Definition: The residue Res(f, c) of a function f(z) at c is the coefficient of (z c) 1 in the Laurent series expansion of f at c.

Transcription

1 efnton: The resdue Res(f, c) of a functon f(z) at c s the coeffcent of (z c) n the Laurent seres expanson of f at c. Computaton: To compute Res(f, c), consder (z c)f(z) : g(z) g(c) + +(z c) n g (n) (c)/n! +. Then f(z)(z c) g(c) + + (z c) n g (n) (c)/n!. Thus Res(f, c) g(c) (z c)f(z) zc. If f s analytcal n a neghborhood of c, then Res(f, c) (z c)f(z) zc. The converse s not generally true. At a smple pole c, the resdue of f s gven by: More generally, f c s a pole of order n, then f(z)h(z)/(z c) n, and so Res(f, c) s gven by: (z c)f(z) zc (z c) h(z)/(z c) n zc h(z)/(z c) n zc h (n ) (z)/(n )! zc [(z c) n f(z)] (n ) / (n )! zc, where the 3 rd equalty s obtaned by dfferentatng the numerator and denomnator (n ) tmes. In other words, f c s a pole of order n, then resdue of f at c can be determned by the formula: Ths formula can be very useful n determnng the resdues for low order poles. For hgher order poles, the calculatons can become unmanageable, and a drect seres expanson may be easer. Applcaton: Accordng to resdue theorem: Here, s a closed curve, and I(, a k ) counts the number of tmes wnds around a k (k th pont nsde the closed curve where f s not analytc) n a counter clockwse manner. Snce resdues can be computed qute easly (as dscussed above), they can be used to easly determne a contour ntegral va the above resdue theorem. As an example, the above theorem s used n dervng the Nyqust crteron for stablty (#RHP poles of closed loop #RHP poles of open loop + #encrclements of (,) by plot of open loop freq. response).

2 The resdue theorem and ts applcatons Olver Knll Caltech, 996 Ths text contans some notes to a three hour lecture n complex analyss gven at Caltech. The lectures start from scratch and contan an essentally self-contaned proof of the Jordan normal form theorem, I had learned from Eugene Trubowtz as an undergraduate at ETH Zürch n the thrd semester of the standard calculus educaton at that school. My text also ncludes two proofs of the fundamental theorem of algebra usng complex analyss and examples, whch examples showng how resdue calculus can help to calculate some defnte ntegrals. Except for the proof of the normal form theorem, the materal s contaned n standard text books on complex analyss. The notes assume famlarty wth partal dervatves and lne ntegrals. I use Trubowtz approach to use Greens theorem to prove Cauchy s theorem. [ When I had been an undergraduate, such a drect multvarable lnk was not n my complex analyss text books (Ahlfors for example does not menton Greens theorem n hs book).] For the Jordan form secton, some lnear algebra knowledge s requred. The resdue theorem efnton Let C be open (every pont n has a small dsc around t whch stll s n ). enote by C () the dfferentable functons C. Ths means that for f(z) f(x + y) u(x + y) + v(x + y) the partal dervatves u x, u y, v x, v y are contnuous, real-valued functons on. efnton. Let : (a, b) be a dfferentable curve n. efne the complex lne ntegral b f(z) dz f((t)) (t) dt If z x + y, and f u + v, we have f(z) dz b a a (uẋ vẏ) + (uẏ + vẋ) dt The ntegral for pecewse dfferentable curves s obtaned by addng the ntegrals of the peces. We always assume from now on that all curves are pecewse dfferentable. Example. { z < r} wth r >, : [, k 2π], (t) (cos(t), sn(t)), f(z) z n wth k N,n Z. If n, we get z n dz k2π If n, we have zn dz k2π dt k 2π. e nt e t dt k2π z n dz n + e(n+2)t k2π e (n+)t dt We recall from vector calculus the Green formula for a vector feld (u, v) n R 2 (uẋ + vẏ) dt (v x u y ) dx dy, where s the open set enclosed by closed curve δ and where dx dy dxdy s the volume form n the plane. Wrte dz dx + dy, dz dx dy and dz dz 2dx dy. efne for f C ()

