hpter 4 Integrls Exercises Pges 5-6. () Use the corresponding rules in clculus to estblish the following rules when w (t) = u (t) + iv (t) is complex-vlued function of rel vrible t nd w (t) exists d dt w ( t) = w ( t) ; where w ( t) denotes the derivtive of w(t) with respect to t, evluted t -t. (b) Use the corresponding rules in clculus to estblish the following rules when w (t) = u (t) + iv (t) is complex-vlued function of rel vrible t nd w (t) exists d dt [w (t)] = w (t) w (t) Solution A function! (t) is sid to be di erentible t t when d!(t+t)!(t) dt! (t) = lim t! (t+t) t eplcing t by -t we get d dt! (![ (t+t)]![ t] t) = lim [ (t+t) ( t)] = lim t! = lim t! = lim t! t!![ (t+t)]![ t] ( t t+t)![ (t+t)]![ t] t![ (t+t)]![ t] t =! (t) Solution b! (t) = u (t) + iv (t) = u + iv [! (t)] = (u + iv) = u + iv + (iv) d dt [! (t)] = d dt u + iuv + i v u + i (uv + u v) + i (vv ) = u +iuv + iu v + i vv = (u + iv) (u + iv) =! (t)! (t) () We hve w (t)=u (t)+iv (t) nd so w (-t)=u (-t)+iv (-t). d dt w ( t) = d dt [u ( t) + iv ( t)] = d dr [u ( t)] + i d dt [v ( t)] = u ( t) d dt ( t) + iv ( t) d dr ( t) = u ( t) iv ( t) = [u ( t) + iv ( t)] = w ( t) (b) We hve [w (t)] = [u (t) + iv (t)] = [u (t)] [v (t)] = [u (t) + iv (t)] = [u (t)] [v (t)] n + iu(t)v(t) Therefore o d dt [w (t)] = d dt [u (t)] [v (t)] + iu(t)v(t) Now we hve
= d dt [u (t)] d dt [v (t)] + i [u(t)v(t)] = u(t)u (t) v(t)v (t) + i [u (t)v(t) + u(t)v (t)] Agin we hve w (t) = u (t) + iv (t). Therefore w(t)w (t) = [u(t)+iv(t)][u (t)+iv (t)] = u(t)u (t)-v(t)v (t)+i[u (t)v(t)+u(t)v (t)]. Thus we hve seen tht d dr [w (t)] = w(t)w (t). Evlute the following integrls () (b) (c) t =6 e it dt; i dt; e t dt (e > ) Solution (b) The derivtive of e it =i = e it. Therefore i = + i p 3= = i=4 + p 3=4 =6 e it dt = Solution () t i dt = t t j = i ln (t) j = =6 Solution (b) p 3 4 + i 4 Solution (c) lim N!+ i t t dt = t ( ) + i ln () = e it dt = i eit j =6 = i e i=3 = i N e t dt = lim e t dt = N! lim N!+ e t j N = dt + i t dt = + i ln () + i p 3 = e N As N! +, we hve e ( N) = N e! ; so e N = e e( N)! Thus e N! s N! +. So we hve e t dt = lim N!+ e N = = 3. Show tht if m nd n re integers, e im e in d = when m 6= n; when m = n. When n 6= m then n ntiderivtive is e (m n)i = (m n) i nd when one plugs in i nd we get the sme thing nmely, (m n)i nd so when we subtrct we get. When n = m then the integrnd is nd so integrl is Solution (3) e im e in d = e i(m n) d If m = n, then the right hnd side is = m n ei(m n) j = Solution If m 6= n, then the right hnd side is equl to m n ei(m n) e = m n ( ) =
This is very importnt integrl tht occurs frequently in nlysis. It gives, e.g., wy to de ne the Dirc delt function if you re fmilir with tht from previous studies. The computtion is not prticulrly di cult. We simply use tht d d eim = ime im ; m Z Thus, if m 6= n; e im e in d = e i(m n) d = n) ei(m i(m n) j = i(m n) e i(m n) However, since m n Z; we hve e i(m n) = nd thus this integrl vnishes. If, however, m = n, then the integrl we did bove is incorrect s we re dividing by ero. Indeed, in this cse, things re even simpler e im e in d = d = 4. According to de nition (), Section 37, of integrls of complex-vlued functions of rel vrible, e (+i)x dx = e x cos xdx+i e x sin xdx Evlute the two integrls of the right here by evluting the single integrl on the left nd then using the rel nd imginry prts of the vlue found. This is nice trick s you my remember tht the integrls on the right re rther involved (you hve to integrte by prts twice nd recognie tht you return to where you begn but with the opposite side... thus you cn solve the desired eqution). The formul follows esily from the de nition of complex exponents nd Euler s formul. Evluting the left side, we hve e (+i)x dx = (+i) e(+i)x j = (+i) e (+i) = (+i) (e cos + ie sin ) ( i) Since (+i) = ; we hve tht this is = ( i) ( e ) Thus, e x cos xdx = (+e ) nd e x sin xdx = (+e ) 7. Apply the inequlity b w (t) dt b jw (t)j dt ( b) to show tht for ll vlues of x in the intervl x ; the functions P n (x) = x + i p x cos n d (n = ; ; ; ) stisfy the inequlity jp n (x)j P n (x) = jp n (x)j = jp n (x)j Z Z Z x + i p x + i p x + i p x cos n d x cos n d x cos n d onsider x + i p q x cos = e i = x + p x cos p = x + ( x ) cos 3
