Rouché's theorem

In complex analysis, Rouché's theorem tells us that if the complex-valued functions f and g are holomorphic inside and on some closed contour C, with |g(z)| < |f(z)| on C, then f and f + g have the same number of zeros inside C, where each zero is counted as many times as its multiplicity. This theorem assumes that the contour C is simple, that is, without self-intersections, and that it is oriented counter-clockwise.

Proof

Denote h = f + g which is holomorphic, being the sum of two holomorphic functions. From the argument principle, we have that

where N_h is the number of zeroes of h inside C, P_h is the number of poles, and I_h(C,0) is the winding number of h(C) about 0. Since h is analytic inside and on C, it follows that P_h is zero, and

One has that h′/h = D log h(z), where D denotes the complex derivative. Keeping in mind that h=f+g, we find <math>N_h={1\over 2\pi i}\oint_C {h'(z) \over h(z)}\,dz<math>

<math>={1\over 2\pi i}\oint_C D \left(\log{\left(f(z)\left(1+{g(z)\over f(z)}\right)\right)}\right)\,dz<math>

<math>={1\over 2\pi i}\oint_C D \left(\log{f(z)}+\log{\left(1+{g(z)\over f(z)}\right)}\right)\,dz<math>

<math>={1\over 2\pi i}\oint_C D \left(\log{f(z)}\right)+D\left(\log{\left(1+{g(z)\over f(z)}\right)}\right)\,dz<math>

<math>={1\over 2\pi i}\oint_C D \left(\log{f(z)}\right)\,dz+{1\over 2\pi i}\oint_C D\left(\log{\left(1+{g(z)\over f(z)}\right)}\right)\,dz<math>

<math>={1\over 2\pi i}\oint_C {f'(z) \over f(z)}\,dz+{1\over 2\pi i}\oint_C { D \left(1+{g(z)\over f(z)}\right) \over 1+{g(z)\over f(z)} }\,dz<math>

The winding number of 1+g/f over C is zero. This because we supposed that |g(z)| < |f(z)|, so g/f is constrained to a circle of radius 1, and adding 1 to g/f shifts it away from zero, and thus 1 + g/f is constrained to a circle of radius 1 about 1, and C under 1 + g/f cannot wind around 0.

The above then equals

which is N_f or the number of zeros of f. <math>\square<math>

Categories: Complex analysis