Complex integration. L3: Cauchy s Theory.

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1 MM Vercelli. L3: Cuchy s Theory. Contents: Complex integrtion. The Cuchy s integrls theorems. Singulrities. The residue theorem. Evlution of definite integrls. Appendix: Fundmentl theorem of lgebr. Discussions bout the vlue of nd best proof for mthemticl proposition will (perforce) come up gin nd gin s long s mthemtics is done by humns beings, nd yet to mny mthemticins the ensuing polemics re s hrd to understnd s the disputtions of the Byzntines bout the gender of the ngels. R. Remmert pge 99 of Theory of Complex functions, GTM 22, Springer (998). Complex integrtion Let f : I = [, b] C be function from n intervl to the complex numbers. Tht is to sy, f(x) = A(x) + i B(x) The integrl of f on I is defined s follows: I f := b A(x)dx + i b B(x)dx It is importnt to remember the stndrd estimte: f f I I Mthemticl Methods, L3 Mthemticl Methods

2 . Integrtion long curves. MM Vercelli.. Integrtion long curves. Let : [, b] C be curve. The integrl fdz := b f((t)). (t)dt is clled curviliner integrl or pth integrl of f long. Notice tht rel integrl b f(x)dx cn be regrded s the curviliner integrl of f long the pth (x) = x + i. Tht is to sy, we regrd the intervl [, b] s curve in C. Exmple.. Let B r (c) be the disc of rdius r centered t c C. Denote by B r (c) the boundry prmeterized counter-clockwise. Then { (z c) n, for n dz = 2π i, for n = B r(c) The bove exmple shows tht integrls long closed curves do not lwys vnish. In prticulr, the integrl long curves with sme initil nd terminl points need not to be equl. The stndrd estimte for curviliner integrls is fdz L(). f where f := mx f((t)), t [, b] nd L() is the length of. Primitives nd pth independence. Let F, f O(D) be holomorphic functions such tht F (z) = f(z). Then f is sid to be integrble nd F is clled primitive of f. Two primitives of the sme function re equl up to constnt. Mthemticl Methods, L3 2 Mthemticl Methods

3 MM Vercelli. An importnt observtion is tht curviliner integrl of n integrble functions is independent of the pth. Nmely the integrl fdz long ny closed pth D must vnish. Indeed, b b b fdz = f((t)). (t)dt = F ((t)). df ((t)) (t)dt = dt = F ((b)) F (()) dt so fdz = since () = (b). Theorem.. (Integrbility Criterion) The following sttements bout function f O(D): (i) f is integrble in D, (ii) fdz = long ny closed pth D. Proof. Tht (i) (ii) ws explined bove. To see (ii) (i) fix point z D nd set F (z) = fdz z where z is ny curve strting t z nd ending t z. The hypothesis (ii) mens tht F (z) is well-defined, i.e. do not depends on the pth z. Then for z D F (z + h) F (z ) = fdz [z,z +h] where [z, z + h] is the pth t z + t.h, t [, ]. So F (z + h) F (z ) f(z ) = f(z) f(z [z,z +h] )dz f(z) f(z ) [z,h+z h h ] nd we get F (z ) = f(z ) for ll z D. Since the function z B r() is integrble for z < we get B r() dz z = for r <. Insted for r > the bove integrl gives 2π i. Indeed, dz z = dz z( ) = ( z + z + )dz = z 2 B r() B r() B r() where we cn exchnge the integrtion becuse of the uniform convergence. dz z = 2π i Mthemticl Methods, L3 3 Mthemticl Methods