h p i = tn x cos = tn h Let = p i x cos x x + i p x cos n = e i n = jj n e in = jj n ) e in = Z ) The eqution jp n (x)j p x + ( ) jp n (x)j p x + ( x ) cos n d Z p x sin + cos n d = Z = d = jp n (x)j x ) cos n d Exercises Pges -. Show tht if w(t) = iv(t) is continuous on n intervl t b, then () (b) b b w ( t) dt = w (t) dt = b w () d; w [ ()] () d, where () is the function in eqution (9), Section 38. Suggestion These identities cn be obtined by noting tht they re vlid for rel-vlued functions of t. Strt by writing I = b w ( t) dt = b u ( t) dt + i b v ( t) dt The substitution s = -t in ech of these two integrls on the right then yields b b I = u (s) ds i v (s) ds = w (s) ds Tht is, I = w ( t) dt = w (s) ds b b) Strt with I = b b w ( t) dt = b u (t) dt + i b b v (t) dt; nd this time the substitution t = (s) in ech of the integrls on the right gives the result.. Let denote the right-hnd hlf of the circle jj =, in the counterclockwise direction, nd note tht two prmetric representtions for re = () = e i p nd = Z (y) = 4 y +iy ( y ) y Verify tht Z (y) = [ (y)], where (y) = rctn p 4 y < rctn t <. Also show tht this function hs positive derivtive, s required in the condition following eqution (9), Section 38. We check 4
[ (y)] = e i rctn y p 4 y = y cos rctn p + i sin rctn p y 4 y 4 y Notice tht rctn y 4 y is the ngle mde from the right tringle with bse p 4 y nd height y. Thus, the hypotenuse is. Using this tringle, we my evlute the cos nd sin to see tht [ (y)] = p 4 y + i y = Z (y) s p 4 y desired. Noting tht + p y 4 y (y) = 4 y = p we see 4 y + p y 4 y tht the derivtive is indeed positive for y ( ; ) 5. Suppose tht function f() is nlytic t point = (t ) lying on smooth rc = (t) ( t b). Show tht if w (t) = f [ (t)], then w (t) = f [ (t)] (t) when t = t. Suggestion Write f() = w(x,y)+iv(x,y) nd (t) = x(t)+iy(t), so tht w (t) = u [x (t) ; y (t)] + iv [x (t) ; y (t)] Then pply the chin rule in clculus for functions of two rel vribles to write w = (u x x + u y y ) + i (v x x + v y y ), nd use the uchy-iemnn equtions. f () = u (x; y) + iv (x; y) nd (t) = x (t) + iy (t) so tht w (t) = u [x (t) ; y (t)] + iv [x (t) + iy (t)]. The chin rule sys tht w (t) = u x x (t) + u y y (t) + i [v x x (t) + v y y (t)]. Since u x = v y nd u y = v x we get w (t) = u x x (t) v x y (t) + i [v x x (t) + u x y (t)] = u x (t) + iv x (t) = f [ (t)] (t) 6. () Let y(x) be rel-vlued function de ned on the intervl x x by mens of the equtions y (x) = 3 sin x when < x when x =. Show tht the eqution = x + iy (x) ( x ) represents n rc tht intersects the rel xis t the points = =n (n = ; ; ) nd =. (b) Let y(x) be rel-vlued function de ned on the intervl x by x mens of the equtions y (x) = 3 sin x when < x when x =. Verify tht the rc in prt () is, in fct, smooth rc. Suggestion To estblish the continuity of y(x) t x =, observe tht x 3 sin x x 3 when x >. A similr remrk pplies in nding y () nd showing tht y (x) is continuous t x =. () For complex vrible we tke the x-xis s the rel xis nd the y-xis s the imginry xis. If n rc intersects the rel xis. If n rc intersects the rel xis, then y =. Given = x + iy (x) nd y (x) = x 3 sin x When y =, we hve x 3 sin x ) x 3 = or sin x = ) x = or x = n; for n = ; ; 5
) x = n When x =, the eqution = x + iy (x) becomes = + iy () ) = When x = n ; the eqution = x + iy (x) becomes = n + i () ) = n (b) y (x) = x 3 sin( x );<x when x =. y (x) = 3x sin x x cos x = x 3x sin x cos x jy (x)j x 3x sin x cos x jxj 3x sin x cos x jxj or 3x sin x cos x ) 3x sin x cos x 3x when x > ) sin x nd cos x Exercises Pges 8-3 For the function f nd contours in Exercises through 6, use prmetric representtions for, or legs of, to evlute f () d. f () = (+) nd is () the semicircle = e i ( ) ; (b) the semicircle = e i ( ) ; (c) the circle = e i ( ) Solution () = e i so () = ie i d Thus the integrl is e i +ie i d = i e i + d = e i( ) + i = 4 + i e i Solution () f () d = e i + e i ie i d = ie i + i d = e i + i j = e i e + i = 4 + i e Solution (b) f () d = i +ie i d = ie i + i d = e i e i + i j = e i e i + 4i i = 4 + i Solution (c) This integrtion is the sum of the integrtions in () nd (b). So it is equl to ( 4 + i) + (4 + i) = 4i. f () = nd is the rc form = to = consisting of () the semicircle = + e i ( ) ; (b) the segment x of the rel xis. Solution () = ie i so the integrl is f () d = e i ie i The ntiderivtive is e i = nd when we plug in theterms cncel so weget. when y <, 4. f() is de ned by the equtions f () = nd is 4y when y >, the rc from = --i to = +i long the curve y = x 3 Here (t) = t + it 3 so (t) = + i3t. Here t Our integrl is + i3t dt + 4t 3 + i3t dt = ( + i) + ( + i) = + 3i 6
