Line Integrals and Entire Functions

Transcription

1 Line Integrls nd Entire Funtions Defining n Integrl for omplex Vlued Funtions In the following setions, our min gol is to show tht every entire funtion n be represented s n everywhere onvergent power series in z. In order to do this, we shll need to develop the onept of line integrl. In vetor lulus, we introdued the ide of line integrl. The onept ws motivted by the ft tht the domin of funtion of two vribles is the whole plne, so when we integrte funtions of two vribles, we n integrte over ny urve in the plne. Sine the domin of omplex funtion is lso plne (the omplex plne), in order to define the integrl for funtion of omplex vrible, we shll hve to use similr ide.. Definition nd Properties of the Line Integrl.. The Definition of Line Integrl. Before we define line integrl nd onsider how to lulte it, we need some preliminry definitions nd results. First we onsider the more stright forwrd se where the rel nd imginry prts of funtion of omplex vribles both depend on some fixed vrible t. Definition.. Let f(t) = u(t) + iv(t) be ny ontinuous omplexvlued funtion of the rel vrible t with t b. Then we define b f(t)dt = b u(t)dt + i b v(t)dt. lultion of integrls suh s these re stright forwrd single vrible lultions. In order to define generl line integrl, we need wy to represent urves in the line. As with vetor lulus, this n be done through the use of prmeteriztion. The following re importnt definitions regrding prmeterized urves. Definition.2. (i) Let z(t) = x(t) + iy(t) t b. The urve determined by z(t) is lled pieewise differentible nd we set ż(t) = x (t) + iy (t) if x nd y re ontinuous on [, b] nd ontinuously differentible ( ) on eh subintervl [, x ], [x, x 2 ],...,[x n, b] of some prtition of [, b]. (ii) The urve is sid to be smooth if in ddition ż(t) exept t finite number of points. For the rest of the ourse, unless otherwise stted, ll urves will be ssumed to be smooth. We n now define line integrl.

2 2 Definition.3. Let be urve prmeterized by z(t), t b nd suppose f is ontinuous t ll points of z(t). Then the integrl of f long is b f(z)dz = f(z(t))ż(t)dt. Note tht this is now single vrible omplex vlued funtion of t so n be lulted in the stndrd wy..2. Independene of Pth Prmeteriztion. The diretion in whih we trvel long will hnge the vlue of the integrl, but it seems tht how we trvel long the urve (one diretion is speified) should not ffet the vlues of the integrl i.e. if z(t) nd w(t) re two different prmeteriztions of, then the integrl long should be the sme regrdless of the prmeteriztion we use. Though this n sometimes fil, under ertin dditionl onditions imposed on the prmeteriztions, we n gurntee tht this will lwys be the se. Definition.4. The two urves prmeterized by z(t) with t b nd 2 prmeterized by w(t) with t d re sid to be smoothly equivlent if there exists mpping λ: [, d] [, b] suh tht λ (t) for ll t nd w(t) = z(λ(t)). Proposition.5. If nd 2 re smoothly equivlent, then f(z)dz = f(z)dz. 2 Proof. Suppose tht f(z) = u(z) + iv(z), nd 2 re prmeterized by z(t) = x(t) + iy(t) nd w(t) respetively, nd λ: [, d] [, b] is the mpping with w(t) = z(λ(t)) = x(λ(t)) + iy(λ(t)) nd λ (t). Then we hve = nd b = u(z(t))x (t)dt = d 2 f(z)dz = d f(z)dz = b d b v(z(t))y (t)dt+i f(w(t)) w(t)dt = f(z(t)) z(t)dt b d u(z(t))y (t)dt+ b f(z(λ(t))) z(λ(t))dt [u(z(λ(t))) + iv(z(λ(t)))][x (λ(t)) + iy (λ(t))]λ (t)dt [u(z(λ(t)))x (λ(t))λ (t)]dt d [v(z(λ(t)))y (λ(t))λ (t)]dt v(z(t))x (t)dt

