Chapter Six More Integration 6.1. Cauchy s Integral Formula. Suppose f is analytic in a region containing a simple closed contour C with the usual positive orientation and its inside , and suppose z0 is inside C. Then it turns out that fÅ»ðzÞð 1 fÅ»ðz0Þð =ð Xð z ?ð z0 dz. 2^ði C This is the famous Cauchy Integral Formula. Let s see why it s true. Let Pð >ð 0 be any positive number. We know that f is continuous at z0 and so there is a number Nð such that fÅ»ðzÞð ?ð fÅ»ðz0Þð <ð Pð whenever z ?ð z0 <ð Nð. Now let _ð >ð 0 be a number | | | | such that _ð <ð Nð and the circle C0 =ð áðz : z ?ð z0 =ð _ðâð is also inside C. Now, the function | | fÅ»ðzÞð is analytic in the region between C and C0; thus z?ðz0 fÅ»ðzÞð fÅ»ðzÞð Xð Xð z ?ð z0 dz =ð z ?ð z0 dz. C C0 1 We know that Xð dz =ð 2^ði, so we can write z?ðz0 C0 fÅ»ðzÞð fÅ»ðzÞð 1 Xð Xð Xð z ?ð z0 dz ?ð 2^ðifÅ»ðz0Þð =ð z ?ð z0 dz ?ð fÅ»ðz0Þð z ?ð z0 dz C0 C0 C0 fÅ»ðzÞð ?ð fÅ»ðz0Þð =ð Xð z ?ð z0 dz. C0 For zOðC0 we have fÅ»ðzÞð ?ð fÅ»ðz0Þð | | fÅ»ðzÞð ?ð fÅ»ðz0Þð z ?ð z0 =ð |z ?ð z0| Pð ²ð . _ð Thus, 6.1 fÅ»ðzÞð fÅ»ðzÞð ?ð fÅ»ðz0Þð Xð Xð z ?ð z0 dz ?ð 2^ðifÅ»ðz0Þð =ð z ?ð z0 dz C0 C0 Pð ²ð 2^ð_ð =ð 2^ðPð. _ð But Pð is any positive number, and so fÅ»ðzÞð Xð z ?ð z0 dz ?ð 2^ðifÅ»ðz0Þð =ð 0, C0 or, fÅ»ðzÞð fÅ»ðzÞð 1 1 fÅ»ðz0Þð =ð Xð Xð z ?ð z0 dz =ð 2^ði z ?ð z0 dz, 2^ði C0 C which is exactly what we set out to show. Meditate on this result. It says that if f is analytic on and inside a simple closed curve and we know the values fÅ»ðzÞð for every z on the simple closed curve, then we know the value for the function at every point inside the curve quite remarkable indeed. Example Let C be the circle z =ð 4 traversed once in the counterclockwise direction. Let s evaluate | | the integral cos z dz. Xð z2 ?ð 6z +ð 5 C We simply write the integrand as fÅ»ðzÞð cos z cos z =ð =ð , z ?ð 1 Å»ðz ?ð 5ÞðÅ»ðz ?ð 1Þð z2 ?ð 6z +ð 5 where cos z fÅ»ðzÞð =ð . z ?ð 5 Observe that f is analytic on and inside C, and so, 6.2 fÅ»ðzÞð cos z dz =ð dz =ð 2^ðifÅ»ð1Þð Xð Xð z ?ð 1 z2 ?ð 6z +ð 5 C C cos 1 i^ð =ð 2^ði =ð ?ð cos 1 1 ?ð 5 2 Exercises 1. Suppose f and g are analytic on and inside the simple closed curve C, and suppose moreover that fÅ»ðzÞð =ð gÅ»ðzÞð for all z on C. Prove that fÅ»ðzÞð =ð gÅ»ðzÞð for all z inside C. 2. Let C be the ellipse 9x2 +ð 4y2 =ð 36 traversed once in the counterclockwise direction. Define the function g by s2 +ð s +ð 1 gÅ»ðzÞð =ð ds. Xð s ?