3 f(a + ǫe t ) f(a) C ǫ Theorem. (Complex Green Formula) f C (), C, δ. f(z)dz f z dz dz. Proof. Green s theorem appled twce (to the real part wth the vector feld (u, v) and to the magnary part wth the vector feld (v, u)) shows that b f(z) dz (uẋ vẏ) + (uẏ + vẋ) dt concdes wth f z dz dz a 2 (u x v y ) + 2 (u y + v x ) 2dx dy ( u y v x )dx dy + (u x v y )dx dy We check that f z u x v y, v x u y. The rght hand sde are called the Cauchy-Remann dfferental equatons. efnton. enote by C ω () the set of functons n C () for whch f z for all z. Functons f Cω () are called analytc or holomorphc n. Corollary.2 (Theorem of Cauchy) f C ω (), C, δ. f(z) dz. Proof. f(z) dz f dz dz z Corollary.3 (Cauchy s Integral formula) f C ω (), smply connected, δ. For any a f(a) f(w) dw 2π w a. Proof. efne for small enough ǫ > the set ǫ \ { z a ǫ} and the curve ǫ : t z + ǫe t. (Because s open, the set ǫ s contaned n for small enough ǫ). Because ǫ δ ǫ we get by Cauchy s theorem.2 f(w) w a dw f(w) w a dw. We compute ǫ f(w) 2π w a dw ǫ f(z + ǫ e t ) d a + ǫ e t a dt (a + ǫet ) dt 2π The rght hand sde converges for ǫ to 2πf(a) because f C () mples f(a + ǫ e t ) dt

4 Corollary.4 (Generalzed Cauchy Integral formulas) Assume f C ω () and C smply connected, and δ. For all n N one has f (n) (z) C ω () and for any z / f (n) (z) n! f(w) dz 2π (w z) n+. Proof. Just dfferentate Cauchy s ntegral formula n tmes. It follows that f C ω () s arbtrary often dfferentable. efnton Let f C ω ( \ {a}) and a wth smply connected C wth boundary. efne the resdue of f at a as Res(f, a) : f(z) dz. 2π By Cauchy s theorem, the value does not depend on. Example. f(z) (z a) and { z a < }. Our calculaton n the example at the begnnng of the secton gves Res(f, a). A generalzaton of Cauchy s theorem s the followng resdue theorem: Corollary.5 (The resdue theorem) f C ω ( \ {z } n ), open contanng {z } wth boundary δ. f(z) dz 2π n Res(f, z ). Proof. Take ǫ so small that { z z ǫ} are all dsjont and contaned n. Applyng Cauchy s theorem to the doman \ n leads to the above formula. 2 Calculaton of defnte ntegrals The resdue theorem has applcatons n functonal analyss, lnear algebra, analytc number theory, quantum feld theory, algebrac geometry, Abelan ntegrals or dynamcal systems. In ths secton we want to see how the resdue theorem can be used to computng defnte real ntegrals. The frst example s the ntegral-sne S(x) x a functon whch has applcatons n electrcal engneerng. It s used also n the proof of the prme number theorem whch states that the functon π(n) {p n p prme} satsfes π(n) x/log(x) for x. sn(t) t dt, S( ) sn(x) x dx π 2 Proof. Let f(z) ez z whch satsfes f C ω (C \ {}). For z x R, we have Im(f(z)) sn(x) x. efne for R > ǫ > the open set enclosed by the curve 4, where

