# Complex Analysis: #21 The Logarithmic Derivative

This is an interesting equation, particularly so in the ﬁeld of analytical number theory. But the theorem we will look at here concerns path integrals. For simplicity, we will assume that G = ℂ, and that the function f is meromorphic, so that it is analytic everywhere in ℂ, except possibly for some set of isolated singularities. At each of these singularities, f has a pole of some order.

So let us say that f has a pole of order k at the point a ∈ ℂ. Then we can write

Combining this with the residue theorem, we have a method of counting the zeros and poles. Namely:

**Theorem 40**Let f be a meromorphic function, defined in a region G ⊂ ℂ, and let Ω be some cycle which avoids all zeros and poles of f and which is such that all points of ℂ \ G have index zero with respect to Ω. Let N

_{Ω}be the number of zeros of f which have non-vanishing index with respect to Ω, counted according to their orders, and let P

_{Ω}be the poles, again counted with their orders. Then

Another way to look at this is to remember that f is, after all, just a mapping of a region of ℂ back into ℂ. Thus if γ is a closed path in the region G (again, with winding number zero with respect to all points in the compliment of G in ℂ), it follows that f ◦ γ, given by f ◦ γ(t) = f(γ(t)), is itself another closed path in ℂ. If we assume that γ passes through no zero or pole of f, then the path f ◦ γ avoids both 0 ∈ ℂ and also the special point ∞. Therefore we can think about the index of 0 with respect to this path.

**Theorem 41**ν

_{f◦γ}(0) = N

_{γ}− P

_{γ}, where the numbers N

_{γ}and P

_{γ}have been defined in the previous theorem (theorem 40).

*Proof*

As we saw in theorem 25, we have

where we use the result of the previous theorem. Here, we have just imagined that the path γ is defined on some interval of the form [a, b]. Of course it is a trivial matter to see that we could replace the single path γ with a cycle Ω.

**Theorem 42 (Roché)**Let f, g : G → ℂ be analytic functions, and let Ω be a cycle in G. Assume A = {w ∈ ℂ : ν

_{Ω}(w) ≠ 0} ⊂ G, and assume furthermore that |g(z)| < |f(z)| for all z which lie directly on Ω. Then both functions f and f + g have the same number of zeros in A.

*Proof*

Let z be a point of Ω. Then since 0 ≤ |g(z)| < |f(z)|, we certainly do not have f(z) = 0. But also, for all τ ∈ [0, 1], we have

|f(z) + τg(z)| ≥ |f(z)| − τ|g(z)| > 0.

Therefore the cycle (f + g) ◦ Ω is homotopic to the cycle f ◦ Ω in ℂ \ {0}, and the result then follows from theorem 41.

This theorem gives us a very quick proof of the Fundamental Theorem of Calculus. For let

P(z) = z

^{n}+ a_{n−1}z^{n}+ · · · + a_{1}z + a_{0}be some polynomial of degree n ≥ 1 in ℂ. Let R = |a

_{n−1}| + · · · + |a

_{0}| + 1. Then on the circle given by |z| = R, we have

0 ≤ |a

_{n−1}z^{n}+ · · · + a_{1}z + a_{0}| < |z^{n}|.But then theorem 42 implies that P(z) must have n zeros (counted with their multiplicities) within the circle of radius R.

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