22 Cauchy Integral Formula Cauchy’s Integral Formula If f is a holomorphic function defined in a domain D ⊂ C and ∆(a; r) ⊂ D is a disk, the ∂∆(a;r) f(z)dz = 0. R What happens if the function f has a “singularity” inside the disk ∆(a; r) ? We first take look at a special case. Lemma 22.1 1 dz =2πi. Z∂∆(a;r) z − a Proof: Write φ : [0, 2π] → C, t 7→ a + reiθ as a parametric function for the circle ∂∆(a; r). iθ iθ 1 2π 1 iθ 2π Since de = ie dθ, we have ∂∆(a;r) z−a dz = 0 reiθ ire dθ = i 0 dθ =2πi. R R R For general case, we have Theorem 22.2 (Cauchy’s Integral Formula) Let D ⊂ C be a domain and ∆(a; r) ⊂ D. Let f be a holomorphic function defined in D. Then 1 f(ζ) f(z)= dζ, ∀z ∈ ∆(a; r). 2πi Z∂∆(a;r) ζ − z Proof: Assume a = 0. Fix a point z = |z|eiθ0 ∈ ∆(r). To show: f(ζ) dζ =2πif(z). Z∂∆(r) ζ − z ∂∆(0; r) A C Take a disk ∆(z; δ) ⋐ ∆(r). Take a point B at G r B r r r the circle ∂∆(z; δ) and another point A at the circle ♥D ∂∆(z; δ) r E ∂∆(r). We use a line segment to join these two r points. Then we get a closed curve C by connecting F the points A,B,C,D,B,A,E,F,G,A in order. r This closed curve encloses a domain which excludes 143 f(ζ) the point z. By Cauchy Integral Theorem, C ζ−z dζ = 0. In other words, by cancellation of the part of the integral along the line segmentsR AB and BA, we have f(ζ) f(ζ) dζ − dζ =0. Z∂∆(r) ζ − z Z∂∆(z;δ) ζ − z Then f(ζ) f(ζ) dζ = dζ Z∂∆(r) ζ − z Z∂∆(z;δ) ζ − z 2π f(z + δeiθ) δieiθdθ = iθ Z0 (z + δe ) − z 2π = f(z + δeiθ)idθ Z0 2π = f(z)+(f(z + δeiθ) − f(z)) idθ Z0 2π 2π = f(z)idθ + (f(z + δeiθ) − f(z)) idθ Z0 Z0 2π 2π = f(z)idθ + lim (f(z + δeiθ) − f(z)) idθ Z0 δ→0 Z0 2π = f(z) idθ +0=2πif(z) Z0 by applying Lemma 22.1. In the above theorem, the C1-smoothness condition is not needed. It implies C∞- smoothness. z2−4z+4 [Example] Evaluate C z+i dz where C is the circle ∂∆(2) with countclockwise orien- tation. H Solution: Let f(z)= z2 − 4z + 4 and z = −i. Then by Cauchy Integral Formula: z2 − 4z +4 dz =2πif(−i)=2πi(3+4i)=2π(−4+3i). IC z + i z [Example] Evaluate C z2+9 dz where C is the circle ∂∆(2i, 2) with countclockwise orien- tation. H 144 Solution: Applying Cauchy Integral Formula, z z z 3i dz dz z+3i dz πif i πi πi 2 = = =2 (3 )=2 = IC z +9 IC (z +3i)(z − 3i) IC z − 3i 6i z where f(z) := z+3i . Corollary 22.3 Let f, ∆(a; r) and D be as in Theorem 22.2. Then f ′ exists and 1 f(ζ) f ′ z dζ, z a r . ( )= 2 ∀ ∈ ∆( ; ) 2πi Z∂∆(a;r) (ζ − z) More generally, n! f(ζ) f (n) z dζ, z a r . ( )= n+1 ∀ ∈ ∆( ; ) 2πi Z∂∆(a;r) (ζ − z) Consequently, any holomorphic function is C∞-smooth. Proof: Fixing z ∈ ∆(a; r), to show: f(z+h)−f(z) 1 f(ζ) h ? → 2πi ∂∆(a;r) (ζ−z)2 dζ k R 1 f(ζ) − f(ζ) dζ 2πhi ∂∆(a;r) ζ−(z+h) ζ−z R ↓ h → 0 1 f(ζ) 2πi ∂∆(a;r) (ζ−z)2 dζ R because 1 f(ζ) f(ζ) f(ζ)h f(ζ − = → . hζ − (z + h) ζ − z h[ζ − (z + h)](ζ − z) (ζ − z)2 z+1 [Example] Evaluate C z4+4z3 dz where C = ∂∆(1) with counterclockwise orientation. Solution By the corollaryH above, z +1 z+1 f(z) 2πi 3πi dz z+4 dz dz f ′′ , 4 3 = 3 = 3 = (0) = − IC z +4z IC z IC z 2! 