Rouché's theorem

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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.

[edit] Geometric explanation

Since the distance between the curves is small, h(z) does exactly one turn around like f(z) does.

It is possible to provide an informal explanation on why the Rouche's theorem holds.

First we need to rephrase the theorem a little bit. Let h(z) = f(z) + g(z). Notice that f, g holomorphic implies h holomorphic too. Then, with the conditions imposed above, Rouche's theorem says that

If |f(z)| > |h(z) − f(z)| then f(z) and h(z) have the same number of zeros on the interior of C.

Notice that the condition |f(z)| > |h(z) − f(z)| means that for any z, the distance of f(z) to the origin is larger than the length of h(z) − f(z), which in the follow picture means that for each point on the blue curve, the segment joining to the origin is larger than the green segment associated to it. Informally we can say that the red curve g(z) is always closer to the blue curve f(z) than to the origin.

But the previous paragraph shows that since f(z) winds exactly once around 0, so must h(z), and by the argument principle, the index of both curves around zero is the same, which means that f(z) and h(z) have the same number of zeros.

[edit] Proof

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

$N_h-P_h=I_h(C,0)={1\over 2\pi i}\oint_C {h'(z) \over h(z)}\,dz$

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

$N_h=I_h(C,0)={1\over 2\pi i}\oint_C {h'(z) \over h(z)}\,dz.$

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

$={1\over 2\pi i}\oint_C D (\log{h(z)})\,dz$

$={1\over 2\pi i}\oint_C D (\log{(f(z)+g(z))})\,dz$

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

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

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

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

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

$=I_f(C,0)+I_{1+{g(z)\over f(z)}}(C,0).$

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.