4 MM Vercelli. 2 The Cuchy s theorems Notice tht n nlytic function f : Ω C, i.e. given by power series convergent on ll Ω, is integrble. So the integrl round closed curve is zero. In prticulr, if z, z Ω re two points the integrl z z f(z)dz is well-defined i.e. does not depend on the pth from z to z. The following theorem of Cuchy shows tht the independence of the integrl lso holds for holomorphic functions if the pth re so clled homotopic. Theorem 2.. (Cuchy s deformtion Theorem) Let Ω C be n open set nd let f O(Ω). Assume tht f is lso C (Ω). Let, : [, ] Ω to curves with the sme initil nd terminl point. Assume lso tht, re homotopic i.e. there exists : [, ] [, ] Ω continuous with (, t) = (t), (, t) = (t) nd the curves s (t) := (s, t) hve the sme initil nd terminl points s nd. Then f(z)dz = f(z)dz Proof. Let i(s) := s f(z)dz be the integrl long s. We will see tht i(s) is constnt. It is possible to show tht (s, t) cn be ssumed to be smooth. So we cn tke derivtives respect to s. Then nd so i (s) = i(s) = f(z)dz = s f((s, t)) (s, t) dt t ( ) (s, t) (s, t) df((s, t))( ) + f((s, t)) 2 (s, t) dt s t s t nd by using Cuchy-Riemnn conditions we get ( f((s, t)) i (s, t) (s, t) (s) = z s t ( ) d (s, t) = f((s, t)) dt dt s = = ( f((s, t)) (s, t) s ) t= t= ) + f((s, t)) 2 (s, t) dt s t Here the derivtive goes under the integrl becuse f ws ssumed to be C. Gourst s proof void the hypothesis C on f see s_integrl_theorem Mthemticl Methods, L3 4 Mthemticl Methods

5 MM Vercelli. since (s, ) z nd (s, ) z. The importnce of the bove theorem is tht in order to compute n integrl we cn deform the pth of integrtion. As corollry we get the following result. Theorem 2.2. (Cuchy s integrl theorem) Under the bove hypothesis f(ξ)dξ = if is null-homotopic curve in Ω. Theorem 2.3. (Cuchy s integrl formul) Under the sme hypothesis of the bove theorem let Ω be closed curve which is prmetriztion of the boundry D of simply connected open subset D Ω. Then for z D f(z) = 2π i f(ξ) ξ z dξ. Proof. By using the bove theorem to the function ξ f(ξ) ξ z the bove integrl is equl to 2π i B z(ɛ) f(ξ) ξ z dξ where B z (ɛ) is bll of rdius ɛ centered t z. By exmple (.) we hve f(z) = f(z) 2π i B z(ɛ) ξ z dξ. So it is enough to show lim ɛ 2π i B z(ɛ) f(z) f(ξ) dξ =. ξ z Figure : Integrl formul. Since f is holomorphic t z there exists M such tht f(z) f(ξ) M ner z. So ξ z f(z) f(ξ) dξ M dξ = ɛm 2π i B z(ɛ) ξ z 2π B z(ɛ) This shows tht the bove limit is indeed zero. Mthemticl Methods, L3 5 Mthemticl Methods

6 2. Cuchy-Tylor representtion theorem MM Vercelli. 2. Cuchy-Tylor representtion theorem Here we prove the following theorem due to Cuchy. Theorem 2.4. Let f O(D) then f is nlytic in D. More precisely, round ech point z D, f hs convergent power series n n(z z ) n where the coefficients re given by n = f n (z ) n! = 2π i B z(ɛ) f(ξ) dξ (ξ z ) n+ Proof. Expnd the kernel (ξ z) in Cuchy s integrl formul. The min ppliction is the following identity theorem. Theorem 2.5. The following sttements bout pir f, g O(D), D open nd connected, re equivlent: (i) f(z) = g(z) for ll z D, (ii) there is n open subset G D such tht f(z) = g(z) for ll z G, (iii) there is point p D nd sequence p n convergent to p such tht f(p n ) = g(p n ). (iv) there is point p D such tht f (n) (p) = g (n) (p) for ll n N. Proof. Assume g =. If f (n) (p) = then there exists n open bll round p such tht f =. This show tht (iv) (ii) nd so (iv) (iii). Assume (iii). Then the power series of f round p must be identiclly zero otherwise f(z) = (z p) m ψ(z) ner p with ψ(p). But this implies tht p is n isolted zero of f contrdicting (iii). So (iii) (iv). Thus, (ii), (iii), (iv) re equivlent. Tht (i) (ii) is obvious. To show (ii) (i) let Ω D be the set of points p D such tht f vnish in neighborhood of p. Then Ω is open. Let p n Ω sequence whose limit p D. By the proof of (iii) (iv) it follows tht p Ω. Thus, Ω is lso closed. Hence Ω = D since D is connected. Mthemticl Methods, L3 6 Mthemticl Methods

