Calculus Solutions for Special Assignment 8

Solutions for Special Assignment 8 Problem 1: Find a Möbius transformation that involves an inversion (so c 6= 0) and has only one fixed point. Solution: Assuming the standard form for Möbius transformations, fixed points occur when p a − d ± (d − a)2 + 4bc az + b 2 = z ⇔ cz + (d − a)z − b = 0 ⇔ z = , cz + d 2c assuming that c 6= 0 and we are careful evaluating the square root in the set of complex numbers. This equation has only one distinct root when (d − a)2 = −4bc. Here are two possibilities: 3z − 1 i. the constants a = 3, d = 1, b = −1, c = 1 give f (z) = , which has 1 as the only fixed point; z+1 ii. the constants a = 1 + 2i, d = 1, b = 1, c = 1 give f (z) = (1 + 2i)z + 1 , which has i as the only fixed point. z+1 Of course, there are infinitely many other options. Problem 2: Find a Möbius transformation that has two fixed points, namely 3i and 1 + i. Solution: Referring to the solution for Problem 1, the distinct points α and β are fixed points when (z − α)(z − β) = cz 2 + (d − a)z − b which occurs when c = 1, a − d = α + β, b = −αβ. Let k be any constant other than α or β. Then one possible set of solutions is a = k and d = k − α − β, and these values yield T (z) = kz − αβ . z+k−α−β (Note what happens if k = α or k = β.) For the given problem, we can let k = 1 and obtain T (z) = z + 3 − 3i . z − 4i As a check, we note that T (3i) = 3 = 3i −i and T (1 + i) = 4 − 2i 1 1 = (2 − i)(1 + 3i) = (5 + 5i) = 1 + i. 1 − 3i 5 5 As with the previous problem, there are many other options. Problem 3: Find a Möbius transformation that maps the interior of the circle {z : |z − (−1 + i)| = 1} to the lower half plane. Solution: Sketching a graph (omitted here) and traversing the boundaries so that the appropriate region is on the left, we find that one option is to map the points −1, i, and −1 + 2i on the circle to the points 0, −1, and ∞ on the real axis. Using the cross-ratio notation, the desired Möbius transformation T is T (z) = −(z, −1, i, −1 + 2i) = − (z + 1)(1 − i) z+1 (1 − i)2 iz + i = · = . (z + 1 − 2i)(1 + i) z + 1 − 2i −2 z + 1 − 2i There are, of course, many other options. In addition, it is a good idea to check the final form of the mapping to ensure that it has the desired properties, but we do not include these details. However, note that T (∞) = i, which indicates that the exterior of the circle maps to the upper half plane. 1 Problem 4: Find the image of the first quadrant under the Möbius transformation f (z) = (2z + i)/(z + 1). Solution: We make note of the following functional values: f (∞) = 2, f (i) = 3i 3 3 3 = i(1 − i) = + i, 1+i 2 2 2 f (0) = i, f (1) = 1 + 1 i, 2 f (∞) = 2. Tracing these points in the given order makes the first quadrant be the left region. Since the blue line on the left does not pass through the pole of the function, its image will be a circle. Since the red line on the left does pass through the pole of the function, its image will be a line. These curves, and their corresponding directions, are represented on the graph at the right. y y < • ∨ • • • • • > ∗ x x The colored dots match the inputs with the outputs, with the asterisk indicating the image of ∞. Hence, the image √ of the first quadrant under this mapping is the upper half of the circle centered at 1 + 12 i with radius 5/2. We can also check some sample points to see where their images go. Choosing values that make the numbers come out nice, we find that f (−i/2) = 0, f (−1) = ∞, f ((1 + i)/2) = 1 + 2i 1 = (1 + 2i)(3 − i) = 1 + i. (3 + i)/2 5 The first two values show that points outside the first quadrant map to points outside the circle and the third value shows that a point in the first quadrant maps into the interior of the circle, as expected. 