Solution (4) is prmeteried by (t) = t + it 3 ; t If t <, Im (t) = t 3 < ; so f ( (t)) = If t >, Im (t) = t 3 >, so f ( (t)) = 4 Im (t) = 4t 3 We compute (t) = + i3t Thus f () d = f ( (t)) (t) dt = f ( (t)) (t) dt + f ( (t)) (t) dt = 4t 3 + i3t dt = t + it 3 j + t 4 + it 6 j ( + i) + ( + i) = + 3i + i3t dt + Problem sttement Evlute when y <, f () d for f () = nd is the rc from 4y when y >, = --i to = +i long the curve y = x 3 The curve cn be represented prmetriclly by (x (t) ; y (t)) = t; t 3 ; t We cll the portion of where t nd the portion for t Then f () d = f () d + f () d = = i3t dt + 4t 3 + i3t dt = t + it 3 j + t 4 + it 6 j = ( + i) + ( + i) = + 3i 6. f() is the brnch +i = exp [( + i) log ] (jj > ; < rg < ) of the indicted power function, nd is the positively oriented unit circle jj = Let = re i Then +i = e ( +i)(log r+i) = e log r +i(log r ) = e r [cos (log r ) + i sin (log r )] The correct brnch is obtined by ssuming < < The circle cn be prmeteried s e i ; < Therefore f () d = e (cos i sin ) ie i d (since r = on contour) = i e d = i e (by Euler s formul) Problem sttement Evlute f () d for f() for the brnch +i = exp [( + i) log ] (jj > ; < rg < ) of the indicted power function, nd is the positively oriented unit circle jj = First, (t) = e it ; t Then f () d = exp ( + i) log e it ie it dt 7
= exp ( = i e t dt = + i) log e it ie it dt ie t j = i e. Let denote the circle j j =, tken counterclockwise. Use the prmetric representtion = + e i ( ) for to derive the following integrtion formuls d () d = i; (b) ( ) n d = (n = ; ; ) Let = + e i so tht d = ie i nd = e i. Then the rst integrl is id = i nd the second integrl is i n e ni = () d ( d = ) e i ie i d = i (b) First possibility use the ntiderivtive of ( ) n Second possibility prmeterie s in ).. Use the prmetric representtion in Exercise for the oriented circle there to show tht ( ) n d = i sin (), where is ny rel number other thn ero nd where the principl brnch of the integrnd nd the principl vlue of re tken. [Note how this generlies Exercise (b).] ( ) d = exp [( ) Log ( )] d = = i exp [( h exp ( ) Log e ii i e i d ) (ln + i)] i e i d = i e ( )i e i d e i d = ei j = e i e i( ) = [i sin ()] At the lst step, we re using sin = (ei e i ) i Exercises Pges 33-34. Without evluting the integrl, show tht d 3 when is the sme rc s the one in Exercise, Section 4. On the circle 3 so the integrl is bounded by jdj The integrl is rclength which is 4=4 = 3 8
Solution () On the contour, jj =, so = = 3, which implies =3 We lredy know tht L () =, so d 3 L () = 3. Let denote the line segment from = i to =. By observing tht, of ll the points on tht line segment, the midpoint is the closest to the origin, show tht d 4 p without evluting the integrl. 4 The closest point, the midpoint, is ( + i) = t distnce p wy from. Thus jj 4 =4 nd so the integrl is bounded by 4 jdj = 4 p 3. Show tht if is the boundry of the tringle with vertices t the points, 3i, nd -4, oriented in the counterclockwise direction (see Figure 47), then (e ) 6 The tringle is 3-4-5 right tringle, nd thus, its length is. Note tht je j je j + jj je j + 4 on since it is esy to see tht -4 is the point on the tringle frthest from the origin. Note lso tht e for ll points on the tringle. Thus, je j = je x j e iy on the tringle. Thus, the integrnd is 5 on the tringle. Hence, (e ) d 5 () = 6 4. Let denote the upper hlf of the circle jj = ( > ), tken in the counterclockwise direction. Show tht 4 +5 +4 d +) ( ( )( 4). Then, by dividing the numertor nd denomintor on the right here by 4, show tht the vlue of the integrl tends to ero s tends to in nity. Solution (4) If, then jj = ; so + j j = jj + = +, nd 4 + 5 + 4 = + + 4 jj jj 4 = 4 > Thus j 4 +5 +4 = j 4 +5 +4 + ( )( 4) We my prmetrie by () = e i ; Then L ( ) = j ()j d = ie i d = d = Thus 4 +5 +4 d + ( )( 4) L ( ) = ( +) ( )( 4) The contour in the upper hlf of the circle jj = On the contour, + = + Similrly, 4 + 5 + 4 4 = 5 4 = 4 5 4 We hve 9