3 +i d [u(z(λ(t)))y (λ(t))λ (t)]dt + d [v(z(λ(t)))y (λ(t))λ (t)]dt ompring eh of the four integrls, we see it is diret onsequene of the hnge of vrible formuls for rel integrls (i.e. substitution) tht these re equl. For exmple, if we tke s = λ(t) in the integrl d [u(z(λ(t)))x (λ(t))λ (t)]dt we hve ds = λ (t)dt nd when t =, s = nd when t = d, s = b, so we get d [u(z(λ(t)))x (λ(t))λ (t)]dt = b u(z(s))x (s)ds This shows tht line integrl is independent of prmeteriztion. However, s we stted before, it does depend upon diretion. In order to illustrte this ft, we need the following definition. Definition.6. Suppose is given by z(t) with t b. Then is defined by z(b + t) with t b (so is tred in the opposite diretion. Proposition.7. Proof. We hve f(z)dz = f(z)dz = f(z)dz b f(z(b + z))ż (b + t)dt Similr to the lst question, we n distribute to obtin four different rel integrls nd then mke the substitution s = b + z to eh of the rel funtions to prove the equlity. We onsider some exmples. Exmple.8. (i) lulte z dz where z(t) = os (t) + i sin (t) with t 2π. First observe tht z = x x 2 + y iy 2 x 2 + y 2. 3

4 4 z dz = Then we hve 2π (ii) lulte (os (t) i sin (t))( sin (t)+i os (t))dt = z dz 2π idt = 2πi where is the squre of side length oriented ounterlokwise entered t the origin. We need to lulte the integrl over ll four sides. Sine ll lultions re similr, we lulte only over, the top side prmeterized by z(t) = t+i with t. Here we hve ( z dz = = ln (t2 + ) 2 t t i rtn(t) i ) ()dt t 2 + = πi 2. It n be shown tht the integrl over eh side is lso iπ/2, so we get dz = 2πi z (observe tht this is the sme s the previous question - we shll see why lter)..3. Preliminry Results for Line Integrls. We now onsider generlizing some results for line integrls of rel vribles to line integrls of omplex vribles. The first result shows tht the opertion of integrtion is liner nd n be proved by onsidering the orresponding rel integrls nd using the rel result. Proposition.9. Let be smooth urve, let f nd g be ontinuous funtions of z nd let α be ny omplex number. Then (i) (ii) [f(z) + g(z)]dz = f(z)dz + αf(z)dz = α f(z)dz g(z)dz The next result we onsider generlizes the ide tht the bsolute vlue of definite integrl of funtion is bounded by the integrl of the bsolute vlue of the sme funtions.

5 Lemm.. Suppose G(t) is ontinuous nd omplex vlued funtion of t. Then b b G(t)dt G(t) dt. Proof. Suppose tht b G(t)dt = Re iϑ for some fixed ϑ nd R. Then observe tht b G(t)dt = R so we just need to show tht R b G(t) dt. In order to do this, we onsider the funtion e iϑ G(t). First, using the linerity properties of the integrl, we hve e iϑ G(t)dt = R so if e iϑ G(t) = A(t) + ib(t) (s rel nd imginry prts), lso using linerity of the integrl, we must hve But R = b A(t)dt. A(t) = Re(e iϑ G(t)) Re(e iϑ G(t)) e iϑ G(t) = G(t) so we hve so the result follows. R = b A(t)dt b G(t) dt This result now llows us to impose n upper bound on ny line integrl (similr to rel vrible where n upper bound is given by the length of the urve multiplied by the mximum vlue of the funtion on the urve). Theorem.. M L Formul - Suppose tht is smooth urve of length L on G nd tht f(z) M throughout. Then f(z)dz ML. 5