2 Selected Solutions for Third Exam Problem 2b: Determine the function represented by the Maclaurin series ∞ X 2k + 1 k=0 k! z 2k . z Solution: Using properties of Maclaurin series and the series for e , we find that ∞ X 2k + 1 k=0 k! z 2k = ∞ 2 2 d X 1 2k+1 d z = zez = (2z 2 + 1)ez . dz k! dz k=0 This function can also be found by breaking the sum into parts and reindexing. Problem 4b: Consider the function f (z) = 4 . Find its Laurent series for |z − 5| > 5. z(z − 4) Solution: Using properties of the geometric series, we find that 1 1 − z−4 z 1 1 = − z−5+1 z−5+5  1  1 1 =  −1  −  −5  z−5 1− 1− z−5 z−5 ∞ ∞  X (−1)k X (−5)k  1 − = z−5 (z − 5)k (z − 5)k f (z) = k=0 = ∞ X k=0 = ∞ X k=0 k (−1) − (−5) (z − 5)k+1 k
(−1)j−1 − (−5)j−1 (z − 5)−j . j=2 The last version is not necessary; it just indicates another way to write the series. 3 Z ∞ Problem 8: Use techniques in Section 6.3 to show that −∞ x8
1 π dx = sin(π/8) + cos(π/8) . +1 2 1 . The zeros of the denominator in the upper half plane +1 are z1 = eiπ/8 , z2 = e3iπ/8 , z3 = e5iπ/8 , and z4 = ei7π/8 . Using Theorem 4 in the Chapter 6 extra notes and our Solution: Consider the rational function R(z) = z8 quick way to find the residues of simple poles (Example 2 in Section 6.1), we find that Z ∞ −∞  1 1 1 1  1 dx = 2πi + + + x8 + 1 8z17 8z27 8z37 8z47 = = = = =
πi −i7π/8 e + e−i21π/8 + e−i35π/8 + e−i49π/8 4
πi −eiπ/8 − ei3π/8 + e−i3π/8 + e−iπ/8 4 π  eiπ/8 − e−iπ/8 ei3π/8 − e−i3π/8  + 2 2i 2i
π sin(π/8) + sin(3π/8) 2
π sin(π/8) + cos(π/8) . 2 The last step follows from the fact that cos θ = sin( 12 π − θ). Problem 10: Find a Möbius transformation that maps the interior of the circle {z : |z − (1 + i)| = 1} onto the lower half plane {z : Im z < 0}. Solution: Sketching a graph (omitted here) and traversing the boundaries so that the appropriate region is on the left, we find that one option is to map the points i, 1, and 2 + i on the circle to the points 0, −1, and ∞ on the real axis. Using the cross-ratio notation, the desired Möbius transformation T is T (z) = −(z, i, 1, 2 + i) = − (z − i)(−1 − i) z−i (1 + i)2 iz + 1 = · = . (z − 2 − i)(1 − i) z−2−i 2 z−2−i It is easy to check that this function has the correct three values. In addition, the fact that T (∞) = i confirms that the exterior of the circle maps to the upper half plane. 4 Solutions for Special Assignment 7 Z 2π Problem 1: Use the techniques from Section 6.2 to evaluate 0 1 dθ. (2 + cos θ)2 Solution: We know that this integral exists since the integrand is continuous on the interval [0, 2π]. Using results from Section 6.2, we find that Z 2π 0 √ I dz 1  2 · 1 1 iz |z|=1 2 + z+ 2 z I 1 4z = dz i |z|=1 z 2 + 4z + 1   z , z , = 8π Res 1 (z − z1 )2 (z − z2 )2 1 dθ = (2 + cos θ)2  √ 3 are the two roots of the polynomial z 2 + 4z + 1. As noted in the last step √ above, only the root z1 lies within the unit circle. Using the fact that z1 − z2 = 2 3, we find that where z1 = −2 + 3 and z2 = −2 −  z z d Res , z1 = 2 2 (z − z1 ) (z − z2 ) dz (z − z2 )2  It follows that 2π Z 0 z=z1 √ √