( +) ( 4 +5 +4) +) ( for on the contour. ( 4 5 4) Therefore the contour integrl is bounded by The limit of this bound is s! ( +) ( 4 5 4) 5. Let be the circle jj = ( > ), described in the counterclockwise direction. Show tht Log d < +ln, nd then use l Hospitl s rule to show tht the vlue of this integrl tends to ero s tends to in nity. Let us compute the integrl. numbers belonging to We note tht e i d = Z Z [Log()] d = i [ln + i ( + k)] e i d jj= = i (ln + ki) Z Z = = 4i e i d e i d We use the expression = e i for complex Z Z e i d Z e i d 4i The lim! = 6. Let denote the circle jj = ( < < ), oriented in the counterclockwise direction, nd suppose tht f() is nlytic in the disk jj. Show tht if = represents ny prticulr brnch of tht power of, then there is non- negtive constnt M, independent of, such tht = f () d Mp. Thus show tht the vlue of the integrl here pproches s tends to. Suggestion Note tht since f() is nlytic, nd therefore continuous, throughout the disk jj, it is bounded there (Section 7). Since f() is nlytic, nd therefore continuous throughout the disk jj ; it is bounded there. This implies the Z existence of constnt M such tht jf ()j M for ll jj We hve = f () d M = d The ltter integrl Z is clculted s follows. = d = = e [i(+k)=] ie i d Z = i = e ik e i= d
= = e ik e i = 4 p e ik Here k = or k = depending on the brnch of the function = In both cses we hve = f () d M = d p M Hence, s! the vlue of the integrl tends to ero s w Exercises Pges 4-4. By nding n ntiderivtive, evlute ech of these integrls, where the pth is ny contour between the indicted limits of integrtion i= () e d; (b) (c) i +i cos d; 3 ( ) 3 d Solution (c) An ntiderivtive is ( ) 4 =4 so tht the integrl is evluted by evluting t the endpoints which gives /4-/4 =. Solution () Since e is n ntiderivtive of e, so i= e d = ei= e i = ( + i) = i Solution (b) +i cos d = sin Since sin (=) is n ntiderivtive of cos (=), so j +i = sin (= + i) = i e i(=+i) e i(=+i) = i ie ( ie) = =e + e Solution (c) Since 4 ( )4 is n ntiderivtive of ( ) 3, so 3 ( ) 3 d = 4 ( )4 j 3 = 4 (3 ) ( ) 4 = 3. Use the theorem in Section 4 to show tht ( ) n d = (n = ; ; ) when is ny closed contour which does not pss through the point. [ompre Exercise (b), Section 4.] The fundmentl theorem of clculus implies tht the integrls re. For n = ; ; ; ; the integrls re even if the contour psses through For n = ; ; ; the integrnd is unde ned t = Therefore the contour must not pss through 4. Find n ntiderivtive F () of the brnch f () of = in Exmple 4, Section 43, to show tht the integrl (6) there hs vlue p 3 ( + i). Note tht the vlue of the integrl of the function (5) round the closed contour in tht exmple is, therefore, 4 p 3
We wnt to nd p d with p = p re i= for = re i with < The brnch cut in the de nition of p cn be moved to the positive imginry xis in such wy tht p does not chnge on the contour De- ne p = p re i= for = re i with = < 5= The ntiderivtive of this brnch of p is given by 3 3= = 3 r3= e i3= for = re i with = < < 5= The integrl over the contour cn be obtined s follows 3 3= j 3 3 = 3 33= e i3= j = = = 3 33= e i3 e i3= = p 3 ( + i) 5. Show tht i d = +e ( i), where i denotes the principl brnch i = exp (i Log ) (jj > ; < Arg < ) nd where the pth of integrtion is ny contour from = - to = tht, except, for its end points, lies bove the rel xis. Suggestion Use n ntiderivtive of the brnch i = exp (i Log ) jj > ; < Arg < 3 of the sme power function. An ntiderivtive is i+ i+. Evluting t the endpoints we get i+ = ( + i) ( ) i+ = (i + ) The rst term is = (i + ). The second is e (i+) log( ) = ( + i) = e (i+)i = ( + i) = e = ( + i). Putting it together we get e +i = (+e ) ( i) The ntiderivtive is given by (+i) (+i) = (+i) e(+i)log ; with Log () being the principl brnch of the logrithm. Of course, Log () = Since the contour vries on the upper hlf of the complex plne before terminting t -, we cn tke Log ( clculus, the integrl is equl to (+i) Exercises Pges 53-56 ) = i By the fundmentl theorem of e e i(+i) = (+e ) ( i). Apply the uchy-gourst theorem to show tht f () d = when the contour is the circle jj =, in either direction, nd when () f () = 3 ; (b) f () = e ; (c) f () = ++ ; (d) f () = sec h ; (e) f () = tn ; (f) f () = Log ( + )
