Lecture-16 : Cauchy's Integral Formula and Analyticity

LECTURE-16 : CAUCHY’S INTEGRAL FORMULA AND ANALYTICITY

VED V. DATAR∗

The aim of this chapter is to establish a beautiful formula due to Cauchy which will at the heart of the local study of holomorphic functions.

Cauchy’s integral formula Theorem 0.1. Let f(z) be holomorphic on a domain Ω, and let D be a disc whose closure is contained in Ω. Then for any z ∈ D, 1 Z f(ζ) f(z) = dζ. 2πi ∂D ζ − z Proof. One way to prove this formula is to use generalized Cauchy’s theorem to reduce the integral, to integrals on arbitrarily small circles. Since our proof of the generalized Cauchy’s theorem was not complete, and keep- ing in view the importance of this theorem in what follows, we provide an independent proof replying only on the homotopy theorem. Recall that if z ∈ D, then 1 Z 1 dζ = 1, 2πi ∂D ζ − z and so 1 Z f(ζ) 1 Z f(ζ) − f(z) dζ − f(z) = dζ 2πi ∂D ζ − z 2πi ∂D ζ − z Now deﬁne a new function ( f(ζ)−f(z) , ζ 6= z, g(ζ) = ζ−z f 0(z), ζ = z. Then g(ζ) is continuous on D, and holomorphic on D \{z}, since f is holomorphic on D. Consider the contour in the ﬁgure. The curves Γ1 and Γ2 lie in C\Im(ζ) < z and C\Im(ζ) > z respectively, which are simply connected domains where g(ζ) is holomorphic. Hence by the homotopy theorem, the integrals on each of the curves is zero. But since the orientations on the cuts are opposite, this shows that Z Z Z Z 0 = g(ζ) dζ + g(ζ) dζ = g(ζ) dζ − g(ζ) dζ, Γ1 Γ2 ∂D ∂Dε(z)

Date: 24 August 2016. 1 and hence Z Z g(ζ) dζ = g(ζ) dζ, ∂D ∂Dε(z) for all ε > 0 small. Now since g(ζ) is continuous, in particular it is bounded, and hence there exists an M > 0 such that |g(ζ)| < M for all ζ ∈ D. In particular, Z ε→0 g(ζ) dζ < Mlen(Dε(z)) = 2πMε −−−→ 0. ∂Dε(z) Hence 1 Z f(ζ) 1 Z f(ζ) − f(z) 1 Z dζ − f(z) = dζ = g(ζ) dζ = 0, 2πi ∂D ζ − z 2πi ∂D ζ − z 2πi ∂D which completes the proof of the theorem.

Analyticity A complex valued function is called analytic in a disc D centered at a if for all z ∈ D, ∞ X n f(z) = an(z − a) . n=0 By the theorem on differentiating power series, it follows that any analytic function has complex derivatives of all orders, and in fact by term-wise differentiation, it is then clear that the coefficients are given by f (n)(a) a = . n n! Although the class of analytic functions appears a priori to be smaller than the class of holomorphic functions, the next theorem asserts the remarkable fact, that the two classes of functions are in fact identical.

Theorem 0.2. Let Ω ⊂ C open, and f :Ω → C a holomorphic function. Then for any disc D whose closure is contained in Ω, f(z) is analytic on D. Moreover, if the center of the disc is a and the radius is R, then the coeﬃcients are given by 1 Z f(ζ) (0.1) an = n+1 dζ. 2πi |ζ−a|=R (ζ − a)

Proof. By CIF, if D is a of radius R centered at a with boundary circle CR, then for any z ∈ D, 1 Z f(ζ) f(z) = dz. 2πi |ζ−a|=R ζ − z 2 Writing ζ − z = (ζ − a) − (z − a), we see that 1 1 z − a−1 = 1 − . ζ − z ζ − a ζ − a

For z ∈ D and ζ ∈ CR, |z−a| < R = |ζ−a|, or equivalently |(z−a)/(ζ−a)| < 1, and hence using the geometric series ∞ 1 1 X z − an = . ζ − z ζ − a ζ − a n=0 Since power series converge uniformly, we can also integrate term-wise (see Appendix), we get that ∞ Z f(ζ) X z − an 2πif(z) = dζ ζ − a ζ − a CR n=0 ∞ X Z f(ζ) = dζ · (z − a)n (ζ − a)n+1 n=0 CR ∞ X n = 2πi an(z − a) , n=0 where an is given by the formula (0.1), and this completes the proof of the theorem. Example 0.1. We have seen that there is a holomorphic branch of log z deﬁned on C \{z ≤ 0}. Then by the theorem above, there is a power series expansion on the unit ball centered at z = 1. In fact we saw in the lecture on the exponential function, that on |z − 1| < 1, ∞ X (−1)n+1 log z = zn. n n=1 As an immediate corollary, we have the following.

Corollary 0.1. Any holomorphic function f :Ω → C is infinitely complex differentiable. Moreover, if D is a disc with boundary C whose closure is contained in Ω, then for any z ∈ D, we have Z (n) n! f(ζ) f (z) = n+1 dζ. 2πi C (ζ − z) Proof. Since f(z) is analytic in the neighborhood of any point z ∈ Ω, it is automatically holomorphic at that point. Given a point z ∈ D, there exists an ε (depending on z) such that Dε(z) ⊂ D. We denote the boundaries of D and Dε(z) by C and Cε respectively. Now f has a power series expansion in Dε(z) by Theorem 0.2 with coefficients given by formula (0.1) (with a replaced by z and R replaced by ε). On the other hand, by the relation 3 between the coefficients of a power series and derivative, we see that Z (n) n! f(ζ) f (z) = n!an = n+1 dζ. 2πi Cε (ζ − z) But then by the generalized Cauchy’s theorem (see figure), since f(ζ)/(ζ − n+1 a) is holomorphic on D \ Dε(z), Z f(ζ) Z f(ζ) n+1 dζ = n+1 dζ, C (ζ − z) Cε (ζ − z) and this completes the proof of the corollary. Another consequence of the analyticity is the following criteria for holo- morphicity. Corollary 0.2 (Morera). Any continuous function on an open set Ω that satisfies Z f(z) dz = 0 T for all triangles T ⊂ Ω is holomorphic. Proof. The condition is equivalent to f having a primitive F (z) on Ω. But 0 then by Corollary 0.1, F (z) and hence f(z) has to be holomorphic.

∗ Department of Mathematics, UC Berkeley E-mail address: [email protected]