√ 12 − 4 3 −2 + 3 (z1 − z2 )2 − 2z1 (z1 − z2 ) 3 = = = . (z1 − z2 )4 144 18 √ √ 3 1 4 3 dθ = 8π · = π. (2 + cos θ)2 18 9 Z ∞ Problem 2: Use the techniques from Section 6.3 to evaluate −∞ x4 dx. x6 + 1 Solution: Since the integrand is a rational function and the degree of the numerator is two less than the degree z4 of the denominator, the improper integral exists. Consider the rational function R(z) = 6 . The zeros of the z +1 denominator in the upper half plane are z1 = eiπ/6 , z2 = eiπ/2 , and z3 = ei5π/6 . Using Theorem 4 in the Chapter 6 notes and our quick way to find the residues of simple poles (Example 2 in Section 6.1), we find that Z ∞ −∞  z4 x4 z24 z34  1 dx = 2πi + + x6 + 1 6z15 6z25 6z35 πi  1 1 1 = + + 3 z1 z2 z3
πi −iπ/6 = e + e−iπ/2 + e−i5π/6 3 √ √ πi  3 1 3 1  = − i−i− − i 3 2 2 2 2 2π = . 3 5 Z ∞ Problem 3: Use the techniques from Section 6.4 to evaluate −∞ x3 sin x dx. (x2 + 16)2 Solution: This problem is very similar to Example 9 in the Chapter 6 notes. Since the integrand is the sine function multiplied by a rational function for which the degree of the numerator is one less than the degree of the denominator, the improper integral exists. We first note that Z ∞ −∞ x3 sin x dx = (x2 + 16)2 Z ∞ −∞ x3 sin x dx − i (x2 + 16)2 Z ∞ −∞ x3 cos x dx = (x2 + 16)2 Z ∞ −∞ −ix3 eix dx (x2 + 16)2 since x3 cos x is an odd function. Let f (z) = −iz 3 eiz /(z 2 + 16)2 and note that the only 0 of the denominator that lies in the upper half plane is 4i. Applying Theorem 7 from the Chapter 6 notes, we find that Z ∞ −∞ x3 sin x dx = (x2 + 16)2 Z ∞ −∞ −ix3 eix dx (x2 + 16)2 = 2πi Res (f, 4i) = 2π · d z 3 eiz dz (z + 4i)2 z=4i
(8i) (4i) (ie−4 ) + 3(4i)2 e−4 − (4i)3 e−4 2(8i) = 2π · (8i)4  
2π = 4 4 (8i)2 i(4i)3 + 3(4i)2 − (4i)3 (16i) 8 e 
2π  = 4 4 (−64) 64 − 48 − 64 · 16 8 e  2π  = 4 4 −2 · 64 · 16 8 e 2π(4 · 83 ) =− 84 e4 π = − 4. e 2 Z ∞ Problem 4: It is known that −∞ 3 sin x dx = π. Use this fact and integration by parts to evaluate x Z ∞ −∞ 1 − cos x dx. x2 Solution: Using integration by parts with u = 1 − cos x and dv = dx/x2 , we find that Z 1 − cos x dx = uv − x2 Z v du = − 1 − cos x − x Z −1 cos x − 1 · sin x dx = + x x Z sin x dx. x For each positive number b, we find that Z b −b 1 − cos x cos x − 1 dx = x2 x Z b −b b + −b sin x cos b − 1 cos(−b) − 1 dx = − + x b −b Z b −b sin x dx. x Since 1 − cos b is nonnegative and bounded above by 2, it follows that Z ∞ −∞ 1 − cos x dx = lim b→∞ x2 Z b −b  2(cos b − 1) 1 − cos x dx = lim + 2 b→∞ x b 6 Z b −b Z ∞ sin x  sin x dx = dx = π. x −∞ x I Exercise 6.1.3e: Use the Cauchy Residue Theorem to evaluate |z|=1 z2 1 dz. sin z Solution: Since the function z/ sin z is analytic at 0 and has a value of 1 at 0, the function z/ sin z 1 = z 2 sin z z3 has a pole of order 3 at 0. With f (z) = z/ sin z, the Cauchy Residue Theorem yields I |z|=1 z2  1  f 00 (0) 1 πi dz = 2πi Res 2 , 0 = 2πi · = πif 00 (0) = . sin z z sin z 2 3 The somewhat tedious part of the problem is showing that f 00 (0) = 13 . There are multiple ways to do this; two of these are presented below. Consider the function g(z) = (sin z)/z and note that (using the Maclaurin series for sin z) g(z) = 1 − 1 4 1 1 2 z + z − z6 + · · · . 