Rouche's theorem
/ roo-SHAY /
Suppose you want to count the zeros of a complicated function inside a region, but the function is too messy to attack directly. Rouche's theorem offers a shortcut: if you can find a simpler function that dominates it on the boundary, then the two functions have the same number of zeros inside. You count the zeros of the easy function and conclude that the hard one has just as many — a perturbation argument for roots.
The statement: if f and g are analytic on and inside a closed contour C, and on the boundary the inequality |g(z)| is strictly less than |f(z)| holds everywhere, then f and f + g have the same number of zeros (counted with multiplicity) inside C. The intuition is that g is a perturbation too small, all along the boundary, to drag any zero across the boundary or to wind the image differently; since |g| never reaches |f| on C, the sum f + g can never vanish on C and its image winds around the origin exactly as f's does. It is, in essence, a clean corollary of the argument principle: equal winding numbers mean equal zero counts.
Rouche's theorem is the standard, elementary route to the fundamental theorem of algebra — a degree-n polynomial p(z) is dominated on a large circle by its leading term z^n, which has n roots at the origin, so p has exactly n roots. It is used to locate roots of transcendental equations region by region, to show stability margins in control and numerical analysis, and to prove that small perturbations of a system do not change its number of unstable modes. Its power is that it converts a hard counting problem into an easy inequality check on a curve.
How many roots of p(z) = z^4 + 6z + 3 lie inside the unit circle? On |z| = 1 take f(z) = 6z and g(z) = z^4 + 3. There |f| = 6, while |g| is at most 1 + 3 = 4, so |g| is less than |f| on the boundary. Hence p has the same number of zeros inside as f(z) = 6z, namely one.
Pick the dominant term on the boundary; its zeros inside count the zeros of the whole function.
The dominance inequality must be strict and must hold at every point of the contour; a single boundary point where |g| equals |f| can break the conclusion. Choosing which term to call f and which g is an art — pick the wrong split and the inequality fails even though a valid split exists.