z(t) = z 1 (t) + t(z 2 z 1 ) z(t) = 1 + i + t( 2 3i (1 + i)) z(t) = 1 + i + t( 3 4i); 0 t 1

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1 (4.): ontours. Fnd an admssble parametrzaton. (a). the lne segment from z + to z 3. z(t) z (t) + t(z z ) z(t) + + t( 3 ( + )) z(t) + + t( 3 4); t (b). the crcle jz j 4 traversed once clockwse startng at z 4 +. z(t) + 4e t ; t (c). the arc of the crcle jzj R lyng n the second quadrant, from z R to z R. z(t) Re t ; t (d). the segment of the parabola y x from the pont (; ) to the pont (3; 9). z(t) t + t ; t 3 3. Show that the ellpse x a admssble parametrzaton. + y b s a smooth curve by producng an Let x a cos t; y b sn t. Substtute these values n the ellpse to obtan cos t + sn t, a true statement for all t. 5. Identfy the nteror of the curve. Is t postvely orented? Yes. 7. Parametrze the contour consstng of the permeter of the square wth vertces ; ; + ; and +. What s the length of ths contour? : z(t) + t ; t : z(t) + (t ) ; t 3 : z(t) + (t ) ; t 3 4 : z(t) + (t 3) ; 3 t 4 l( ) Parametrze the contour. Also parametrze.

2 The contour conssts of of two smooth curves - the straght lne pece and the sem-crcle : : z (t) ( + ) + t( ( + )) + + t( ) ; t : z (t) e t ; t : z (t) e t ; t : z (t) + (t )( + ) ; t 9. Parametrze the barbell-shaped contour wth ntal pont and termnal pont. rcle : z(t) + e t ; t Straght lne : z(t) + (t ) ; t rcle 3 : z(t) e (t ) ; t 3 Straght lne 4 : You can reduce the above parametrzaton to the one gven n the back of the text. Namely, 8 < + e 6t f t 3 z(t) + 6(t 3) f 3 t 3 : e 6(t 3) f 3 t : Follow Wen s soluton dscussed n class.. Fnd the length of the contour parametrzed by z(t) 5e 3t ; t. z (t) 5e 3t ) jz (t)j 5: l( ) R jz (t)j dt 5: (4.): ontour Integrals 3. Evaluate the followng ntegrals. (a). (t + t )dt t + 3 t3 + 3

3 (b). ( + ) ( + ) cos(t)dt ( + ) snh sn(t) ( + ) sn() (c). ( + t) 5 dt ( + t) 6 ( + ) ( + ) 6 3 (d). t (t + ) dt 4+ u 4+ u dt ; u t + ) du tdt lmts: u to u Evaluate where : jz (4 + ) (z ) + + 3(z ) dz z j 4 traversed once counterclockwse. 6 (z ) + + 3(z ) dz z 6() + () + + 4: 7. ompute R Re zdz along the drected lne segment from z to z +. The lne can be parametrzed as z(t) ( + )t t + t for t ) z (t) ( + )dt R R Re zdz (+) t ( + )dt [t ] (+) : 8. Let be the permeter of the square wth vertces at the ponts z ; z ; z +; and z traversed once n that order. Show that R ez dz. 3

4 Refer to the class notes for a soluton that uses parametrzaton. For an alternate soluton consder the followng: e z dz e z dz + + e z dz + e z dz + + e z dz [e z ] + [e z ] + + [e z ] + + [e z ] (e ) + (e + e) + (e e + ) + ( e ) : 9. Evaluate R (x xy)dz over the contour : z t + t ; t ) z ( + t)dt. R R (x xy)dz (t t3 )( + t)dt R [(t + 4t4 ) + (t t 3 )]dt h t + 4t5 5 + t 3 t : I I. True or False: zdz z dz. jzj jzj The statement s true snce we can use the fact that all z satsfy jzj. Indeed, I jzj zdz I jzj z z z dz I jzj I jzj z dz jzj z dz. 3. ompute R (jz + j z)dz along the semcrcle z + e t ; t. (jz + j z)dz ( + e t ( + e t ) e t dt e t + ( + e t )) e t dt ( e t e t )dt e t [ ]. (4.3): Independence of Path. alculate the followng. 4