6 120 7! It is easy to see that g(0) = 1, g 0 (0) = 0, and g 00 (0) = − 13 . Since f is the reciprocal of g, we find that f 0 (z) = −g 0 (z) (g(z))2 It then follows that f 00 (0) = and f 00 (z) = −(g(z))2 g 00 (z) + g 0 (z)2g(z)g 0 (z) (g(z))4 −1(− 31 ) + 0 −(g(0))2 g 00 (0) + g 0 (0)2g(0)g 0 (0) 1 = = . 4 (g(0)) 1 3 For a second method, we note that f has a Maclaurin series at 0 with radius of convergence π. (Why is this true?) Since f (0) = 1 and f is an even function, we know that f (z) = 1 + a2 z 2 + a4 z 4 + · · ·. Using the fact that f (z)g(z) = 1, we see that    1 4 1 1 z − z6 + · · · 1 = 1 + a2 z 2 + a4 z 4 + a6 z 6 + · · · 1 − z 2 + 6 120 7!   1 2  1 a2 = 1 + a2 − z + − + a4 z 4 + · · · . 6 120 6 Since Taylor series at a given center are unique, all of the coefficients on powers of z must be 0. It follows that a2 = 1 6 and thus f 00 (0) = 31 . For the record, we also have a4 = a2 1 1 1 7 − = − = 6 120 36 120 360 and thus f (4) (0) = Knowing the value of a4 , we find that I |z|=1  f (z)  1 7πi dz = 2πi Res , 0 = 2πi a4 = . z 4 sin z z5 180 7 7 15 Solutions for Special Assignment 6 Problem 1: Find the sum of the series ∞ P 2+(−1)k
k z k and use the result to determine the radius of convergence k=0 of the series. Solution: For each positive integer n, we find that (combining the even and odd terms, respectively) 2n+1 X
k k k 2 + (−1) z = k=0 n X
2k 2k 2k z 2 + (−1) + k=0 = n X = 2 + (−1)2k+1
2k+1 z 2k+1 k=0 32k z 2k + k=0 n X n X n X z 2k+1 k=0 (9z 2 )k + z k=0 n X (z 2 )k . k=0 Using the formula for the sum of a geometric series, it then follows that ∞ X 2 + (−1)k
k zk = ∞ X (9z 2 )k + z k=0 k=0 ∞ X (z 2 )k = k=0 z −9z 3 − z 2 + z + 1 1 + = , 2 2 1 − 9z 1−z (1 − z 2 )(1 − 9z 2 ) where the first sum is valid when |9z 2 | < 1 and the second sum is valid for |z 2 | < 1. Hence, the series converges for all values of z that satisfy |z| < 31 , that is, the radius of convergence for the power series is distance from the center (z0 = 0) to the the nearest point where the sum is not analytic (z = 1 3. ± 31 ) Note that the is 13 . Problem 2: Let c be a real number that satisfies 0 < c < 1. A frog jumps one meter (from z = 0 to z = 1) on its first jump, c meters on its second jump, c2 meters on its third jump, and so on, each time turning exactly an angle α to its left. Show that the frog lands on a circle, where the circle depends on the value of c but does not depend on the value of α. Solution: Given the nature of the frog’s jumps, the limiting position of the frog is given by S = 1 + ceiα + c2 e2iα + c3 e3iα + · · · = 1 , 1 − ceiα where we have used the formula for the sum of a geometric series. (We have convergence since |ceiα | = |c| < 1.) We next note that (1 − c2 )S − 1 = 1 − c2 −c2 + ceiα 1 − ce−iα iα − 1 = = |ce | · = c, 1 − ceiα 1 − ceiα 1 − ceiα where we have used the fact that 1 − ce−iα is the complex conjugate of 1 − ceiα and thus has the same modulus. Writing this equation in the form S− 1 c = 1 − c2 1 − c2 reveals that the frog lands on the circle with center z0 = the answer given in the textbook for Exercise 5.1.21. 