6 6 Proof. We hve b f(z)dz = b f(z(t))ż(t)dt f(z(t)) ż(t) dt sine = M b b (x (t)) 2 + (y (t)) 2 dt = ML (x (t)) 2 + (y (t)) 2 dt = L b M ż(t) dt is the formul for the length of urve prmeterized by z(t) (rell Vetor lulus nd lulus 2). Exmple.2. Show tht z 2dz 2π on the unit irle (oriented in either diretion). For this we simply observe tht /z 2 on the unit irle, so z 2dz 2π. We n use he ML-formul to show tht the integrl of sequene of uniformly integrls onverge. Proposition.3. Suppose tht {f n } is sequene of funtions nd f n f uniformly on smooth urve. Then f(z)dz = lim f n (z)dz. n Proof. First by linerity of the integrl, we hve f(z)dz f(z)dz = (f(z) f n (z))dz. Sine f n f uniformly, for n suffiiently lrge, we hve f f n < ε for ny ε >. Then we hve (f(z) f n (z))dz εl where L is the length of the urve. Sine ε n be s smll s we plese, it follows tht f n (z)dz = f(z)dz. lim n The lst preliminry result we prove is generliztion of the fundmentl theorem of lulus for rel vribles.

7 Proposition.4. Suppose tht F = f where F is n nlyti funtion whih is smooth on. Then f(z)dz = F(z(b)) F(z()). Proof. We shll prove this by showing tht the derivtive of F(z(t)) with respet to t is f(z(t))ż(t) nd then use the Fundmentl Theorem from rel vrible lulus to finish. First note tht sine F is nlyti, F(z(t)) will be smooth urve. It follows tht F(z(t)) F(z(t + h)) F(z(t)) = lim h h h rel i.e. sine F(z(t)) is smooth, the derivtive is defined so n be lulted using the differene quotient tking h in ny diretion towrd. Next we note tht sine ż(t), (exept for finite number of points), nd is ontinuous, we n find δ suh tht h < δ implies z(t + h) z(t). Thus we hve F(z(t)) F(z(t + h)) F(z(t)) = lim h h h rel F(z(t + h)) F(z(t)) z(t + h) z(t) = lim h z(t + h) z(t) h h rel = F (z(t)) ż(t) = f(z(t))ż(t). Then by definition nd using the Fundmentl Theorem of lulus for rel vribles, we hve b f(z)dz = f(z(t))ż(t)dt = F(z(b)) F(z()), hene the result. 2. The losed urve Theorem for Entire Funtions In this setion we onsider the generliztion of the result in multivrible lulus whih sttes tht the line integrl of grdient vetor over losed urve is. In order to prove this result, we shll hve to prove the result for ertin speil urves first. 2.. The Retngle Theorem. We shll first onsider integrls over retngles. We need the following definitions. Definition 2.. A urve is losed if its initil points nd terminl points oinide. 7

8 8 Definition 2.2. By the boundry of retngle, we men simple losed urve prmeterized so tht the retngle it tres is on the left ofs the urve is tred (see piture). Theorem 2.3. Suppose f is n entire funtion nd Γ is the boundry of retngle R. Then f(z)dz =. Γ Proof. In order to prove this result, we first note tht by the fundmentl theorem of lulus, the result holds for liner funtions i.e. f(z) = z + b is everywhere the derivtive of z 2 /2 + bz nd the endpoints of the urve R re the sme. Thus the result holds for liner funtions. Now suppose tht f(z) is n rbitrry entire funtion nd let I = f(z)dz. Γ We n brek up the retngle R into 4 eqully sized retngles with boundries Γ, Γ 2, Γ 3 nd Γ 4 s illustrted below. Observe tht the sum of the integrls round eh of these retngles is equl to the integrl round R sine the integrls over the sides inside R nel eh other out. Thus we hve 4 f(z)dz = f(z)dz. Γ i Γ i= It follows tht for t lest one of these integrls, whih we shll denote by Γ (), we hve f(z)dz I Γ 4. () Let R denote the retngle with boundry Γ (). We n do extly the sme thing with the retngle R () nd ontinue this proess until we obtin sequene of retngles with R () R (2)...