5 (a). R (3z 5z + )dz : lne z to z : (3z 5z + )dz (3z 5z + )dz [z 3 5 z + z] 3 + ( : ) (b). R e z dz : upper half of crcle jzj from z to z : e z dz [e z ] z e e (only the end ponts matter) (e e ) snh : (e). R sn z dz : contour n the gure. sn z dz 3 sn 3 z z 3 [sn ]3 snh 3 : 3 (f). R e z dz : contour n the gure. e z dz (use ntegraton by parts) [ez (sn z + )] z [e (sn + cos ) e ( )] e + e (cosh + snh ): (h). R (Log(z)) dz : lne z to z : 5

6 We rst need to nd the ntegral of R (log(x)) dx: I (log(x)) dx ; u log x ) du x dx dv log xdx ) v x[log x ] I x log(x)[log x ] [log x ]dx x log(x)[log x ] x[log x ] + x x x log(x) + x(log(x)) : (Log(z)) dz z zlog(z) + z(log(z)) z Log() + (Log()) [] [log jj + ] + [log jj + ] + 4 (). R +z dz : lne z to z + : + ( 4): We wll use the de nton h of the complex nverse tangent functon: tan (z) log +z z : + z dz tan z + z tan ( + ) tan () + + log log( ) log + log( ) log p 5 + ( + tan ()) 3 4 log(5) tan () + 4 :. If P (z) s a polynomal and s any closed contour, explan why R P (z)dz : P (z) s entre and therefore has an antdervatve. Thus, f contour, R P (z)dz : s any closed 6

7 4. True or False: If f s analytc at each pont of a closed contour, then R f(z)dz : Ths s false, auchy s Theorem requres that f be analytc on and nsde. For example, the functon f(z) z s analytc on jzj ; but not nsde and dz 6 : R z 5. Explan why f(z) z has no antdervatve n the punctured plane rfg: Suppose f(z) z has an antdervatve n r fg. Then R z dz for every closed curve r fg. But ths s not true for any loop whch encloses the pole z : For nstance f : jzj, R jzj z dz. (4.4): auchy s Integral Theorem. Let D be the doman gven on page 99 and the contour shown. Decde f the followng gven contours can be deformed to. (a). Yes (b). No (c). Yes (d). No 3. Let D : < jzj < 5 and : jz 3j traversed once postvely startng at z 4. Decde whch of the followng contours are contnuously deformable to n D. (a). Yes (b). Yes (c). No (d). Yes (e). Yes. Determne the doman of analytcty and explan why R f(z)dz : jzj (a). f(z) z z +5 RD fz j z 6 5g. f(z)dz snce f s analytc on and nsde : jzj. jzj (b). f(z) e z (z + ) Ths s an entre functon, so R f(z)dz. jzj 7

8 (c). f(z) z 6z+ RD fz j z 6 3 g. f(z)dz snce f s analytc on and nsde : jzj. jzj (d). f(z) sec z cos(z) ) z 6 (n ) ) z 6 (n ); n ; ; ; D R fz j z 6 (n ); n ; ; ; g. f(z)dz snce f s analytc on and nsde : jzj. jzj (e). f(z) Log(z + 3) z + 3 x + y + 3 x y. We need to avod y and x+3. D R r fx + y j y ; x 3g. f(z)dz snce f s analytc on and nsde : jzj. jzj 3. Evaluate R z + (a). R z + dz R (b). Deform nto a barbell. R z + dz R (c). R z + dz R dz along each of the three contours. (z+)(z ) dz R (z+)(z ) dz R (z+)(z ) dz R h (z+) h (z+) h (z+) (z ) dz (z ) dz (z ) : : dz : 5. Evaluate R z (z+)(z ) dz; where : jzj 4 traversed twce n the clockwse drecton. R z (z+)(z ) dz R jzj4 h 3 (z+) + 3 (z ) dz : 7. Evaluate R z z+ (z+)(z ) dz; where s the contour shown n the gure. Frst get the partal fractons. z z + (z + )(z ) A () (z + ) + A() (z ) + A() (z ) over-up rule gves, A () ; A (). To nd A () d z z + lm z!! dz (z + ) d (z + )(4z ) (z z + ) lm z! dz (z + ) 6 : 4 8