5 Fgure : t [ǫ, R] t +. 2 : t [, π] R e t. 3 : t [ R, ǫ] t +. 4 : t [π, ] ǫ e t. R R By Cauchy s theorem f(z) dz R ǫ e x π x dx + e Ret ǫ e x Re t Ret dt + R x dx + e ǫet π ǫe t ǫet dt. The magnary part of the frst and the thrd ntegral converge for ǫ, R both to S( ). The magnary part of the fourth ntegral converges to π because lm ǫ π e ǫet dt π. The second ntegral converges to zero for R because e Ret e R sn(t) e Rt for t (, π/2]. + x 2 dx π Proof. Take f(z) + z 2 whch has a smple pole a n the upper half plane. efne for R > the half-dsc wth a hole whch has as a boundary the curve 2 wth : t [ R, R] t +. 2 : t [, π] R e t. f s analytc n \ {} and by the resdue theorem R π f(z) dz + x 2 dx Re t dt + R 2 2π Res(f(z), ) 2π lm(z ) f(z) π e2t z R The second ntegral from the curve 2 goes to zero for R. Let f, g be two polynomals wth n deg(g) 2 + deg(f) such that the poles z of h : f/g are not on R + {}. h(x) dx n Res(h(z) log(z), z ) Proof. efne for R > r > the doman enclosed by the curves R r + wth + : t [r, R] t +. : t [R, r] t +. R : t [, 2π] R e t. r : t [, 2π] r e t. and apply the resdue theorem for the functon h(z) log(z): h(z) log(z) dz h(z)log(z) dz + h(z)log(z) dz + h(z)log(z) dz + h(z) log(z) dz

6 Because of the degree assumpton, R h(z)log(z) for R. Because h s analytc near and log(z) goes slower to than z we get also r h(z)log(z) dz as r. The sum of the last two ntegrals goes to (2π) h(x) dx because h(z)log(z) dz h(z)(log(z) + 2π) dz. + π dθ a + cos(θ) dθ π a2, a > Proof. Put z e θ. Then Let : θ e θ π a + cos(θ) a + z + z 2 2az + z2 + 2z dθ a + cos(θ) dθ 2π dθ 2 a + cos(θ) dθ 2dz 2 2az + z 2 +. From the two zeros a ± a 2 of the polynomal 2az + z 2 + the root λ + s n the unt dsc and λ outsde the unt dsc. From the resdue theorem, the ntegral s 2π Res( 2az + z 2 +, λ +) 3 Jordan normal form for matrces 2π λ + λ. π a2. As an other applcaton of complex analyss, we gve an elegant proof of Jordan s normal form theorem n lnear algebra wth the help of the Cauchy-resdue calculus. Let M(n,R) denote the set of real n n matrces and by M(n,C) the set n n matrces wth complex entres. For A M(n, C) the characterstc polynomal s det(λ A) k (λ λ ) µ. We smply wrte λ A nstead of λi A, where I s the dentty matrx. The complex numbers λ are called the egenvalues of A and µ denote ther multplctes. Clearly k µ n. Two matrces A, B M(n,C) are called smlar f there exsts an nvertble matrx S such that A S BS. Theorem 3. (Jordan normal form theorem) [A ]... A M(n,C) s smlar to a matrx [A 2 ] where A [A k ] λ λ λ are called normal blocks. Remark. It follows f all egenvalues of A are dfferent, then A s dagonalzable. enote for λ λ the resolvent matrx R(λ) (λ A). The functon λ R(λ) s analytc n C \ {λ } n the sense that for all, j, the functons λ [R(λ)] j are

