g = z^5 - 1$, has $|g| \le 1 + 1 = 2$.
- $2 < 6$ on the whole circle, so $p$ has as many zeros inside as $6z^2$: two.
Respuesta: $2$
Ejemplo resuelto: A larger circle
How many zeros does $p(z) = z^5 + 6z^2 - 1$ have in $|z| < 2$?
- On $|z| = 2$ take $f = z^5$: $|f| = 32$; $|6z^2 - 1| \le 24 + 1 = 25$.
- $25 < 32$, so $p$ has $5$ zeros in $|z| < 2$.
- Combined with the previous example: $5 - 2 = 3$ zeros lie in the annulus $1 \le |z| < 2$ (none on $|z| = 1$, where $|p| \ge 6 - 2 > 0$).
Respuesta: $5$
Ejemplo resuelto: No zeros
Show that $z^4 + z + 5$ has no zeros in $|z| < 1$.
- On $|z| = 1$ take $f = 5$: $|g| = |z^4 + z| \le 2 < 5$.
- The constant $5$ has no zeros, so neither does $z^4 + z + 5$ inside the circle.
Respuesta: no zeros
Error común
Checking the inequality at one point, or picking the wrong main term. $|g| < |f|$ must hold at every point of $C$, so bound $|g|$ by its worst case (the triangle inequality adds the largest sizes of its terms). And choose $f$ as the term that wins on that circle: on $|z| = 1$ the term $6z^2$ wins for $z^5 + 6z^2 - 1$, but on $|z| = 2$ the top power $z^5$ does. If no single term beats the sum of the others, try a different circle.