INDIAN INSTITUTE OF TECHNOLOGY BOMBAY MA205 Complex Analysis Autumn PDF Free Download

Lecture 6: Line Integrls INDIAN INSTITUTE OF TECHNOLOGY BOMBAY MA205 Complex Anlysis Autumn 2012 August 8, 2012

Lecture 6: Line Integrls Lecture 6: Line Integrls

Lecture 6: Line Integrls Integrls of complex vlued functions Definition Given f : [, b] C continuous. We define b f (t) dt := b b Re (f (t)) dt +ı Im (f (t)) dt. (1)

Lecture 6: Line Integrls Integrls of complex vlued functions Definition Given f : [, b] C continuous. We define b f (t) dt := b b Re (f (t)) dt +ı Im (f (t)) dt. (1) Stndrd properties of Riemnn integrls of rel vlued functions ll hold for the bove integrl of complex vlued function. For instnce, linerity properties re esy to verify.

Lecture 6: Line Integrls Integrls of complex functions However, you hve to be cutious bout those properties which involve inequlites. Here is something which my be new for you nd which is indeed most fundmentl for us now.

Lecture 6: Line Integrls Integrls of complex functions Then w = r = e ıθ w. Tht is, b b f (t) dt = r = e iθ f (t) dt = = b Re (e iθ f (t)) dt b b e iθ f (t) dt f (t) dt.

Lecture 6: Line Integrls Let U be n open subset in C. By smooth prmeterised curve in U, we men function γ : [, b] U which hs continuous derivtive γ(t) 0, throughout the intervl.

Lecture 6: Line Integrls Let U be n open subset in C. By smooth prmeterised curve in U, we men function γ : [, b] U which hs continuous derivtive γ(t) 0, throughout the intervl. Here the dot on the top denotes differentition with respect to t. This just mens tht γ(t) = (x(t), y(t)) U nd ẋ, ẏ exist nd re continuous, nd (ẋ(t), ẏ(t)) (0, 0).

Lecture 6: Line Integrls Exmple The curve γ : t (t 2, t 3 ) (OR t t + ιt 3 ) is given by function which hs continuous derivtive. However, t t = 0, γ = (0, 0). Therefore, if the domin of the function is llowed to include the point 0 then it is not smooth curve. Otherwise it is smooth curve.

Lecture 6: Line Integrls Exmple Consider the curves C 1 (t) = e 2πıt, C 2 (t) = e 4πıt, C 3 (t) = e 2πıt, 0 t 1. Ech of them hve its imge equl to the unit circle. However, they re ll different curves s prmetrized curves.

Lecture 6: Line Integrls Sense in prmeterized curve Remrk Geometriclly, by curve we often men the imge set of curve s given bove. A prmetrized curve is much refined notion thn tht. For instnce, observe tht the prmetriztion utomticlly defines sense of orienttion on the curve, the wy in which the geometric curve is being trced.

Lecture 6: Line Integrls We shll fix the following nottion for certin prmeterized curves: Given z 1, z 2 C, write [z 1, z 2 ] for the curve given by t (1 t)z 1 + tz 2, 0 t 1.}

Lecture 6: Line Integrls We shll fix the following nottion for certin prmeterized curves: Given z 1, z 2 C, write [z 1, z 2 ] for the curve given by t (1 t)z 1 + tz 2, 0 t 1.} The circle with centre w nd rdius r trced exctly once in the counterclockwise sense will be denoted by z w = r := {t w + re 2πıt, 0 t 1.}

Contour integrtion Lecture 6: Line Integrls Let γ be smooth curve in U. Then for ny continuous function f : U C we define the contour integrl, or line integrl of f long γ to be γ f dz := b f (γ(t)) γ(t) dt. (3)

Contour Integrtion Lecture 6: Line Integrls Observe tht γ(t) is complex number for ech t, sy, γ(t) = x(t) + ıy(t), then γ(t) = ẋ(t) + ıẏ(t).

Lecture 6: Line Integrls The expressions dx, dy etc. Similrly if we write f (z) = u(z) + ıv(z), then f (γ(t)) = u(γ(t)) + ıv(γ(t)).