7 MM Vercelli. 3 Singulrities Let D be domin nd let c D. Assume we hve function f defined nd holomorphic in D \ {c}. In such sitution c is clled n isolted singulrity of f. Wht hppens t c? There re three possibilities: ) Removble singulrities which upon closer exmintion turn out to be not singulrities t ll, i.e. f cn be extended s holomorphic function round c. 2) Poles this mens tht (z c) n f(z) hs removble singulrity t c for some n N. The smllest n such tht (z c) n f(z) hs removble singulrity t c is clled the order of the pole c of f For exmple f(z) = hs pole of order t z =. A z pole of order is clled simple pole. 3) Essentil singulrity this is the worst cse. An exmple is f(z) = e z t z =. Theorem 3.. If f is bounded round c then c is removble singulrity of f. Proof. We cn ssume tht c =. Set h() = nd h(z) = z 2 f(z) for z. Then h is holomorphic t. Indeed, h(z) h() lim z z = lim z z 2 f(z) z since f ws ssumed to be bounded ner z =. Let h(z) = z + 2 z z 3 + be the convergent series representing h ner zero. Notice tht the bove limit sy tht =. So the power series z + is convergent round nd gree with f(z) ner. Thus by putting f() = 2 we get tht f is holomorphic lso t z = nd z + is its Tylor power series. A meromorphic function on D is function which is holomorphic up to subset P D of points in which f hs poles. = Mthemticl Methods, L3 7 Mthemticl Methods

8 MM Vercelli. 4 Residues nd rel integrls Let Ω be closed curve which is prmetriztion of the boundry D of simply connected open subset D Ω, where Ω is domin. Let f O(Ω \ A), where A = {c,, c n } D is finite set of points of D. It is cler tht by deforming the pth of integrtion we cn reduce the computtion of n integrl f(ξ)dξ = f(ξ)dξ + + f(ξ)dξ B ɛ(c ) where B ɛ (c i ) is smll disk round c i, i =,, n. The residue of f t the point c is the integrl f(ξ)dξ 2π i B ɛ(c) B ɛ(c n) nd is usully denoted by res s f It is importnt to remrk tht res c cn be used to decide if f hs primitive in smll disk round c. Indeed, if f hs such primitive then the integrl do not depend on the pth, so the residue must be zero. Reciproclly, if such residue is zero the integrl do not depend on the pth. As corollry we get the following result. Theorem 4.. The residue res c f is the unique complex number such tht the function f(z) z c hs primitive in smll disk round c. Thus, if f hs n expnsion with h(c) holomorphic round c then + 2 (z c) 2 + (z c) + h(z) Here is the residue theorem 2. res c f = 2 For more see Mthemticl Methods, L3 8 Mthemticl Methods

9 4. Rel integrls MM Vercelli. Theorem 4.2. (Residue theorem ) In the bove ssumptions n f(ξ)dξ = res ci f 2π i Remember: If f hs simple pole t c its residue is given by the limit: i= res c f = lim z c (z c)f(z) 4. Rel integrls The improper integrl f(x)dx is the limit M lim f(x)dx M + M Theorem 4.3. Let f be holomorphic in the closed upper hlf-plne, except possibly t finitely mny points none of which is rel. Suppose tht f(x)dx exists nd tht lim z + zf(z) =. Then f(x)dx = 2π i res w f w H Figure 2: Improper integrl. where H is the upper hlf-plne. Proof. Since f hs finite number of singulrities they re contined in circle of rdius m centered t. So if M > m we hve: M f(x)dx + f(z)dz = 2π i res w f w R M where R is the red domin contining ll singulrities of f. The curve cn be prmeterized by Me i θ with θ [, π]. Then f(z)dz f(z) dz πm f(z ) = π z f(z ) So s M + we hve f(z)dz. Mthemticl Methods, L3 9 Mthemticl Methods