32 z+1 where f(z) := z+4 . 145 Historic Remarks on Cauchy’s Integral Formula In 1831, Cauchy obtained his famous Cauchy Integral Formula during his exile in Turin. This integral formula became generally accessible in 1841, when Cauchy back in Paris. It was for a circle. 1 f(ζ) f(z)= dζ. 2πi Z|z|=r ζ − z As he did for Cauchy Integral Theorem, Cauchy only put “finite and continuous” condition on f. But in 1839, Cauchy changed his mind about the correct statement of his theorem. When he presented a paper to the Paris Academy, Cauchy required not only finiteness and continuity on f, but also on f ′. (cf. Hans Niels Jahnke (editor), A History of Analysis, AMS, 2003, p.226.) Equivalence of Holomorphic and Analytic Functions Recall we have proved: f analytic =⇒ f holomorphic. Now we’ll prove: f holomorphic =⇒ f analytic. Theorem 22.4 A holomorphic function f defined on a domain D ⊂ C is analytic, i.e., ∀a ∈ D, we have ∞ n f(z)= an(z − a) , ∀z ∈ ∆(a; r) ⋐ D. Xn=0 z−a Proof: We have ζ−a < 1, for any z ∈ ∆(a; r) and ζ ∈ ∂∆(a; r). By using the sum formula of geometric series, 1 1 1 1 1 ∞ z−a n ζ−z = ζ−a−(z−a) = ζ−a · z−a = ζ−a n=0 ζ−a . z 1− ζ−a ✬✩ P a q q ζ q ✫✪ Then 146 1 f(ζ) f(z) = dζ (Cauchy′s integral formula) 2πi Z∂∆(a;r) ζ − z 1 ∞ (z − a)n = f(ζ) dζ (by the above formula) 2πi Z (ζ − z)n+1 ∂∆(a;r) Xn=0 ∞ 1 f(ζ)dζ = (z − a)n 2πi Z (ζ − a)n+1 Xn=0 ∂∆(a;r) ∞ n = an(z − a) , Xn=0 where 1 f(ζ) a . n = n+1 (105) 2πi Z∂∆(a;r) (ζ − a) (n) n! f(ζ)dζ Remark Since f (a) = 2πi ∂∆(a;r) (ζ−z)n+1 , we obtain the usual Tylor coefficient for- mula: R f (n)(a) a = . n n! Corollary 22.5 Let f be holomorphic function on C − E, where E is a subset, but f is not holomorphic at any point in E. Let a ∈ C − E and R := sup{r | ∆(a; r) ⊂ C − E}. Then ∞ n R is the radius of convergence of the power series f(z)= n=0 an(z − a) . P In fact, Cauchy thought about this statement for some twenty years. sin(x2+x) [Example] Let f(x)= (x−1)(x−2) . Find the radius of convergence of f centered at x = 5. Solution: Consider the extended complex-valued function sin(z2 + z) f(z)= . (z − 1)(z − 2) Centered at z = 5, the biggest disk ∆(5,r) in which the function f is holomorphic must be ∆(5, 3). By Corollary above, the radius of convergence of f centered at 5 is 3. 69 69This is an example how we use complex analysis to solve some problems in Calculus. 147 Remarks: Cauchy Integral Formula for C1 smooth functions A version of Cauchy’s integral formula is the Cauchy-Pompeiu formula, and holds for smooth functions. Let D be a disc in C and suppose that f is a complex-valued C1-smooth function on the closure of D. Then 1 f(z)dz 1 ∂f dz ∧ dz¯ f(ζ)= + (z) . 2πi Z∂D z − ζ 2πi ZZD ∂z¯ z − ζ In particular, if f is holomorphic, i.e., ∂f = 0, the above formula gives 1 f(z)dz f(ζ)= 2πi Z∂D z − ζ which is the Cauchy Integral Formula. 70 70For the proof of the Cauchy-Pompeiu formula, see Principles of Algebraic Geometry, by Griffiths and Harris, John Wiley & Sons, Inc., 1994, p.3. 148.