Lecture 6: Line Integrls The expressions dx, dy etc. Similrly if we write f (z) = u(z) + ıv(z), then f (γ(t)) = u(γ(t)) + ıv(γ(t)). Hence the of the bove definition cn lso be expressed s b f (z)dz := (u(γ(t))ẋ(t) v(γ(t))ẏ(t)) dt γ +ı b (u(γ(t))ẏ(t) + v(γ(t))ẋ(t)) dt.

Lecture 6: Line Integrls The expressions dx, dy etc. Therefore it follows tht dx + ıdy = dz. There expressions re clled 1-forms. For us they re good for crrying out integrtion: indictors of which vrible is being integrted. This is the only justifiction for the nme complex integrls which mny uthors use.

Exmples Lecture 6: Line Integrls (1) Compute the vlue of [0,1+i] x dz.

Exmples Lecture 6: Line Integrls (1) Compute the vlue of [0,1+i] x dz. Sol: Here the curve γ is the line segment from 0 to 1 + i. Recll tht this curve is given by: γ(t) = (1 + ı)t, 0 t 1.

Exmples Lecture 6: Line Integrls (1) Compute the vlue of [0,1+i] x dz. Sol: Here the curve γ is the line segment from 0 to 1 + i. Recll tht this curve is given by: γ(t) = (1 + ı)t, 0 t 1. Then γ(t) = 1 + i for ll t nd hence by definition γ x dz = 1 0 x(γ(t)) γ(t) dt = 1 0 t(1+ı) dt = 1 + ı 2.

Exmples Lecture 6: Line Integrls (2) Let us compute C z n dz, where C is ny circle with origin s centre nd oriented counter-clockwise.

Exmples Lecture 6: Line Integrls (2) Let us compute C z n dz, where C is ny circle with origin s centre nd oriented counter-clockwise. Sol: We hve C : γ(t) = re ı2πt, 0 t 1.

Exmples Lecture 6: Line Integrls (2) Let us compute C z n dz, where C is ny circle with origin s centre nd oriented counter-clockwise. Sol: We hve C : γ(t) = re ı2πt, 0 t 1. By definition, we hve, 1 z n dz = r n e 2nπıt (2πı)re 2πıt dt C 0 1 1 = r n+1 e 2πı(n+1)t dt = 2πır n+1 e 2πı(n+1) dt. 0 0

Exmples Lecture 6: Line Integrls This is esily seen to be = 0 if n 1 nd = 2πı if n = 1.

Exmples Lecture 6: Line Integrls This is esily seen to be = 0 if n 1 nd = 2πı if n = 1. Shifting the origin to z =, tking n = 1 we obtin z =r dz z = 2πı. (4)

Lecture 6: Line Integrls Some bsic properties of the integrl: 1. Invrince Under Chnge of Prmeteriztion Let τ : [α, β] [, b] be continuously differentible function with τ(α) =, τ(β) = b, τ(t) > 0, t. Then γ τ f (z) dz γ f (z) dz (5)

Lecture 6: Line Integrls Chnge of prmeteriztion: LHS := β α RHS = f (γ τ(t)) b d(γ τ) (t) dt. dt f (γ(s)) γ ds ds

Lecture 6: Line Integrls Chnge of prmeteriztion: LHS := β α RHS = f (γ τ(t)) b d(γ τ) (t) dt. dt f (γ(s)) γ ds ds Now mke the substitution s = τ(t) nd use the fct ds = τdt.

Lecture 6: Line Integrls (2) Linerity For ll α, β C γ (αf + βg)(z) dz = α f (z) dz + β g(z) dz (6) γ γ

Lecture 6: Line Integrls (3) Additivity Under Sub-division or Conctention If < c < b nd γ 1 = γ [,c], γ 2 = γ [c,b], re the restrictions to the respective sub-intervls of prmeterized curve γ : [, b] C, then γ f (z) dz := f (z) dz + γ 1 f (z) dz γ 2 (7)

Lecture 6: Line Integrls (4) Orienttion Respecting We lso hve, f (z) dz = γ 1 γ f (z) dz (8) where γ 1 is the curve γ itself trced in the opposite direction, viz., γ 1 (t) = γ( + b t).