ð z C Find a) gÅ»ðiÞð b) gÅ»ð4iÞð 3. Find e2z dz, Xð z2 ?ð 4 C where C is the closed curve in the picture: e2z 4. Find Xð dz, where @ð is the contour in the picture: z2?ð4 @ð 6.3 6.2. Functions defined by integrals. Suppose C is a curve (not necessarily a simple closed curve, just a curve) and suppose the function g is continuous on C (not necessarily analytic, just continuous). Let the function G be defined by gÅ»ðsÞð GÅ»ðzÞð =ð ds Xð s ?ð z C for all z 6ð C. We shall show that G is analytic. Here we go. Consider, GÅ»ðz +ðAðzÞð ?ð GÅ»ðzÞð 1 1 1 =ð ?ð gÅ»ðsÞðds Xð s ?ð z Aðz Aðz s ?ð z ?ð Aðz C gÅ»ðsÞð =ð ds. Xð Å»ðs ?ð z ?ð AðzÞðÅ»ðs ?ð zÞð C Next, GÅ»ðz +ðAðzÞð ?ð GÅ»ðzÞð gÅ»ðsÞð 1 1 ?ð ds =ð ?ð gÅ»ðsÞðds Xð Xð Aðz Å»ðs ?ð z ?ð AðzÞðÅ»ðs ?ð zÞð Å»ðs ?ð zÞð2 Å»ðs ?ð zÞð2 C C Å»ðs ?ð zÞð ?ð Å»ðs ?ð z ?ð AðzÞð =ð gÅ»ðsÞðds Xð Å»ðs ?ð z ?ð AðzÞðÅ»ðs ?ð zÞð2 C gÅ»ðsÞð =ð Aðz ds. Xð Å»ðs ?ð z ?ð AðzÞðÅ»ðs ?ð zÞð2 C Now we want to show that 6.4 gÅ»ðsÞð lim Aðz ds =ð 0. Xð Å»ðs ?ð z ?ð AðzÞðÅ»ðs ?ð zÞð2 Aðz¸ð0 C To that end, let M =ð maxáð gÅ»ðsÞð : s 5ð Câð, and let d be the shortest distance from z to C. | | Thus, for s 5ð C, we have s ?ð z Å‚ð d >ð 0 and also | | s ?ð z ?ð Aðz Å‚ð s ?ð z ?ð Aðz Å‚ð d ?ð Aðz . | | | | | | | | Putting this all together, we can estimate the integrand above: gÅ»ðsÞð M ²ð Å»ðs ?ð z ?ð AðzÞðÅ»ðs ?ð zÞð2 Å»ðd ?ð Aðz Þðd2 | | for all s 5ð C. Finally, gÅ»ðsÞð M Aðz ds ²ð Aðz lengthÅ»ðCÞð, Xð | | Å»ðs ?ð z ?ð AðzÞðÅ»ðs ?ð zÞð2 Å»ðd ?ð Aðz Þðd2 | | C and it is clear that gÅ»ðsÞð lim Aðz ds =ð 0, Xð Å»ðs ?ð z ?ð AðzÞðÅ»ðs ?ð zÞð2 Aðz¸ð0 C just as we set out to show. Hence G has a derivative at z, and gÅ»ðsÞð GvðÅ»ðzÞð =ð ds. Xð Å»ðs ?ð zÞð2 C Truly a miracle! Next we see that Gvð has a derivative and it is just what you think it should be. Consider 6.5 GvðÅ»ðz +ðAðzÞð ?ð GvðÅ»ðzÞð 1 1 1 =ð ?ð gÅ»ðsÞðds Xð Aðz Aðz Å»ðs ?ð z ?ð AðzÞð2 Å»ðs ?ð zÞð2 C Å»ðs ?ð zÞð2 ?ð Å»ðs ?ð z ?ð AðzÞð2 gÅ»ðsÞðds 1 =ð Xð Aðz Å»ðs ?ð z ?ð AðzÞð2Å»ðs ?ð zÞð2 C 2Å»ðs ?ð zÞðAðz ?ð Å»ðAðzÞð2 gÅ»ðsÞðds 1 =ð Xð Aðz Å»ðs ?ð z ?ð AðzÞð2Å»ðs ?ð zÞð2 C 2Å»ðs ?ð zÞð ?ð Aðz =ð gÅ»ðsÞðds Xð Å»ðs ?ð z ?ð AðzÞð2Å»ðs ?ð zÞð2 C Next, GvðÅ»ðz +ðAðzÞð ?ð GvðÅ»ðzÞð gÅ»ðsÞð ?ð 2 ds Xð Aðz Å»ðs ?ð zÞð3 C 2Å»ðs ?