8 c 1 and radius ρ = . When c = 12 , we obtain 1 − c2 1 − c2 ∞ X (−1)k 2k z satisfies the differential equation zf 00 (z) + f 0 (z) + zf (z) = 0. Problem 3: Show that f (z) = 4k (k!)2 k=0 Solution: We first note that f (z) = ∞ X (−1)k 2k z , 4k (k!)2 f 0 (z) = k=0 ∞ X (−1)k 2k 4k (k!)2 k=1 z 2k−1 , f 00 (z) = and ∞ X (−1)k 2k(2k − 1) k=1 4k (k!)2 z 2k−2 . (Make note of the starting index for each series.) It follows that 00 0 zf (z) + f (z) = ∞ X (−1)k 2k(2k − 1) 4k (k!)2 k=1 = 4k (k!)2 ∞ X k=1 = = −z 4k (k!)2 z 2k−1 z 2k−1 (−1)k z 2k−1 − 1)!)2 4k−1 ((k ∞ X (−1)k+1 k=0 + ∞ X (−1)k 2k k=1 ∞ X (−1)k 2k(2k − 1 + 1) k=1 = z 2k−1 4k (k!)2 z 2k+1 ∞ X (−1)k 2k z 4k (k!)2 k=0 = −zf (z). Hence, the function f satisfies the differential equation zf 00 (z) + f 0 (z) + zf (z) = 0. Problem 4: Consider the function g(z) = 1 . Find the Laurent series for g in each of the domains (z + 2)(z − 3) 0 < |z| < 2, 2 < |z| < 3, and |z − 3| > 5. Solution: Since f is analytic except at the points −2 and 3, we know the series in the domain 0 < |z| < 2 will be a Maclaurin series. Using the geometric series formula, we find that g(z) = 1 1 1 1  = − (z + 2)(z − 3) 5 z−3 z+2 1  1  − 31 2 = − 1 1 5 1 − 3z 1 − (− 2 z) ∞ ∞ 1 X −1 k X (−1)k k  = z − z 5 3k+1 2k+1 k=0 = k=0 = k=0 ∞ X 1  (−1)k+1 5 2k+1 − 1  3k+1 ∞  X (−3)k+1 − 2k+1  k=0 30 · 6k zk zk . The final form (or the one right above it) is preferred since it represents the standard form of a power series and clearly shows the formula for the coefficients. 9 For z that satisfy 2 < |z| < 3, we can use the same series above for 1/(z − 3) (since it converges for |z| < 3) but we need to handle the other fraction differently: ∞ ∞ k=0 k=1 X X (−1)k 2k 1 1/z 1 = = = (−1)k−1 2k−1 · k k+1 z+2 1 − (−2/z) z z (Note that this series converges for | − 2/z| < 1 ⇔ |z| > 2.) It follows that g(z) = ∞ ∞ k=0 k=1 X (−1)k 2k 1 1 1 1  X −1 k = z − + · k 5 z−3 z+2 15 · 3k 10 z This gives the Laurent series for the domain 2 < |z| < 3. Finally, for the domain |z − 3| > 5, we first note that 1 1 = = z+2 z−3+5 1 ∞ 1 z − 3 = X(−1)k 5k · , −5 (z − 3)k+1 k=0 1− z−3 which converges for |z − 3| > 5. It follows that g(z) = ∞ ∞ k=0 k=2 X X (−1)k 5k 1 1 1 1 · = (−1)k 5k · = · . k+2 z−3 z−2 (z − 3) 25 (z − 3)k This is the desired Laurent series. Note that we could have obtained this last series using the partial fraction decomposition. In the expansion of 1/(z + 2) just above, note that the first term is 1/(z − 3). We then have g(z) = ∞ ∞ k=1 k=2 X (−1)k 5k 1 1 1 1  1X 1 (−1)k+1 5k · − = · , = k+1 5 z−3 z+2 5 (z − 3) 25 (z − 3)k the same as the previous answer. 10 Solutions for Special Assignment 5 Z Problem 1: Evaluate Γ (z 2 eiz dz, where Γ is the circle |z| = 3 traversed once counterclockwise. + 1)2 Solution: We present a slightly different approach to this problem than the one suggested by the textbook. We first note that  1  1 1 1 = − . 2 2 2 + 1) 4iz (z − i) (z + i) (z 2 It then follows that Z Γ eiz dz = (z 2 + 1)2 Z Γ f (z) dz − (z − i)2 Z Γ f (z) dz, (z + i)2 where f (z) = eiz /(4iz). Using Theorem 4.19, we find that Z Γ Since (z 2   eiz  z + i  (4iz)(ieiz ) − (eiz )(4i) iz −4z − 4i = e = , (4iz)2 −16z 2 4 z2 f 0 (z) = we find that f 0 (i) = It follows that Z Γ
eiz dz = 2πi f 0 (i) − f 0 (−i) . 2 + 1) e−1  2i  1 = 4 −1 2ei and f 0 (−i) = 0.  1  π