Lecture 6: Line Integrls To see this, put t = + b s. Then, L.H.S. = = b = b b f (γ 1 (s)) dγ 1 (s) ds ds f (γ(t))( γ(t))( dt) f (z) dz = R.H.S.

Lecture 6: Line Integrls (5) Interchnge of order of integrtion nd limit If {f n } is sequence of continuous functions uniformly convergent to f then the limit nd integrtion cn be interchnged viz., lim f n (z) dz = f (z) dz. (9) n γ γ This follows from the corresponding property of Riemnn integrtion.

Lecture 6: Line Integrls (6) Term-by-term Integrtion From (5) it lso follows tht whenever we hve uniformly convergent series of functions then term-by-term integrtion is vlid. γ ( ) f n (z) dz = n n ( γ ) f n (z) dz (10)

Lecture 6: Line Integrls (7)Fundmentl Theorem of integrl clculus Suppose g is complex differentible in U. Then for ll smooth curves γ : [, b] U we hve g (z)dz = g(γ(b)) g(γ()). (11) γ For the composite function g γ is differentible in [, b]. Therefore b g d (z)dz = (g γ(t))dt = g(γ(b)) g(γ()). dt γ

Contours Lecture 6: Line Integrls Definition By contour, we men the conctention (composite) γ = γ 1 γ 2 γ k of finite number of smooth prmeterized curves γ i tken in fixed order.

Lecture 6: Line Integrls Contours Definition By contour, we men the conctention (composite) γ = γ 1 γ 2 γ k of finite number of smooth prmeterized curves γ i tken in fixed order. Observe tht γ is continuously differentible except t finitely mny points of the intervl, where even the continuity lso my brek.

Lecture 6: Line Integrls Contours Property (7) comes to our help nd sys tht the only nturl wy to define the integrls over rbitrry contours is by the formul γ f (z) dz := k j=1 γ j f (z) dz. (12)

Lecture 6: Line Integrls Contours Property (7) comes to our help nd sys tht the only nturl wy to define the integrls over rbitrry contours is by the formul γ f (z) dz := k j=1 γ j f (z) dz. (12) Verify directly tht ll the bsic properties mentioned bove for line integrls is vlid for contour integrls s well.

Length of countor Lecture 6: Line Integrls Definition Length of contour: Let γ : [, b] R 2, γ(t) = (x(t), y(t)) be continuously differentible rc. Then the rc-length of γ is obtined by the integrl L(γ) := b γ(t) dt = b [(ẋ(t)) 2 + (ẏ(t)) 2 ] 1/2 dt (13)

Length of countor Lecture 6: Line Integrls It is esily checked tht L(γ) is independent of the choice of prmeteriztion of γ s discussed erlier.

Length of countor Lecture 6: Line Integrls It is esily checked tht L(γ) is independent of the choice of prmeteriztion of γ s discussed erlier. Sometimes we use the following complex nottion for (13): If γ(t) = z(t) = x(t) + ıy(t), this becomes L(γ) := dz (14) γ

Exmples Lecture 6: Line Integrls As simple exercise, let us compute the length of the circle C r := z(θ) == (r cos θ, r sin θ) 0 θ 2π. L(C r ) = dz C r = = r 2π 0 2π 0 (r 2 sin 2 θ + r 2 cos 2 θ) 1/2 dθ dθ = 2πr.

Lecture 6: Line Integrls A nottion nd consequence We now introduce the nottion: b f (z) dz := f (γ(t)) γ(t) dt. (15) γ for ny continuous function f nd ny contour γ.

Lecture 6: Line Integrls A nottion nd consequence We now introduce the nottion: b f (z) dz := f (γ(t)) γ(t) dt. (15) γ for ny continuous function f nd ny contour γ. Note tht s consequence of (2), it follows tht f (z)dz f (z)dz (16) γ γ

M-L Inequlity Lecture 6: Line Integrls Theorem M-L Inequlity Let U be n open set in C, f be continuous function on U nd γ : [, b] U be contour in U. Let M = sup{ f (γ(t)) : t b}. Then γ f (z) dz ML(γ). (17)

M-L Inequlity Lecture 6: Line Integrls Proof: This is n immedite consequence of (2) b f (z) dz = f (γ(t)) γ(t) dt γ M b γ(t) dt = ML(γ).