ð zÞð ?ð Aðz 2 =ð ?ð gÅ»ðsÞðds Xð Å»ðs ?ð z ?ð AðzÞð2Å»ðs ?ð zÞð2 Å»ðs ?ð zÞð3 C 2Å»ðs ?ð zÞð2 ?ð AðzÅ»ðs ?ð zÞð ?ð 2Å»ðs ?ð z ?ð AðzÞð2 gÅ»ðsÞðds =ð Xð Å»ðs ?ð z ?ð AðzÞð2Å»ðs ?ð zÞð3 C 2Å»ðs ?ð zÞð2 ?ð AðzÅ»ðs ?ð zÞð ?ð 2Å»ðs ?ð zÞð2 +ð 4AðzÅ»ðs ?ð zÞð ?ð 2Å»ðAðzÞð2 gÅ»ðsÞðds =ð Xð Å»ðs ?ð z ?ð AðzÞð2Å»ðs ?ð zÞð3 C 3AðzÅ»ðs ?ð zÞð ?ð 2Å»ðAðzÞð2 gÅ»ðsÞðds =ð Xð Å»ðs ?ð z ?ð AðzÞð2Å»ðs ?ð zÞð3 C Hence, GvðÅ»ðz +ðAðzÞð ?ð GvðÅ»ðzÞð gÅ»ðsÞð 3AðzÅ»ðs ?ð zÞð ?ð 2Å»ðAðzÞð2 gÅ»ðsÞðds ?ð 2 ds =ð Xð Xð Aðz Å»ðs ?ð zÞð3 Å»ðs ?ð z ?ð AðzÞð2Å»ðs ?ð zÞð3 C C Å»ð 3m +ð 2 Aðz ÞðM | | | | ²ð Aðz , | | Å»ðd ?ð AðzÞð2d3 where m =ð maxáð s ?ð z : s 5ð Câð. It should be clear then that | | GvðÅ»ðz +ðAðzÞð ?ð GvðÅ»ðzÞð gÅ»ðsÞð lim ?ð 2 ds =ð 0, Xð Aðz Å»ðs ?ð zÞð3 Aðz¸ð0 C or in other words, 6.6 gÅ»ðsÞð GvðvðÅ»ðzÞð =ð 2 ds. Xð Å»ðs ?ð zÞð3 C Suppose f is analytic in a region D and suppose C is a positively oriented simple closed curve in D. Suppose also the inside of C is in D. Then from the Cauchy Integral formula, we know that fÅ»ðsÞð 2^ðifÅ»ðzÞð =ð ds Xð s ?ð z C and so with g =ð f in the formulas just derived, we have fÅ»ðsÞð fÅ»ðsÞð 1 2 fvðÅ»ðzÞð =ð ds, and fvðvðÅ»ðzÞð =ð ds Xð Xð 2^ði 2^ði Å»ðs ?ð zÞð2 Å»ðs ?ð zÞð3 C C for all z inside the closed curve C. Meditate on these results. They say that the derivative of an analytic function is also analytic. Now suppose f is continuous on a domain D in which every point of D is an interior point and suppose that Xð fÅ»ðzÞðdz =ð 0 for every closed C curve in D. Then we know that f has an antiderivative in D in other words f is the derivative of an analytic function. We now know this means that f is itself analytic. We thus have the celebrated Morera s Theorem: If f:D ¸ð C is continuous and such that Xð fÅ»ðzÞðdz =ð 0 for every closed curve in D, then f is C analytic in D. Example Let s evaluate the integral ez dz, Xð z3 C where C is any positively oriented closed curve around the origin. We simply use the equation fÅ»ðsÞð 2 fvðvðÅ»ðzÞð =ð ds Xð 2^ði Å»ðs ?ð zÞð3 C 6.7 with z =ð 0 and fÅ»ðsÞð =ð es. Thus, ez ^ðie0 =ð ^ði =ð dz. Xð z3 C Exercises 5. Evaluate sin z dz Xð z2 C where C is a positively oriented closed curve around the origin. 6. Let C be the circle z ?