omplex vribles lecture 5: omplex integrtion Hyo-Sung Ahn School of Mechtronics Gwngju Institute of Science nd Technology (GIST) 1 Oryong-dong, Buk-gu, Gwngju, Kore Advnced Engineering Mthemtics
omplex integrls - Def. (Integrl of f(t)): Let f(t) = u(t) + iv(t), where u nd v re rel-vlued functions of the rel vrible f for t b. Then, f(t)dt = u(t)dt + i - If U (t) = u(t), nd V (t) = v(t), for t b, we hve v(t)dt (1) f(t)dt = [U(t) + iv (t)] t=b t=1 = U(b) U() + i[v (b) V ()] (2) - Ex. Show tht 1 (t i)3 dt = 5. Since f(t) = (t 4 i)3 = t 3 3t + i( 3t 2 + 1), u(t) = t 3 3t nd v(t) = 3t 2 + 1. The integrls of u nd v leds the solution. - Ex. π/2 exp(t+it)dt = π/2 e t e it dt = π/2 e t (cos t+i sin t)dt = π/2 e t cos tdt+ i π/2 e t sin tdt. Since π/2 e t (cos t)dt = 1 2 (eπ/2 1) nd π/2 e t (sin t)dt = 1 2 (eπ/2 + 1), the solution is 1 2 (eπ/2 1) + i 2 (eπ/2 + 1).
omplex integrls - omplex integrls hve similr properties to those of rel integrls. [f(t) + g(t)]dt = f(t)dt = c f(t)dt + f(t)dt + (c + id)f(t)dt = (c + id) f(t)dt = f(t)g(t)dt = b f(t)dt c g(t)dt (3) f(t)dt (4) f(t)dt (5) [u(t)p(t) v(t)q(t)]dt + i (6) [u(t)q(t) + v(t)p(t)]dt (7) f (t)dt = f(b) f() (8) - Ex. π/2 exp(t+it)dt = 1 1+i et(1+i) t=π/2 t= = 1 1+i (ieπ/2 1) = 1 2 (eπ/2 1)+ i 2 (eπ/2 +1).
ontours nd contour integrls - Define nd evlute integrls of the form c f(z)dz. - A curve: : z(t) = x(t) + iy(t), for t b (9) - A simple curve (): It it does not cross itself, which mens tht z(t 1 ) z(t 2 ). - A closed curve: A curve with the property z(b) = z(). - If z(b) = z() is the only point of intersection, then is simple closed curve. - The complex-vlued function z(t) = x(t) + iy(t) is sid to be differentible on [, b] if both x(t) nd y(t) re differentible for t b.
ontours nd contour integrls - From (9), the derivtive z is z (t) = x (t) + iy (t). The curve defined by eq. (9) is sid to be smooth curve if z is continuous nd nonzero on the intervl. If is smooth curve, then hs nonzero tngent vector t ech point z(t), which is given by the vector z (t). If x (t ) =, then the tngent vector z (z ) = iy (t ) is verticl. If x (t ), then the slope dy of the tngent dx line to t the point z(t ) is given by y (t ). x (t ) - If is smooth curve, then ds, the differentil of rc length, is given by ds = [x (t)] 2 + [y (t)] 2 dt = z (t) dt - The function s(t) = [x (t)] 2 + [y (t)] 2 is continuous, s x nd y re continuous functions, so the length L() of the curve is L() = [x (t)] 2 + [y (t)] 2 dt = z (t) dt (1) - A curve tht is constructed by joining finitely mny smooth curves end to end is clled contour.