9 R z z+ (z+)(z ) dz R h (z+) + (z ) + (z ) dz [ + ] : (4.5): auchy s Integral Formula and Its onsequences. Let f be analytc nsde and on the smple closed contour. What s the value of f(z) dz z z when z les outsde? Snce f s analytc nsde and on, the nteror of the closed loop s a smply-connected doman (call t D) n whch f s analytc. Gven that z les outsde, the functon f(z) z z s analytc n D as well. By auchy s Integral Theorem, R f(z) z z dz. R f(z) Therefore, z z dz.. Let f and g be analytc nsde and on the smple loop. Prove that f f(z) g(z) for all z on, then f(z) g(z) for all z nsde. Let z be any pont nsde. We need to show that f(z ) g(z ). Applyng auchy s Integral Formula we obtan f(z) g(z) f(z ) g(z ) dz dz z z z z f(z) g(z) dz z z ; snce f(z) g(z) for all z on. 3. Let : jzj traversed once postvely. ompute the followng. (a). (b). sn 3z z dz [sn 3z] z : ze z z 3 dz ze z z 3 dz [zez ] z3 3 e3 3 e3 : 9

10 (c). (d). Let f(z) 5z + z +. (e). Let f(z) e z. (f). Let f(z) sn z z 4. z 3 + 9z dz z(z + 9) dz z(z + 3)(z 3) dz cos (z + 3)(z 3) 9 9 : z 5z + z + (z ) 3 dz! [f () ()] : e z (z + ) dz! [f () ( )] e: sn z z (z 4) dz! [f () ()] : 4. ompute where s (z 4) sn z z + z 3 + z dz; (a). crcle jzj traversed once counterclockwse. z + z 3 + z dz z + z (z + ) dz z +! z + ( ) : z (z 4) (z + ) (z + ) (z + ) z z

11 (b). crcle jz + j traversed once counterclockwse. z + z 3 + z dz z + z (z + ) dz z + + z 4 ( + ) : z (c). crcle jz j traversed once counterclockwse. z + z 3 + z dz z + z (z + ) dz : 6. Evaluate e z (z + ) dz; where s the crcle jzj 3 traversed once counterclockwse. Sketch the contour jzj 3 and reduce the contour to a vertcal barbell conssng of: a crcle traversed counter clockwse : jz j straght lne 3 connectng to a crcle traversed counter clockwse : jz + j and straght lne 3 connectng to Let f(z) ez (z+) and g(z) ez (z ). Then e z (z + ) dz f ()! f(z) (z ) dz + + cosh (z + ) dz + 3 e z 3 e z (z + ) dz + snh 3 e z (z + ) dz + e 3 z (z + ) dz + g ()! (cosh snh ) g(z) (z + ) dz

12 7. ompute z (z 3) dz; where s the ndcated contour. The poles of the ntegrand are z (mul ); and z 3. Snce the small loop does not enclose the poles we can deform to a crcle, say jzj. Therefore, z (z 3) dz! lm z! 9 : d dz (z 3) lm z! (z 3) sn z (z 3). Let f u + v be analytc n a doman D. Explan n D. s harmonc By Theorem 6, all dervatves of f exst and are analytc. In partcular, f () (z) u x + v x ; and f () (z) u xx + v xx. Snce f () (z) u xx + v xx s analytc, ts real ; s harmonc n D.

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