ð i =ð 2 with the positive orientation. Evaluate | | 1 1 a) Xð dz b) Xð dz z2+ð4 Å»ðz2+ð4Þð2 C C 7. Suppose f is analytic inside and on the simple closed curve C. Show that fvðÅ»ðzÞð fÅ»ðzÞð dz =ð dz Xð Xð z ?ð w Å»ðz ?ð wÞð2 C C for every w 6ð C. 8. a) Let Jð be a real constant, and let C be the circle LðÅ»ðtÞð =ð eit, ?ð^ð ²ð t ²ð ^ð. Evaluate eJðz dz. Xð z C b) Use your answer in part a) to show that ^ð eJð cos t cosÅ»ðJð sin tÞðdt =ð ^ð. Xð 0 6.3. Liouville s Theorem. Suppose f is entire and bounded; that is, f is analytic in the entire plane and there is a constant M such that fÅ»ðzÞð ²ð M for all z. Then it must be true | | that fvðÅ»ðzÞð =ð 0 identically. To see this, suppose that fvðÅ»ðwÞð ®ð 0 for some w. Choose R large M enough to insure that <ð fvðÅ»ðwÞð . Now let C be a circle centered at 0 and with radius | | R 6.8 _ð >ð maxáðR, w âð. Then we have : | | fÅ»ðsÞð M 1 <ð fvðÅ»ðwÞð ²ð dz | | Xð _ð 2^ði Å»ðs ?ð wÞð2 C 1 M M ²ð 2^ð_ð =ð , _ð 2^ð _ð2 a contradiction. It must therefore be true that there is no w for which fvðÅ»ðwÞð ®ð 0; or, in other words, fvðÅ»ðzÞð =ð 0 for all z. This, of course, means that f is a constant function. What we have shown has a name, Liouville s Theorem: The only bounded entire functions are the constant functions. Let s put this theorem to some good use. Let pÅ»ðzÞð =ð anzn +ð an?ð1zn?ð1 +ðuð +ða1z +ð a0 be a polynomial. Then an?ð1 an?ð2 a0 pÅ»ðzÞð =ð an +ð +ð +ðuð +ð zn. z zn z2 an?ðj |an| Now choose R large enough to insure that for each j =ð 1, 2, uð , n, we have <ð zj 2n whenever z >ð R. (We are assuming that an ®ð 0. ) Hence, for z >ð R, we know that | | | | an?ð1 +ð an?ð2 +ðuð +ð a0 |z|n pÅ»ðzÞð Å‚ð an ?ð | | | | z zn z2 an?ð1 ?ð an?ð2 ?ðuð ?ð a0 |z|n Å‚ð an ?ð | | z zn z2 an an an | |n | | | | | | >ð an ?ð ?ð ?ðuð ?ð z | | 2n 2n 2n an | |n | | >ð z . 2 Hence, for z >ð R, | | 1 2 2 <ð ²ð . an z an Rn pÅ»ðzÞð | || |n | | 1 1 Now suppose pÅ»ðzÞð ®ð 0 for all z. Then is also bounded on the disk z ²ð R. Thus, | | pÅ»ðzÞð pÅ»ðzÞð is a bounded entire function, and hence, by Liouville s Theorem, constant! Hence the polynomial is constant if it has no zeros. In other words, if pÅ»ðzÞð is of degree at least one, there must be at least one z0 for which pÅ»ðz0Þð =ð 0. This is, of course, the celebrated 6.9 Fundamental Theorem of Algebra. Exercises 9. Suppose f is an entire function, and suppose there is an M such that Re fÅ»ðzÞð ²ð M for all z. Prove that f is a constant function. 