According to the uchy-gourst theorem if f() is n nlytic single-vlued function in the convex domin G thnfor ny regulr closed curve contined in G the following integrl vnishes f () d = () In our cse f () = = ( 3) This function hs the only singulrity t point = 3. Throughout the disk jj the function f() is nlytic nd single-vlued. Hence, ccording to the uchy-gourst theorem, its integrl over the curve jj = is equl to ero d ( 3) = jj= (b) As before, the function f () = e is nlytic nd single-vlued throughout the disk jj Hence, by the uchy-gourst theorem e d = (c) The function f () = = + + = ( + + i) ( + i) Hence, the function f() hs singulrities only t points = --i nd = -+i. Both points lie outside of the disk jj since j ij = p ' 4 Hence, the function is nlytic throughout the disk jj nd by the uchy-gourst theorem d ( ++) = (d) The function f () = sec h does not hve singulrities nd is nlytic in ny nite disk jj Hence, by the uchy-gourst theorem sec hd = jj= (e) The function f () = tn = sin cos is not de ned only if the denomintor is ero; tht is cos =. The generl solution is = (n + ) ; where n is n integer. Whtever the vlues of n my be, the vlue of will not lie inside the unit circle. So tn is nlytic throughout the region. Since f() is nlytic inside nd on, the integrl of f() over is ero by the uchy-gourst theorem. (f) The function f () = Log ( + ) hs singulr point = -. It is nlytic throughout the disk jj 5; nd by the uchy-gourst theorem Log ( + ) d = jj=. Let denote the positively oriented circle jj = 4 nd the positively oriented boundry of the squre whose sides lie long the lines = ; y = (Figure 6). With the id of orollry in Section 46, point out why f () d = f () d when () f () = 3 + ; (b) f () = + sin(=) ; (c) f () = e Solution () The singulrities of this function re t p =3 which re inside. Therefore by uchy for doubly connected domins, f () d = f () d 3
n Solution () Note tht f is nlytic in n p i 3 o, nd p i 3 re ll interior to. Both nd re positively oriented, so fd = fd Solution (b) Note tht f is nlytic in n fn n Ng. When n =, n = lies inside. When n 6= ; jnj > 4, so n lies outside. Thus fd = fd Solution (c) Note tht f is nlytic in n fni n Ng. When n =, ni = lies inside. When n 6= ; jnij > 4, so ni lies outside. Thus fd = fd 3. If denotes positively oriented circle j j =, then ( ) n when n = ; ; ; d = i when n =. ccording the Exercise, Section 4. Use tht result nd orollry in Section 46 to show tht if is the boundry of the rectngle x 3; y, described in the positive sense, then ( i) n when n = ; ; ; d = i when n =. Solution (3) Note tht +i lies inside. Let be positively oriented circle fj ( + i)j = =g. Then lies inside nd ( ( + i)) n is nlytic between nd for ny n N. Thus ( i) n d = ( i) n when n 6= ; d = i when n =. Using the sttement of Exercise 4. nd orollry in section 46 we only hve to show tht we cn tke positively circle centered t = +i such tht f () = ( ) n is closed on this circle, on the contour nd in the region between them. If the exponent n- is positive or equl to, then f is entire nd we cn tke this circle to be ny circle tht is lrge enough. If the exponent is negtive, then f hs singulr point in +i. This point is contined in the interior of the rectngle given by, so gin we my tke our circle to be huge circle enclosing this rectngle, then f hs the desired property of being nlytic outside the interior of the rectngle nd we my pply the corollry. 4. Use the method described below to derive the integrtion formul e x cos bxdx = p e b (b > ) () Show tht the sum of the integrls of exp x long the lower nd upper horiontl legs of the rectngulr pth in Figure 6 cn be written e x dx e b e x cos bxdx nd tht the sum of the integrls long the verticl legs on the right nd left cn be written b ie e y e iy b dy ie e y e iy dy 4