10 4. Rel integrls MM Vercelli. Exmple 4.4. To determine dx we use f(z) = = +x 2 +z 2 (z+i)(z i) dx + x 2 = π. nd we get The function f(x) = whose integrl over the rel line is is clled the π(+x 2 ) stndrd Cuchy distribution 3 4 nd ply n importnt role in the physics of the resonnces. Improper integrls g(x)e i x dx with R cn be integrted by using residues using the domin in the picture. Figure 3: Fourier integrl. Theorem 4.5. Let g be holomorphic in C, except possibly t finitely mny points none of which is rel. Suppose lim z + g(z) =. Then { g(x)e i x 2π i w H dx = res w g(z)e i z if > 2π i w H res w g(z)e i z if < where H is the upper hlf-plne. Proof Assume >. It is enough to show tht the three integrls g(z)e i z dz, g(z)e i z dz, g(z)e i z dz A B C goes to zero for r, s + for q = s + r such tht the singulrities of g re contined in the red domin of the picture. A g(z)e i z dz q where g A is the mximum of g over A. g(s + t i) e t dt g A e q g A Mthemticl Methods, L3 Mthemticl Methods

11 4. Rel integrls MM Vercelli. A similr computtion shows tht g(z)e i z dz g B Finlly B g(z)e i z dz g(z)e i z C q g(z) C e q q C so if s, r + these integrls goes to zero since g ws ssumed to go to zero s z. For < one consider squre in the lower hlf-plne nd the sme rgument yields the result. Exmple 4.6. For R, > nd b C, Rel(b) > we hve e i x x i b dx = 2π i e b. Trigonometric integrls I = 2π R(cos(θ), sin(θ))dθ where R is rtionl function which is finite one the unit circle. Observe tht by using e i θ = cos(θ) + i sin(θ) we cn rewrite this integrl s follows. I = 2π R( ei θ + e i θ 2, ei θ e i θ )dθ 2 i In order to write I s curviliner integrl we need to include the differentil Doing so we get I = 2π R( ei θ + e i θ 2 de i θ = i e i θ dθ., ei θ e i θ ) i e i θ de i θ. 2 i Mthemticl Methods, L3 Mthemticl Methods

12 4. Rel integrls MM Vercelli. Thus, By using the residues we get: I = i B () I = 2π w B () Exmple 4.7. To determine 2π Then R( z+z 2, z z 2 i )z = (z p)( pz) 2π R( z + z 2 res w (R( z + z 2, z z )z dz. 2 i, z z )z ) 2 i dθ 2p cos(θ)+p 2 use R(x, y) = 2px+p 2. dθ 2p cos(θ) + p 2 = nd we get { 2π, if p < p 2, if p >. 2π p 2 Mthemticl Methods, L3 2 Mthemticl Methods

13 4. Rel integrls MM Vercelli. Appendix: Fundmentl theorem of lgebr 5 vi residues Let f(x) R[x] be rel polynomil of degree >. Lemm : There exists complex number c such tht f(c) = Proof. Assume the contrry. Nmely, tht f(z) > for ll complex numbers. We cn ssume f(x) > for rel vlues of x. A simple estimtion shows tht > f(x) dx >. On the other side it is not difficult to see tht f(re i θ ) = cte.r deg(f). This mens tht the integrl round the semicircle θ [, π] re i θ goes to zero s r. So s r + the integrl r r f(x) dx must converge to the sum of the residues of situted on the upper hlf plne. But f(z) is holomorphic everywhere so we get the contrdiction f(z) f(x) dx = As corollry we get Fundmentl theorem of lgebr. The number field C is lgebriclly closed. Nmely, if P (X) = c n X n + c n X n + + c C[X] with n = def(p ) > then P (X) = hs root in C. In prticulr, P (X) cn be written s product: P (X) = c n (X α )(X α 2 ) (X α n ) with α i C. Indeed, the polynomil Q(X) := P (X)P (X) hs rel coefficients nd the bove discussion shows tht Q hs complex root c. Then either c or c is root of P. Once you hve root α of P (X) you cn divide P (X) by X α nd you get P (X) = (X α )P (X) where P hs degree n. If n > then P hs itself root nd you cn repet the process. It is cler tht fter n steps you get the fctoriztion P (X) = c n (X α )(X α 2 ) (X α n ). 5 Mthemticl Methods, L3 3 Mthemticl Methods