MATH 185: COMPLEX ANALYSIS FALL 2009/10 PROBLEM SET 5 SOLUTIONS. cos t cos at dt + i

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1 MATH 85: COMPLEX ANALYSIS FALL 9/ PROBLEM SET 5 SOLUTIONS. Let R nd z C. () Evlute the following integrls Solution. Since e it cos t nd For the first integrl, we hve e it cos t cos t cos t + i t + i. sin t cos t. cos t cos t [cos(t + t) + cos(t t)], sin t sin t [sin(t + t) + sin(t t)], we obtin e it cos t ( ) sin( + )t sin( )t + i ( cos( + )t ( ) sin( + ) sin( ) + i ( cos( + ) ) cos( )t cos( ) For the second integrl, we first pply the formul for complex squre roots in Chpter nd then find prtil frction decomposition. nd so Therefore t + i ( ) t + i (t + t + i ( + i) log i ) α ( + i), β ( + i). ( t + i ) (b) Show tht if Re z >, then the integrl tz exists nd t z + Re z. Solution. Note tht for t, nd so Dte: October 3, 9 (Version.). t z t Re z t z t z α β t + i + t + i ( ( + i) log t + i ) t Re z. )..

2 Since for Re z > the lst integrl converges, tz exists nd the required inequlity follows from (c) Show tht if <, then t Re z tre z+ + Re z cos it t + Re z. nd thus the (improper) integrl t cos it converges bsolutely. Solution. Note tht cos it cosh t nd for t, we hve cosh t <. Hence cos it t cos it t cosh t t t. Since the lst integrl converges for <, we see tht t cos it converges bsolutely.. () For k,, 3, evlute the following integrls Re(z) dz, k z dz, k t dz k z long the curves from the point z to z i in the counter clockwise direction s described in Figure. Figure. Left: is long the boundry of the squre: {x + iy x, y }. Center: is long the boundry of the circle: {e it t π/}. Right: 3 is long the line segment: {( t) + it t }. Solution. Re(z) dz nd For k,, 3, we hve i + Re(z) dz i i i π/ π/ π/ + iπ 4 ; Re(i + t) i Re(e it )e it cos t(cos t + i sin t) cos t π/ cos t sin t t + i;

3 nd Re(z) dz Re[( t) + it](i ) 3 3 ( + i) + i. ( t) The hrd wy to evlute the second integrl would be from definition like bove, i.e. for k,, 3, z dz i t + t 3 i 3 ; nd π/ π/ z dz i e it e it (cos 3t + i sin 3t) 3 i 3 ; nd z dz [( t) + it] ( + i) ( ) ( + i) t + i t i t 3 i 3. The nswers re ll the sme. Why? Becuse of the ntiderivtive theorem tht we proved in lecture f(z) z hs ntiderivtive F (z) z 3 /3, which gives the esy wy to evlute ll three integrls in one go: k z dz F (i) F () 3 i 3. The lst integrl cn be evluted nlogously. (b) Let, b >. By considering pth long the ellipse { cos t + ib sin t t π} or otherwise, show tht π cos t + b sin t π b. (Hint: Recll from lecture tht D(,r) z dz πi nd emulte trick we used in the proof of Cuchy s theorem.) Solution. We shll choose the following prmetriztion for the ellipse z : [, π] C, z(t) cos t + ib sin t. By the trick tht we discussed in lecture, where we dded bck nd forth pth (whose integrls cncel exctly) between nd circle D(, r), we see tht dz z dz πi. (.) z Now dz π z π D(,r) sin t + ib cos t cos t + ib sin t (b ) sin t cos t cos t + b sin t + i π b cos t + b sin. (.) t 3

4 Compring (.) nd (.), we see tht the rel prt of (.) must be nd the imginry prt must equl π, i.e. π b cos t + b sin π. t Dividing by b then gives the required expression. (c) Evlute the following integrls D z z dz nd where D nd D re the curves in Figure. D z dz Figure. Left: D is closed curve in the counter clockwise direction long the semicircle {Re it t π} nd the line segment {z R Re z R, Im z }. Right: D is closed curve in the counter clockwise direction long the semicircle {e it π/ t π/} nd the line segment {z Im z, Re z }. Solution. We hve D z z dz R R π x x dx + i R R 3 πi x dx + R R 3 e it e it π x dx + i R 3 nd D z dz π/ π/ e it ie it + ( it)i πi. 3. Let be smooth curve. () Suppose [, b]. Prove tht for ny integrble function f : [, b] C, b f(t) b f(t). Wht cn you deduce bout the integrbility of f if f is integrble? 4