9 with boundries Γ () Γ (2)..., with the following properties: (i) Side lengthsr (k+) = ( Side lengthsr (k) )/2 (ii) f(z)dz I Γ 4 k. (k) Now observe tht sine f is entire, t ny z R, there exists ε z suh tht f(z) = f(z ) + f (z )(z z ) + ε(z z ) where ε z s z z i.e. it is differentible, so n be pproximted using lol lineriztion. Thus we hve f(z)dz = f(z )+f (z )(z z )+ε(z z )dz = ε z (z z )dz Γ (n) Γ (n) Γ (n) sine the rest is liner. Thus we just need to bound this integrl. Let s denote the length of the lrgest side of Γ. Then we hve length of Γ (n) 4s 2 n (sine s hs the lrgest length) nd 2s z z for ll z Γ (n) (the length from one orner of the squre of side length s/2 n to the digonl opposite - the two furthest points). Next note tht sine ε z s z z, for ny ε, we n find N suh tht 2s z z implies ε z < ε. Then using the ML formul, for n N, we hve f(z)dz 2s ε4s Γ 2 (n) n 2 = 4 2s 2 ε. n 4 n It follows tht I 4 4 2s 2 ε n 4 n or equivlently I 4 2s 2 ε. This is now independent of z nd holds for ll epsilon, so it follows tht I =. 2 n 2 N 9

10 2.2. The Integrl Theorem. Our next tsk is to show tht ny entire funtion f(z) is the derivtive of some nlyti funtion F(z) whih we will then be ble to use to prove tht the integrl of n entire funtion round ny losed urve is. Theorem 2.4. (The Integrl Theorem) If f(z) is entire, then f is everywhere the derivtive of some nlyti funtion i.e. there exists F(z) suh tht f(z) = F (z) for ll z. Proof. We define F(z) = z f(ζ)dζ where z denotes the integrl long the stright lines from to Re(z) nd from Re(z) to Im(z) (see illustrtion below). z Next note tht F(z + h) F(z) = h f(ζ)dζ. We n see this by looking t the lines over whih the integrls re being lulted. Speifilly, F(z + h) F(z) h f(ζ)dζ is the integrl of the funtion f(ζ) from (, ) to Re(z), followed by the retngle with orners Re(h), Re(z + h), z + h nd h + Im(z) in ounterlokwise diretion nd finishing with the integrl from Re(z) to (, ) (see illustrtion). z+h h

11 The retngle theorem tells us the integrl over the retngle will be zero, nd the other two integrls nel (sine they re long the sme line in opposite diretions) so F(z + h) F(z) h Using this equlity, we next observe tht F(z + h) F(z) h f(z) = h f(ζ)dζ =. h sine the integrl does not depend upon z, so h f(z)dz = f(z)h. [f(ζ) f(z)]dζ Sine we re trying to show tht F (z) = f(z), we need to show tht F(z + h) F(z) f(z) h so our observtions imply tht it suffies to show tht h h [f(ζ) f(z)]dζ. Sine f is ontinuous, for suffiiently smll h we n gurntee tht f(z) f(ζ) < ε for ny ζ, so we get F(z + h) F(z) f(z) h = h [f(ζ) f(z)]dζ h ε h = ε. h Letting ε, we hve F(z + h) F(z) f(z) h so F (z) = f(z) The losed urve Theorem. We re now redy to prove the min result of this setion - tht the line integrl of n entire funtion round ny losed urve is. With the results we lredy hve, the proof is now stright forwrd. Theorem 2.5. If f is entire nd is smooth losed urve, then f(z)dz =. Proof. Sine f is entire, we hve f(z) = F (z) for some nlyti funtion F(z). Then f(z)dz = F (z)dz = F(z(b)) F(z()) =

12 2 sine F(z(b)) = F(z()). In this Theorem we stted tht f ws entire, but this theorem n be generlized to ny region on whih f is the derivtive of some nlyti funtion F(z). Speifilly, the more generl losed urve theorem would be the following: Theorem 2.6. If f is entire on some region D (whih is not neessrily the whole omplex plne) ontining the smooth losed urve, then f(z)dz =. Note tht this theorem fils if the funtion is not entire on region tht ontins s we sw when we showed dz = 2πi z for the unit irle entered t the origin. Homework: Questions from pges 54-55; 2,3,4,5,7,9