Thus, with the id of the uchy-gourst theorem, show tht e x cos bxdx = e b e x dx + e ( +b ) b e y sin ydy (b) By ccepting the fct tht e x dx = p nd observing tht b b e y sin ydy < e y d y, obtin the desired integrtion formul by letting tend to in nity in the eqution t the end of prt (). For prt (), we rst consider d + e d; e x where nd re the lower nd upper legs of the rectngulr contour shown in Figure 6. The lower leg cn be prmeteried s = x; x < ; nd the upper leg s = x + ib; with x decresing from to -. For both prmeteritions d = dx. Along the lower leg e = e x ; nd long the upper leg e = e (x+ib) = e x +b ibx = e x +b [cos (bx) + i sin (bx)] Therefore the sum of the integrls long the lower nd upper legs is equl to e x dx + e b e x [cos (bx) + i sin (bx)] dx Notice tht e x sin (bx) is n odd function of x while e x nd e x cos (bx) re even functions of x. Therefore, the sum of the two integrls cn be expressed s e x dx e b e x cos (bx) dx The left nd right legs of the contour in Figure 6 cn be prmeteried s = + iy with y incresing from to b nd s = + iy with y decresing from b to. Use of this prmeterition gives the sum of the integrls long the right nd left legs to be b ie e y e iy b dy ie e y e iy dy If Euler s formul is used for e iy nd e iy, we get b e e y sin (y) dy for the sum of the integrls long the right nd left legs. Since e is nlytic on the contour shown in Figure 6 nd everywhere inside it, uchy s theorem implies tht the sum of the integrls b e x dx e b e x cos (bx) dx nd e e y sin (y) dy is. Therefore e x cos (bx) dx = e b e x dx + e ( +b ) b + e y sin (y) dy For prt (b), we merely hve to note tht the third term in e x cos (bx) dx = 5
e b e x dx + e ( +b ) b + e y sin (y) dy bove tends to s! for ny b > 6. Let denote the positively oriented boundry of the hlf disk r ;, nd let f() be continuous function de ned on tht hlf disk by writing f() = nd using the brnch f () = p re i= r > ; < < 3 of the multiple-vlued function =. Show tht f () d = by evluting seprtely the integrls of f() over the semicircle nd the two rdii which mke up. Why does the uchy-gourst theorem not pply here? The brnch of = which is tken to be f () is nlytic everywhere except on the negtive imginry xis nd t = The contour psses through =, point where f () is not nlytic. Therefore uchy s theorem does not pply. cn be deformed very slightly by bumping it up bit ner the origin. Then f () chnges only slightly s f () is continuous t = uchy s theorem pplies over the deformed contour nd implies tht f () d = By mking the bump smller nd smller nd tking the limit, we cn conclude tht f () d = Exercises Pges 6-64. Let denote the positively oriented boundry of the squre whose sides lie long the lines x = nd y =. Evlute ech of these integrls () e d (i=) ; (b) (c) (d) (e) cos ( +8) d; d + ; cosh d; 4 tn(=) ( ) d ( < x < ) Solution () The function e is nlytic so by IF the integrl is ie i= = i ( i) = Solution (b) The function cos () = + 8 is nlytic inside the contour so the integrl is i cos () =8 = i=4 Solution () Since i= lies inside, nd cos +8 is nlytic inside nd on (it is nlytic throughout except t p 8i, which both lie inside ), so e (i=) = ie i= = Solution (b) Since lies inside, nd cos +8 is nlytic inside nd on (it is nlytic throughout except t p 8i, which both lie outside ), so cos cos() ( +8) d = i +8 = i 4 6
Solution (c) Since -/ is inside nd / is nlytic in, so + d = = ( =) d = i (( =) =) = i= Solution (d) Since lies inside, nd cosh is nlytic in, so cosh d = 4 i 3! cosh(3) () = i 3 sinh () = Solution (e) Since x lies inside, nd tn (=) is nlytic in, so tn(=) d = i d ( x ) d tn (=) j = = i sec (x =). Find the vlue of the integrl of g() round the circle j ij = in the portion sense when () g () = +4 ; (b) g () = ( +4) Solution () We hve +4 = +i i The rst fctor is nlytic inside the contour so the integrl is i 4i = Solution (b) For the second g () = f () where f () = so ( i) (+i) tht by IF for derivtives the integrl is if (i) = 4i = (4i) 3 6 Solution () Let denote the positively oriented circle fj ij = g. Note tht i lies inside, nd -i does not. So (+i) nd re nlytic (+i) inside nd on. +4 d = =(+i) i d = i i+i = = Solution (b) Let denote the positively oriented circle fj ij = g. Note tht i lies inside, nd -i does not. So (+i) nd re nlytic inside nd on. d = (+i) =(+i) d = i d ( +4) ( i) d +i j =i = i = i (i+i) 3 64i = =6 3. Let be the circle jj = 3, described in the positive sense. Show tht if g (w) = w d (jwj 6= 3), then g () = 8i. Wht is the vlue of g() when jwj > 3? Solution (3) Let f () =. By IF g () = if () = 8i For the second prt since the function is nlytic inside the contour when jwj > 3 the integrl is by uchy Theorem. Solution (3) Let f () =. Then f is n entire function. Since lies inside, so by uchy s integrl formul, g () = f() d = if () = i4 = 8i If w is outside, let h w () = f () = ( w). Then h is nlytic on nd inside. Thus by uchy-gourst theorem, g (w) = h () d = 4. Let be ny simple closed contour, described in the positive sense in the plne, nd write g (w) = 3 + d. Show tht g (w) = 6iw when w is ( w) 3 inside nd tht g(w) = when w is outside. Solution (4) Since 3 + is nlytic in, so if w is inside, we hve 7