ontours nd contour integrls
ontours nd contour integrls - The integrl of complex function long contour in the plne with initil point A nd terminl point B. rete prtition P n = {z = A, z 1, z 2,..., z n = B} of points tht proceed long from A to B Form the differences z k = z k z k 1, for k = 1, 2,..., n. Between ech pir of prtition points z k 1 nd z k, select point c k on. Evlute the function f(c k ) t c k - Riemnn sum for the prtition S(P n ) = n f(c k )(z k z k 1 ) = k=1 n f(c k ) z k (11) - Assume tht there exists unique complex number L tht is the limit of every sequence {S(P n )} of Riemnn sums given in eq. (11), where the mximum of z k tends towrd for the sequence of prtitions. We define the number L s the vlue of the integrl of the function f tken long the contour. k=1
ontours nd contour integrls - Def. (omplex integrl): Let be contour. Then, f(z)dt = lim n n f(c k ) z k provided the limit exists in the sense previously discussed. - Ex. Use Riemnn sum to get n pproximtion for the integrl expzdt, where is the line segment joining the point A = to B = 2 + i π. 4 Solution: Set n = 8 nd form the prtition P 8 : z k = k 4 +iπk, for k =, 1, 2,..., 8. 32 For this sitution, we hve uniform increment z k = 1 + i π. For convenience, 4 32 we select c k = z k 1 z k = 2k 1 + i π(2k 1). Then, the Riemnn sum is 2 8 64 S(P 8 ) = 8 f(c k ) z k = k=1 8 k=1 exp[ 2k 1 8 - The true vlue is 4.22485 + 5.22485i. k=1 + i π(2k 1) ]( 1 64 4 + i π ) 4.23 + 5.2i 32
ontours nd contour integrls - Suppose tht we hve prmetriztion of the contour given by the function z(t) for t b. It then follows tht lim n n k=1 f(c k ) z k = lim n = lim n n f(c k )(z k z k 1 ) k=1 n f(z(τ k ))[z(t k ) z(t k 1 )] k=1 where τ k nd t k re the points contined in the integrl [, b] with the property c k = z(τ k ) nd z k = z(t k ).
omplex integrls - By multiplying the kth term in the lst sum by t k t k 1 t k t k 1,then we get lim n n k=1 f(z(τ k )) z(t k) z(t k 1 ) t k t k 1 (t k t k 1 ) = lim n = n k=1 f(z(τ k )) z(t k) z(t k 1 ) t k t k t k 1 f(z(t))z (t)dt - Theorem: Suppose tht f(z) is continuous complex-vlued function defined on set contining the contour. Let z(t) be ny prmetriztion of for t b. Then, f(z)dz = f(z(t))z (t)dt Proof: Out of scope of this lecture.
omplex integrls - Ex. Evlute the integrl expzdt, where is the line segment joining the point A = to B = 2 + i π. 4 Solution: We cn prmetrize by z(t) = (2 + i π )t, for t 1. As 4 z (t) = (2 + iπ/4), by the bove theorem, 1 expzdz = exp[z(t)]z (t)dt = 1 exp[(2 + iπ/4)t](2 + iπ/4)dt = (2 + iπ/4)( 1 e 2t cos πt 4 dt + i 1 e 2t sin πt 4 dt) - Evlute 1 dz. The function z(t) = 2 + 1 + (2) z 2 eit, t π, is prmetriztion for. Here, z (t) = ie it. Hence, 1 z 2 dz = π 1 (2 + e it ) 2 ieit dt = π idt = iπ
omplex integrls - Ex. Show tht zdz = zdz = 4 + 2i 1 2 where 1 is the line segment from 1 i to 3 + i, nd 2 is the portion of the prbol x = y 2 + 2y joining 1 i to 3 + i. 1 : z(t) = 2t + 1 + it nd dz = (2 + i)dt, for 1 t 1. 2 : z(t) = t 2 + 2t + it nd dz = (2t + 2 + i)dt, for 1 t 1.
omplex integrls Along 1, nd using the bove theorem, 1 zdz = (2t + 1 + it)(2 + i)dt = 4 + 2i 1 1 Along 2, nd using the bove theorem, 1 zdz = (t 2 + 2t + it)(2t + 2 + i)dt = 4 + 2i c 1 - Ex. Show tht 1 zdz = πi; but 2 zdz = 4i where 1 is the semicirculr pth from 1 to 1, nd 2 is the polygonl pth from 1 to 1.
omplex integrls 1 : z(t) = cos t + i sin t nd dz = (sin t + i cot t)dt, for t π. Using the previous theorem, π π zdz = ( cos t + i sin t)(sin t + i cot t)dt = i (cos 2 t + sin 2 t)dt = πi 1 2 : There re three segments: (i) z 1 (t) = 1 + it, dz 1 = idt, nd f(z 1 (t)) = 1 it; (ii) z 2 (t) = 1 + 2t + i, dz 2 = 2dt, nd f(z 2 (t)) = 1 + 2t i; (iii) z 3 (t) = 1 + i(1 t), dz 3 = idt, nd f(z 3 (t)) = 1 i(1 t), where t 1. 1 1 1 zdz = ( 1 it)idt + ( 1 + 2t i)2dt + [1 i(1 t)]( i)dt = 4i 2 - Note tht the vlue of the contour integrl long 1 isn t the sme s the vlue of the contour integrl long 2, lthough both integrls hve the sme initil nd terminl points.