5 Solution. Let f(t) u(t) + iv(t). Then b f(t) b b b b u(t) + i u(t) i b b [u(t) iv(t)] f(t). v(t) v(t) Hence if f is integrble over [, b], then so is f. (b) Let f be function tht is continuous on. Consider, the imge of under complex conjugtion z z, i.e. reflected bout the rel xis. Show tht the function z f(z) is lso continuous on nd f(z) dz f(z) dz. Solution. Note tht complex conjugtion, σ : C C, σ(z) z, is bijective (esy) nd continuous function. Continuity follows from the fct tht for ε >, σ(z) σ(z ) z z z z z z < ε for z z < δ ε. Since for z we hve z, nd since f is continuous on, the function z f(z) is continuous on being composition of three continuous functions σ f σ. For the integrl, first note tht if we let w z, then f(z) dz f(w) dw f(w) dw. (3.3) Let be given by w : [, b] C, w(t) x(t) + iy(t) nd let f(w(t)) u(t) + iv(t). Then b f(w) dw f(w(t))w (t) b b b b b (u(t) + iv(t))(x(t) + iy (t)) (u(t) iv(t))(x(t) iy (t)) (u(t)x (t) v(t)y (t)) i (u(t)x (t) v(t)y (t)) + i b b (u(t) + iv(t))(x(t) + iy (t)) Together, (3.3) nd (3.4) imply tht f(z) dz (v(t)x (t) + u(t)y (t)) (v(t)x (t) + u(t)y (t)) f(w) dw. (3.4) f(w) dw. 5

6 Since w is dummy vrible, we could eqully well hve written z in plce of w on the rhs, tking conjugte of both sides then yields the required expression. (c) Suppose is the positiviely oriented (i.e. going counter clockwise) circle z : [, π] C, z(t) e it. Show tht f(z) dz f(z) dz z. Solution. By the proof of (b), we hve f(z) dz π π π f(e it )ie it f(e it )ie it f(e it ) ieit eit f(z) dz z. 4. Let r > nd be the curve z : [, π] C, z(t) re it. Let f : D(, r) C be continuous. () For n N, let n be the curve z n : [, π] C, z n (t) ( /n)re it. Prove tht f(z) dz lim f(z) dz. n n (Hint: The result we discussed in clss is of the form f lim n f n. Find wy to use this.) Solution. We hve π (( f ) ) ( re it ) ire it n n lim f(z) dz lim n n n π π lim n [ (( f ) ) ( re it ) ] ire it n n f(re it )ire it f(z) dz where we hve used the continuity of integrtion (i.e. lim n f n lim n f n ) tht we proved in lecture. (b) Show tht π lim f(re it f(z) ) πf() nd lim dz πif(). r r z (Hint: Emulte the proof of the forementioned result.) Solution. Since the function f is continuous t, put z re it, we hve tht for ny ε > there exists δ > such tht f(re it ) f() < ε π 6

7 for r < δ nd t π. Therefore π f(re it π ) πf() [f(re it ) f()] π < ε π ε f(re it ) f() π for r < δ, t π. Likewise we hve f(z) π dz πif() z f(re it ) π re it ire it i f() π i [f(re it ) f()] for r < δ, t π. π < ε π ε f(re it ) f() π 5. Let R > nd f : D(, R) C be nlytic. () Suppose t lest one of the following four conditions is true (i) Re f (z) > for ll z D(, R); (ii) Re f (z) < for ll z D(, R); (iii) Im f (z) > for ll z D(, R); (iv) Im f (z) < for ll z D(, R). Show tht f is injective on D(, R). Solution. Suppose (i) is stisfied. Let, b D(, R) be two rbitrry points nd b. Let be the line segment connecting nd b, ie. z : [, ] C, z(t) + t(b ). Note tht D(, ). By Proposition 3.9 in our lectures, f(b) f() f (z) dz f (z(t))(b ). So f(b) f() b f (z(t)) Since Re f (z) > for ll z D(, R), nd thus Re f (z(t)) + i Re f (z(t)) f (z(t)), Im f (z(t)). which implies tht f() f(b). For (ii), (iii), (iv), pply the result of (i) to f, if, nd if respectively; note tht these re ll nlytic on D(, R) nd stisfy the condition of (i). 7

8 (b) Suppose [Re f (z)][im f (z)] for ll z D(, R). Show tht f is injective on D(, R). Solution. Since [Re f (z)][im f (z)] for ll z D(, R), it must be the cse tht Re f (z) nd Im f (z) for ll z D(, R). Now suppose there exist z, z D(, R) such tht Re f (z ) < nd Re f (z ) >. Define γ : [, ] D(, R) by γ(t) z + t(z z ) (this line is contined in D(, R) since D(, R) is convex). Observe tht γ is clerly differentible on [, ]. Also observe tht by Theorem.9, since f is nlytic on D(, R), so is f. From these two observtions, conclude tht f γ is differentible on [, ], which implies tht f γ is continuous on [, ]. Then, Re f γ is lso continuous on [, ]. Then by the Intermedite Vlue Theorem, since Re f γ() Re f (γ()) Re f (z ) < nd Re f γ() Re f (γ()) Re f (z ) >, there exists some c (, ) such tht Re f γ(c). But this is contrdiction since for ll z D(, R), Re f (z). Then it must be the cse tht either Re f (z) > for ll z D(, R) or Re f (z) < for ll z D(, R). Then by the result of prt (), since either property (i) or (ii) is stisfied, conclude tht f is injective on D(, R). 8