g (w) = 3 + d = i ( w) 3 d 3 + j =w = 6iw If w is outside, then ( 3 +) is nlytic inside nd on. Thus g (w) = 3 + d = ( w) 3 ( w) 3 5. Show tht if f is nlytic within nd on simple closed contour nd is not on, then d f ()d = f()d ( ) The uchy Integrl Formul pplied to f gives f ( ) = The formul for higher derivtives (p.6, (4)) gives f ( ) =! This implies the clim. i f () ( ) d f() ( ) d 6. Let f denote function tht is continuous on simple closed contour. Prove tht the function g () = i to nd tht g () = i f(s)ds s is nlytic t ech point interior f(s)ds (s ) t such point. In order to prove the nlyticity of the function g() we need to demonstrte tht it is di erentible. To this end we need to nd the liner prt of the increment of the function g () g ( + h) g () for every two points nd +h interior to. We hve [s (+h)] (s ) = h (s )[s (+h)] = h h + (s ) (s ) [s (+h)] From the de nition of function g() nd from the formul bove it follows tht g ( + h) g () = f(s)ds f(s)ds = h f(s)ds i (s ) i [s (+h)] + h i (s ) f(s)ds i (s ) [s (+h)] Now we hve to prove tht the second term is of order O h In other words, we need to limit the second integrl in the right-hnd side by constnt. Since both points nd +h re interior to, there exists constnt such tht js j > ; js ( + h)j > Besides, since f() is continuous on it is bounded throughout, i.e., there exists constnt M such tht jf ()j M for ech point Hence, by the Men Vlue Theorem f(s)ds i ML (s ) [s (+h)] =constnt. 3 Here L is the length of. Thus, we obtin tht the liner prt of the increment of the function g() exists nd is given by g () = f(s)ds i Hence, (s ) the proof tht the function g() is nlyticl is complete, nd its derivtive is given by the required formul. 8
7. Let be the unit circle = e i ( ). First show tht, for ny rel constnt, e d = i Then write this integrl in terms of to derive the integrtion formul e cos cos ( sin ) d = Solution (7) Since e is nlytic in nd lies inside, so we hve e d = ie = i From the de nition of line integrl, we hve e d = exp(e i ) ie i d = i exp e i d Thus exp e i d =. e i Since exp e i = exp ( cos + i sin ) = e cos cos ( sin ) + ie cos sin ( sin ) ; so e cos cos ( sin ) d =! e exp e i d = Since e cos( ) cos ( sin ( )) = e cos cos ( sin ) = e cos cos ( sin ), so e cos cos ( sin ) d Thus e cos cos ( sin ) d = e cos cos ( sin ) d = e cos cos ( sin ) d = 9. Follow the steps below to verify the expression f () = i the lemm in Section 48. () Use the expression for f () in the lemm to show tht f (+) f () f(s)ds = 3(s ) () (s ) 3 i f (s) ds (s ) (s ) 3 i f(s)ds (s ) 3 in (b) Let D nd d denote the lrgest nd smllest distnces, respectively, from to points on. Also, let M be the mximum vlue of jf (s)j on nd L the length of. With the id of the tringle inequlity nd by referring to the derivtion of the expression for f () in the lemm, show tht when < jj < d, the vlue of the integrl on the right-hnd side in prt () is bounded from bove by (3Djj+jj )M L (d jj) d 3 (c) Use the results in prts () nd (b) to obtin the desired expression for f (). In view of the expression for f in this lemm. h f (+) f () = i h i = (s ) i (s ) (s ) Then i f(s)ds (s ) (s ) f (s) ds f (+) f () f() i = (s ) 3 i h i = 3(s ) () i f (s) ds (s ) (s ) 3 h i (s ) (s ) (s ) (s ) 3 9 f (s) ds