omplex integrls - Properties f(z)dz = f(z)dz f(z) + g(z)dz = f(z)dz + g(z)dz ( + ib)f(z)dz = ( + ib) f(z)dz 1 + 2 f(z)dz = 1 f(z)dz + 2 f(z)dz, where the terminl point of 1 coincides with the initil point of 2 (the contour = 1 + 2 is continution of 1 ) - If the contour hs two prmeteriztions : z 1 (t) = x 1 (t) + iy 1 (t), for t b, nd : z 2 (τ) = x 2 (τ) + iy 2 (τ), for c τ d, nd there exists differentible function τ = φ(t) such tht c = φ(), d = φ(b), nd φ (t) >, for < t < b, then we sy tht z 2 (τ) is reprmeteriztion of the contour. If f is continuous on, then we hve f(z 1 (t))z 1(t)dt = d c f(z 2 (τ))z 2(τ)dτ
omplex integrls - Theorem (Absolute vlue inequlity): If f(t) = u(t) + iv(t) is continuous function of the rel prmeter t, then f(t)dt f(t) dt, (12) - Theorem (ML inequlity): If f(z) = u(x, y) + iv(x, y) is continuous on the contour, then f(z)dz ML (13) where L is the length of the contour nd M is n upper bound for the modulus f(z) on ; tht is, f(z) M for ll z.
omplex integrls - Ex. Show tht 1 dz 1 z 2 +1 2, where is the stright-line segment from 5 2 to 2 + i. Solution: Here z 2 + 1 = z i z + i, nd from the below figure, z i 2 nd z + i 5 for z on. Thus, since L = 1, we hve the desired inequlity.
The uchy-gourst theorem - Simple closed contour : Interior of + exterior of. - A domin is connected open set. - A domin is simply connected domin if the interior of ny simple closed contour contined in D is contined in D (i.e., no holes in simply connected domin) - A domin tht is not simply connected is multiply connected domin
omplex integrls - Positive oriented contour vs. negtively oriented contour - Theorem (Green s theorem): Let be simple closed contour with positive orienttion nd let R be the domin tht forms the interior of. If P nd Q re continuous nd hve continuous prtil derivtives P x, P y, Q x, nd Q y t ll points on nd R, then P (x, y)dx + Q(x, y)dy = [Q x (x, y) P y (x, y)]dxdy (14) R
omplex integrls - Theorem (uchy-gourst theorem): Let f be nlytic in simply connected domin D. If is simple closed contour tht lies in D, then f(z)dz = - Ex. Since exp(z), cos z, nd sin z where n is positive integer re ll entire functions. The uchy-gourst theorem implies tht for ny simple closed contour, expzdz = ; cos zdz = ; nd zn dz =.
omplex integrls - Ex. If is simple closed contour such tht the origin does not lie interior to, then there is simple connected domin D tht contins in which f(z) = 1 z is nlytic, s is indicted in the following figure.the uchy-gourst theorem n implies tht 1 dz =. z - Theorem (Deformtion n of contour): Let 1 nd 2 be two simple closed positively oriented contours such tht 1 lies interior to 2. If f is nlytic in domin D tht contins both 1 nd 2 nd the region between them, then f(z)dz = f(z)dz 1 2
omplex integrls - orollry: Let z denote fixed complex vlue. If is simple closed contour with positive orienttion such tht Z lies interior to, then dz z z = 2πi nd dz (z z =, where n is ny integer except n = 1. ) n - Theorem (Extended uchy-gourst theorem): Let, 1,..., n be simple closed positively oriented contours with the properties tht k lies interior to for k = 1, 2,..., n, nd the interior of k hs no points in common with the interior of j if k j. Let f be nlytic on domin D tht contins ll the contours nd the region between nd 1 + 2 + + n. Then, n f(z)dz = f(z)dz k k=1
omplex integrls - Ex. Show tht z 2 dz = 2πi, where is shown in the below. z 2 z Solution: Prtil frctions z 2 z 2 z dz = 2 1 z dz 1 dz (15) z 1 Using the uchy-gourst theorem, the first integrl 1 2 z dz = 2 1 1 z dz + 2 1 2 z dz = 2 1 dz + = 4πi 1 z Similrly, we find tht 1 z 1 dz = 1 1 z 1 dz 2 1 z 1 dz = 2πi = 2πi
The fundmentl theorems of integrtion - Let f be nlytic in the simply connected domin D. Then, n ntiderivtive F cn be constructed by contour integrtion. A consequence will be the fct tht in simply connected domin, the integrl of n nlytic function f long ny contour joining z 1 nd z 2 is the sme, nd its vlue is given by F (z 2 ) F (z 1 ). - Theorem (Indefinite integrls, or ntiderivtives) Let f be nlytic in the simply connected domin D. if z is fixed vlue in D nd if is ny contour in D with initil point z nd terminl point z, then the function z F (z) = f(ξ)dξ = f(ξ)dξ (16) z is well defined nd nlytic in D, with its derivtive given by F (z) = f(z). - Theorem (Definite integrls) Let f be nlytic in the simply connected domin D. If z nd z 1 re ny two points in D joined by contour, then z1 f(z)dz = f(z)dz = F (z 1 ) F (z ) (17) z where F is ny ntiderivtive of f in D.