7 (where A (j) are the matrces obtaned by deletng the th row and the j th column of A), such that [R(λ)] j α j(λ) det(λ A). Lemma 3.2 (The resolvent dentty) R(λ) R(λ ) (λ λ)r(λ)r(λ ) Proof. From A(λ A) (λ A)A follows R(λ)A AR(λ) and we get by fllng n (λ A)R(λ ) I and (λ A)R(λ) I R(λ) R(λ ) R(λ)(λ A)R(λ ) (λ A)R(λ)R(λ ) (λ λ)r(λ)r(λ ). efnton C ω (, M(n,C)) denotes the set of functons f : M(n,C), such that for all, j the map z [f(z)] j s n C ω (). Gven a curve n we defne the complex ntegral f(z) dz by [ f(z) dz] j [f(z)] j dz. efne for δ < mn j λ λ j and : t z + δe t the matrces P R(λ) dλ 2π N (λ λ )R(λ) dλ. 2π Theorem 3.3 (Jordan decomposton of a matrx) ) P P j δ j P j, 2) k P I. 3) N P j δ j N P j N 4) N N j, j, P (A λ ) N, N µ 5) A k λ P + k N Proof. ) For j we have usng the resolvent dentty (2π) 2 P P j R(λ) dλ R(λ ) dλ R(λ)R(λ ) dλdλ j j R(λ) R(λ ) j λ dλ dλ λ R(λ) j λ λ dλ dλ + R(λ ) j λ λ dλ dλ. On the other hand, wth : t λ + δ/2 e t (2π) 2 R(λ) R(λ ) P P λ dλ dλ R(λ) R(λ ) λ λ dλ dλ λ R(λ) λ λ dλ dλ R(λ ) λ dλ dλ λ 2π R(λ )dλ, where we used λ λ dλ and λ λ dλ 2π. 2) Usng Cauchy s theorem we have for any curve R { λ R} enclosng all the egenvalues k k P R(λ) dλ R(λ) dλ 2π 2π R R(λ) A R(λ) dλ + A R(λ) dλ

8 The clam follows from and from the fact that for R 2π R λ dλ A R(λ) dλ R λ snce R(λ) j C λ for a constant C only dependent on A. 3) For j, we get wth the resolvent dentty (2π) 2 N P j (λ λ ) R(λ)R(λ ) dλ dλ j λ λ j λ λ (R(λ) R(λ )) dλ dλ λ λ j λ λ R(λ) λ λ dλ dλ j λ λ R(λ ) dλ dλ. Usng the curve { λ λ δ/2} we have (2π) 2 N P (λ λ ) R(λ)R(λ ) dλ dλ λ λ λ λ (R(λ) R(λ )) dλ dλ (λ λ )R(λ) dλ (2π) 2 N. λ λ dλ + (λ λ )R(λ ) dλ λ λ dλ 4) N N j s left as an exercse. (The calculaton goes completely analogue to the already done calculatons n ) or n 3) (2π)P (A λ ) R(λ)(A λ ) dλ R(λ)(A λ ) I dλ R(λ)(A λ ) R(λ)(A λ) dλ (λ λ )R(λ) dλ (2π)N. Usng 3) and ) we get from the just obtaned equalty (2π)N k (2π)P (A λ ) k R(λ)(A λ ) k dλ R(λ)(A λ ) k R(λ)(A λ ) k (A λ) dλ R(λ)(A λ ) k (λ λ ) dλ R(λ)(A λ ) (k ) (λ λ ) R(λ)(A λ ) k 2 (A λ) dλ R(λ)(A λ ) k 2 (λ λ ) 2 dλ. Repeatng lke ths k 2 more tmes, we get (2π)N k (λ λ ) k R(λ) dλ. The clam N µ follows from the fact that (λ λ ) µ R(λ) s analytc.

9 Remark. It follows that the matrx A leaves nvarant the subspaces H P H of H C n and acts on H as v λ v + N v. There s a bass of H, the matrces N have n the sde dagonal and are everywhere else: by 4), we know that µ s the smallest k such that N k. It mples that all egenvalues of N are. There exsts a vector v H such that {N kv v k} n k form a bass n H. (Ths s an exercse: any nontrval relaton j a jn j v would mply that N had an egenvalue dfferent from ). In that bass, the transformaton N s the matrx N The matrx A s now a drect sum of Jordan blocks A k A, where λ A λ λ Exercses: ) Perform the calculaton whch had been left out n the above proof: show that N N j, j.. 2) Show that f a lnear transformaton N on a µ dmensonal space has the property N µ and N k for k < µ, then there s a bass n whch N s a Jordan block wth zeros n the dagonal. 4 The argument prncple Cauchy s ntegral formula and the resdue formula can be expressed more naturally usng the noton of the wndng number.. Lemma 4. Let be closed curve n C avodng a pont a C. There exsts k Z such that dz z a 2πk. Proof. efne Snce h (t) z (t)/(z(t) a), we get h(t) : t z (s) ds z(s) a. d dt e h(t) (z(t) a) h (t)e h(t) (z(t) a) + e h(t) z (t). and e h(t) (z(t) a) e h() (z() a) s a constant. Therefore e h(t) z(t) a z() a and especally eh(2π) z(2π) a z() a whch means h(2π) 2πk. efnton The ndex or wndng number of a closed curve wth respect to a pont a / s n(, a) dz 2π z a 2πZ. The defnton s legtmated by the above lemma. Cauchy s ntegral formula and the resdue theorem holds more generally for any closed curve n a smply con-