Lecture 6: Line Integrls Continuity of the Integrls Theorem Let Ω be n open set in R n nd g : Ω [, b] C be continuous function. Put φ(p) = b g(p, t) dt, P Ω. Then φ : Ω C is continuous function.

Lecture 6: Line Integrls Continuity of Integrls Proof: Let B be closed bll of rdius, sy δ 1 > 0, round point P 0 Ω such tht B Ω.

Lecture 6: Line Integrls Continuity of Integrls Proof: Let B be closed bll of rdius, sy δ 1 > 0, round point P 0 Ω such tht B Ω. Then B [, b], is closed nd bounded subset of Euciden spce. Hence, g restricted to this set is uniformly continuous.

Lecture 6: Line Integrls Continuity of Integrls This mens tht given ɛ > 0, we cn find δ 2 > 0 such tht g(p 1, t 1 ) g(p 2,, t 2 ) < ɛ/(b ) for ll (P i, t i ) B [, b] whenever (P 1, t 1 ) (P 2, t 2 ) < δ 2. Now let δ = min{δ 1, δ 2 } nd P P 0 < δ.

Lecture 6: Line Integrls Differentition Under Integrl Sign Theorem Differentition Under the Integrl Sign Let U be n open subset of C nd g : U [, b] C be continuous functions such tht for ech t [, b], the function z g(z, t) is complex differentible nd the mp g : U [, b] C is continuous. z Then f (z) = b g(z, t)dt is complex differentible in U nd f (z) = b g (z, t) dt. z

Lecture 6: Line Integrls Differentition Under Integrl Sign Proof: Given z 0 U, we need to show tht [ f (z) f (z0 ) b ] lim z z 0 z z 0 z g(z 0, t)dt = 0.

Lecture 6: Line Integrls Differentition Under Integrl Sign Proof: Given z 0 U, we need to show tht [ f (z) f (z0 ) b ] lim z z 0 z z 0 z g(z 0, t)dt = 0. Put h(z, t) = g(z, t) f (z 0, t)) (z z 0 ) z g(z 0, t).

Lecture 6: Line Integrls Differentition Under Integrl Sign Let r > 0 be such tht B = B r (z 0 ) U. Then B [, b] is closed nd bounded nd hence g z is uniformly continuous on it.

Lecture 6: Line Integrls Differentition Under Integrl Sign Let r > 0 be such tht B = B r (z 0 ) U. Then B [, b] is closed nd bounded nd hence g z is uniformly continuous on it. Hence, given ɛ > 0 we cn choose 0 < δ < r such tht g z (z 1, t) g z (z 2, t) < ɛ (18) b for ll t [, b] nd z 1, z 2 B such tht z 1 z 2 < δ.

Lecture 6: Line Integrls Differentition Under Integrl Sign Now, let 0 < z z 0 < δ. Then h(z, t) is equl to ( g g (w, t) w z (z 0, t)) dw ɛ z z 0, [z 0,z] by M-L inequlity.

Lecture 6: Line Integrls Differentition Under Integrl Sign Now, let 0 < z z 0 < δ. Then h(z, t) is equl to ( g g (w, t) w z (z 0, t)) dw ɛ z z 0, [z 0,z] by M-L inequlity. 1 b h(z, t)dt z z 0 < ɛ. This proves the theorem.

Vnishing derivtive. Lecture 6: Line Integrls Theorem Let U be convex open set, f : U C be C-differentible function such tht f (z) = 0 for ll z U. Then f is constnt function on U.

Vnishing derivtive. Lecture 6: Line Integrls Theorem Let U be convex open set, f : U C be C-differentible function such tht f (z) = 0 for ll z U. Then f is constnt function on U. Proof: Fix z 0 U nd for every point z U define g(t) = f ((1 t)z 0 + tz). Then g : [0, 1] C is differentible function nd g (t) = 0 by chin rule. This implies tht g(1) = g(0) Tht is the sme s sying f (z) = f (z 0 ).

INDIAN INSTITUTE OF TECHNOLOGY BOMBAY MA205 Complex Analysis Autumn 2012