10. Suppose w is a solution of 5z4 +ð z3 +ð z2 ?ð 7z +ð 14 =ð 0. Prove that w ²ð 3. | | 11. Prove that if p is a polynomial of degree n, and if pÅ»ðaÞð =ð 0, then pÅ»ðzÞð =ð Å»ðz ?ð aÞðqÅ»ðzÞð, where q is a polynomial of degree n ?ð 1. 12. Prove that if p is a polynomial of degree n Å‚ð 1, then pÅ»ðzÞð =ð cÅ»ðz ?ð z1Þðk1Å»ðz ?ð z2Þðk2uð Å»ðz ?ð zjÞðkj, where k1, k2, uð , kj are positive integers such that n =ð k1 +ð k2 +ðuð +ðkj. 13. Suppose p is a polynomial with real coefficients. Prove that p can be expressed as a product of linear and quadratic factors, each with real coefficients. 6.4. Maximum moduli. Suppose f is analytic on a closed domain D. Then, being continuous, fÅ»ðzÞð must attain its maximum value somewhere in this domain. Suppose this | | happens at an interior point. That is, suppose fÅ»ðzÞð ²ð M for all z 5ð D and suppose that | | fÅ»ðz0Þð =ð M for some z0 in the interior of D. Now z0 is an interior point of D, so there is a | | number R such that the disk Cð centered at z0 having radius R is included in D. Let C be a positively oriented circle of radius _ð ²ð R centered at z0. From Cauchy s formula, we know fÅ»ðsÞð 1 fÅ»ðz0Þð =ð Xð s ?ð z0 ds. 2^ði C Hence, 2^ð 1 fÅ»ðz0Þð =ð fÅ»ðz0 +ð _ðeitÞðdt, Xð 2^ð 0 and so, 6.10 2^ð 1 M =ð fÅ»ðz0Þð ²ð | | fÅ»ðz0 +ð _ðeitÞð dt ²ð M. | | Xð 2^ð 0 since fÅ»ðz0 +ð _ðeitÞð ²ð M. This means | | 2^ð 1 M =ð | | fÅ»ðz0 +ð _ðeitÞð dt. Xð 2^ð 0 Thus, 2^ð 2^ð 1 1 M ?ð | | Xðßð | |Ä…ð fÅ»ðz0 +ð _ðeitÞð dt =ð M ?ð fÅ»ðz0 +ð _ðeitÞð dt =ð 0. Xð 2^ð 2^ð 0 0 This integrand is continuous and non-negative, and so must be zero. In other words, fÅ»ðzÞð =ð M for all z 5ð C. There was nothing special about C except its radius _ð ²ð R, and so | | we have shown that f must be constant on the disk Cð. I hope it is easy to see that if D is a region (=ðconnected and open), then the only way in which the modulus fÅ»ðzÞð of the analytic function f can attain a maximum on D is for f to be | | constant. Exercises 14. Suppose f is analytic and not constant on a region D and suppose fÅ»ðzÞð ®ð 0 for all z 5ð D. Explain why fÅ»ðzÞð does not have a minimum in D. | | 15. Suppose fÅ»ðzÞð =ð uÅ»ðx, yÞð +ð ivÅ»ðx, yÞð is analytic on a region D. Prove that if uÅ»ðx, yÞð attains a maximum value in D, then u must be constant. 6.11