10 If f C ω () n(, z) f(z) f(w) dw. 2π w z If f C ω ( \ {z } n ) f(z) dz 2π n n(, z ) Res(f, z ). Theorem 4.2 (Argument prncple for analytc functons) Gven f C ω () wth smply connected. Let a be the zeros of f n and a curve n avodng a. Then n(, a ) 2π f (z) f(z) dz Proof. Wrte where g(z) has no zeros n. We compute Cauchy s theorem gves f(z) (z a )(z a 2 )...(z a n )g(z) f (z) f(z) g (z) z a z a 2 z a n g(z). g (z) g(z) dz and the formula follows from the defnton of the wndng number. efnton If f C ω ( \ {z}) for a neghborhood of a, then a s called an solated sngularty of f. If there exsts n N such that lm n (z a)n+ f(z) then a s called a pole of f. The smallest n such that the above lmt s zero s called the order of the pole. If an solated sngularty s not a pole, t s called an essental sngularty. If f C ω ( \ {z }) and each z s a pole then f s called meromorphc n. Theorem 4.3 (Argument prncple for meromorphc functons) Let f be meromorphc n the smply connected set, a the zeros of f, b the poles of f n and a closed curve avodng a, b. n n(, a ) k j n(, b j ) f (z) 2π f(z) dz Proof. The functon g(z) : f(z) (z b )...(z b 2 )...(z b k ) s analytc n and has the zeros a. Wrte ( f (z) 2π f(z) ) dz z b z b 2 z b k 2π g (z) g(z) dz The rght hand sde s by argument prncple for analytc maps equal to n(, a ). The left hand sde s f (z) 2π f(z) dz + n(, b j ).

11 Theorem 4.4 (Generalzed argument prncple) Let f be meromorphc n the smply connected set, a the zeros of f, b the poles of f n and a closed curve avodng a, b and g C ω (). n g(a )n(, a ) k j g(b )n(, b j ) g(z) f (z) 2π f(z) dz Proof. Wrte agan n f(z) h(z) (z a ) k j (z b j) wth analytc h whch s nowhere zero and so g(z) f n (z) f(z) g(a ) z a k j g(b j ) z b j + g(z) h (z) h(z). As an applcaton, take h C ω ( ) wth { z a < r} and h( ). For z put ξ h(z). The functon f(w) h(w) ξ has only one zero n. Apply the last theorem wth g(w) w. We get h (ξ) z wh (w) 2π δ() h(w) ξ dw whch s a formula for the nverse of h. Assume f C ω ( \ {z }) s meromorphc. enote by Z f Z f () the number of zeros of a functon n and wth P f P f () the number of poles of f n. Theorem 4.5 (Rouché s theorem) Gven meromorphc f, g C ω ( \ {z }). Assume for R {z z a < R}, δ R f(z) g(z) < g(z), z. Then Z f P f Z g P g. Proof. The assumpton mples that f(z) <, z g(z) and h : f/g maps therefore nto the rght half plane. We can defne log(f/g) by requrng that Im(log(h(z))) arg(h(z)) ( π/2, π/2). The functon log(h(z)) s a prmtve of h /h (f/g) f/g. We have so Apply the argument prncple. (f/g) 2π f/g dz 2π f f g g dz. Rouché s theorem leads to an other proof of the fundamental theorem of algebra: Theorem 4.6 (Fundamental theorem of algebra) Every polynomal f(z) p(z) z n + a z n a n has exactly n roots. Proof. p(z) z n + a /z +... a n /z n goes to for z. Consder the functon g(z) z n and { z < R}.