b) The tringle inequlity tells us tht 3 (s ) () 3 js j jj + jj 3D jj + jj We lso cn see tht js j d jj > (p.6, line 3). Hence, (s ) (s ) 3 (d jj) d 3 > Together we obtin h i 3(s ) () f (s) ds (3Djj+jj )M (s ) (s ) 3 (d jj) d 3 c) If we let tend to in this inequlity, we nd h i lim 3(s ) () i f (s) ds = (s ) (s ) 3! This, together with the result in prt ), yields the desired expression for f. Exercises Pges 7-73. Let f be n entire function such tht jf ()j A jj for ll, where A is xed positive number. Show tht f () =, where is complex constnt. Suggestion Use uchy s inequlity (Section 49) to show tht the second derivtive f () is ero everywhere in the plne. Note tht the constnt M in uchy s inequlity is less thn or equl to A (j 9 j + ) By uchy inequlity we hve jf ( )j M where M is the mximum of jf ()j on the circle of rdius centered t. In this cse jj j j + so M A (j j + ). Plugging tht in we get jf ( )j A(jj+). As! the right hnd side goes to nd this sys tht f ( ) =. Since this is true for ny we hve f () = for ll nd this mens f is constnt. But this mens tht f () = + b. Since jf ()j A jj we hve b = For >, let denote the positively oriented circle f jj = g Fix Then lies inside if > jj From uchy s formul, f ( ) = f() f() i d For ( ) 3 ; ( A ) 3 ( j j) 3 Thus jf ( )j L( ) A = A This inequlity holds for ll > ( j j) 3 ( j j) 3 A j j. Since! s!, so jf ( ( j j) 3 )j =, i.e., f ( ) = Since is chosen from rbitrrily, so f () = for ll Thus f () is constnt in. Let be the constnt nd g () = f (). Then g () = for ll Thus g() is constnt in. Let the constnt be. Then f () = + for ll From jf ()j A = we get f() =. Thus = nd f () =. Suppose tht f() is entire nd tht the hrmonic function u (x; y) = e [f ()] hs n upper bound u ; tht is, u (x; y) u for ll points (x; y) in the xy plne. Show tht u (x; y) must be constnt throughout the plne. Suggestion Apply Liouville s theorem (Section 49) to the function g () = exp [f ()] Let g () = e f() = e u+iv. Since e f() = e u nd u is ssumed bounded, we hve e f() constnt. But this mens e u constnt which mens tht u is constnt.
Let g() = exp(f()). Then g is entire nd jg ()j = exp (u ()) u. From Liouville s Theorem, g must be constnt. Thus u () = ln (jg ()j) is lso constnt throughout. 4. Let function f be continuous in closed bounded region, nd let it be nlytic nd not constnt throughout the interior of. Assuming tht f () 6= nywhere in, prove tht jf ()j hs minimum vlue m in which occurs on the boundry of nd never in the interior. Do this by pplying the corresponding result for mximum vlues (Section 5) to the function g () = =f () Let g() = /f(). Since f is continuous on, nlytic in the interior of nd f () 6= for ll, so g is continuous in nd nlytic in the interior of. Since is closed nd bounded, nd jgj is continuous, so jgj ttin its mximum on. Tht mens there is such tht jg ()j jg ( )j for ll Thus jf ()j jf ( )j for ll If int then from mximum modulus principle we hve g is constnt in int. Becuse g is continuous, so g is lso constnt in. From f() = /g() we then hve tht f is constnt in which is contrdiction. Thus must lie on the boundry of. 6. onsider the function f () = ( + ) nd the closed tringulr region with vertices t the points =, =, nd = i. Find points in where jf ()j hs its mximum nd minimum vlues, thus illustrting results in Section 5 nd Exercise 4. Suggestion Interpret jf ()j s the squre of the distnce between nd -. We must hve jf ()j tking its mximum on the boundry nd since it is never lso its minimum on the boundry. Since jf ()j is the squre of the distnce between nd - the mximum is tken on t the furthest point from - which is t =. The closest point is t =. 7. Let f () = u (x; y)+iv (x; y) be function tht is continuous on closed bounded region nd nlytic nd not constnt throughout the interior of. Prove tht the component function u (x; y) hs minimum vlue in which occurs on the boundry of nd never in the interior. (See Exercise 4.) Solution??? Suppose u hs minimum t point in the interior sy t u ( ). Let f () = e f() which is never. We hve jf ()j = e u. The minimum of this is t. But this violtes (4).. Let be ero of the polynomil P () = + + + + n n ( n 6= ) of degree n (n ). Show in the following wy tht P () = ( ) Q (), where Q () is polynomil of degree n-. () Verify tht k k = ( ) k + k + + k + k (k = ; 3; ) (b) Use the fctorition in prt () to show tht P () P ( ) = ( ) Q (), where Q() is polynomil of degree n-, nd deduct the desired results from this. Solution () We clculte the right-hnd side of the eqution
( ) k + k + + k + k = k + k + + k k + + k = k k (b) Now P () P ( ) = + + n n ( + + n n ) = ( ) + ( ) + ( ) + + n ( ) n = ( ) h + ( ) + + n ( ) n i k = ( ) Q () +