The fundmentl theorems of integrtion - Ex. Show tht dz = 1 + i, where is the line segment joining 4 to 8 + 6i. 2z 1/2 Solution: We know tht if F (z) = z 1/2, then F (z) = 1. Note tht cn be 2z 1/2 contined in the simply connected domin D 4 (6 + 3i). So, since f(z) = 1 is 2z nlytic in D 1/2 4 (6 + 3i), using the previous theorem, dz 2z = (8 + 1/2 6i)1/2 4 1/2 = 1 + i - Ex. Show tht cos zdz = sin 1 + i sinh 1, where is the line segment between 1 nd i. Solution: The ntiderivtive of f(z) = cos z is F (z) = sin z. Becuse F is entire, we hve cos zdz = i 1 cos zdz = sin i sin 1 = sin 1 + i sinh 1. -
Integrl representtions for nlytic functions - Theorem (uchy s integrl formul) Let f be nlytic in the simply connected domin D nd let be simply closed positively oriented contour tht lies in D. If z is point tht lies interior to, then - Ex. Show tht + 1 () expz z 1 f(z ) = 1 2πi f(z) z z dz (18) dz = i2πe. From e = f(1) = 1 2πi expz dz, the result z 1 is direct. - Theorem (Leibniz s rule) Let G be n open set nd let I : t b be n intervl of rel numbers. Let g(z, t) nd its prtil derivtive g z (z, t) w.r.t z be g(z, t)dt is nlytic for z in G, nd F (z) = g z(z, t)dt. - Theorem (uchy s integrl formul for derivtives) Let f be nlytic in the simply connected domin D nd let be simply closed positively oriented contour tht lies in D. If z is point tht lies interior to, then for ny integer n, continuous functions for ll z in G nd ll t in I. Then, F (z) = f n (z) = n! 2πi f(ξ) dξ (19) ξ z
Theorems of Morer nd Liouville - The existence of n ntiderivtive for continuous function is equivlent to the sttement tht the integrl of f is independent of the pth of integrtion. - Theorem (Morer s theorem): Let f be continuous function in simply connected domin D. If f(z)dz = for every closed contour in D, then f is nlytic in D. - Theorem (Guss s men vlue theorem): If f is nlytic in simply connected domin D tht contins the circle R (z ), then f(z ) = 1 2π 2π f(z + Re iθ )dθ - Theorem (Mximum modulus principle): Let f be nlytic nd nonconstnt in the domin D. Then, f(z) does not ttin mximum vlue t ny point z in D. - Let f be nlytic nd nonconstnt in the bounded domin D. If f is continuous on the closed region F tht consists of D nd ll its boundry points B, then f(z) ssumes its mximum vlue, nd does so only t point(s) z on the boundry B.
Theorems of Morer nd Liouville - Ex. Let f(z) = z+b. If we set our domin D to be D 1 (), then f is continuous on the closed region D 1 () = {z : z 1}. Prove tht mx z 1 f(z) = + b nd tht this vlue is ssumed by f t point z = e iθ on the boundry of D 1 (). Solution: We know f(z) = z + b z + b + b. If we choose z = e iθ, where θ rg b rg, then rg(z ) = rg + rg z = rg + (rg b rg ) = rg b so the vectors z nd b lie on the sme ry through the origin. Thus, f(z) = z + b = z + b = + b. - Theorem (uchy s inequlities): Let f be nlytic in the simply connected domin D tht contins the circle R (z ). If f(z) M holds for ll points z R (z ), then f n (z ) n!m, for n = 1, 2,.... R n
Thnk You!! (Q & A)