eiz 0 0 dz = 2πi f (i) − f (−i) = 2πi − 0 = . (z 2 + 1)2 2ei e To solve this problem using the approach suggested by the textbook, we use the Deformation Theorem (the graph in Figure 4.37 might be helpful) to obtain Z Γ eiz dz = 2 (z + 1)2 Z C1 eiz dz + 2 (z + 1)2 Z C2 (z 2 eiz dz, + 1)2 where C1 is the positively oriented circle of radius 1 centered at i and C2 is the positively oriented circle of radius 1 centered at −i. By the Cauchy integral formula, we find that Z Γ eiz dz = (z 2 + 1)2 Z C1 eiz /(z + i)2 dz + (z − i)2 Z C2 eiz /(z − i)2 dz = 2πi (z + i)2  d eiz dz (z + i)2 d eiz + dz (z − i)2 z=i  . z=−i Since d eiz dz (z + i)2 d eiz dz (z − i)2 = (2i)2 (ie−1 ) − (e−1 )(4i) −4ie−1 − 4ie−1 −i = = ; (2i)4 16 2e = (−2i)2 (ie) − (e)(−4i) −4ie + 4ie = = 0; (−2i)4 16 z=i z=−i (notice how the derivative step and the evaluate step have been combined to save a little work), it follows that Z Γ  −i  π eiz dz = 2πi + 0 = . (z 2 + 1)2 2e e 11 Problem 2: Let f be analytic in a bounded domain D and continuous up to and including its boundary. If f is nonzero in D, then the modulus |f (z)| attains its minimum on the boundary of D. Solution: There are two cases to consider. Suppose first that f (z0 ) = 0 at some point z0 on the boundary of D. Since |f (z0 )| = 0, it is clear that the modulus |f (z)| attains its minimum on the boundary of D. Now suppose that f is nonzero on the boundary of D and consider the function g = 1/f . The function g is analytic in D and continuous up to and including the boundary of D. (This is why we must assume that f is nonzero on the boundary.) By the Maximum Modulus Principle (Theorem 24), the function g attains its maximum modulus on the boundary of D. Choose a point w0 on the boundary of D so that 0 < |g(z)| ≤ |g(w0 )| for all z in D and on the boundary of D. Since |f (z)| = 1 1 ≥ = |f (w0 )| |g(z)| |g(w0 )| for all z in D and on the boundary of D, the modulus |f (z)| attains its minimum on the boundary of D. Let D be the domain {z : |z| < 1} and consider the function f (z) = z. Notice that the minimum modulus of |f (z)| occurs at the point 0 inside D, not on the boundary. This shows that the assumption that f is nonzero inside D is essential. As a second example, one that illustrates the simpler of our two cases, let f (z) = z 4 − 1. Then f is nonzero in D, but 1/f is not continuous up to and including the boundary of D. Problem 3: Show that the maximum modulus of az n + b for |z| ≤ 1 is |a| + |b|. (Given standard notation, it is assumed that n is a positive integer and that the numbers a and b are complex constants.) Solution: We begin by noting that the function az n + b is analytic on C, that is, it is an entire function. By the triangle inequality, we know that |az n + b| ≤ |az n | + |b| = |a| · |z|n + |b| ≤ |a| + |b| for all z that satisfy |z| ≤ 1. To get equality, we must show that there is some value of z for which |az n +b| = |a|+|b|. By the Maximum Modulus Principle, we know that this value of z must satisfy |z| = 1. To find an appropriate value for z, we first write a and b in polar form. Let a = |a|eiα and b = |b|eiβ . Consider the number z0 = ei(β−α)/n . Then |z0 | = 1 and
|az0n + b| = |a|eiα · ei(β−α) + |b|eiβ = |a|eiβ + |b|eiβ = |eiβ | · |a| + |b| = |a| + |b|. We conclude that the maximum modulus of az n + b for |z| ≤ 1 is |a| + |b|. 12 ∞ X k! converges. Exercise 5.1.2d: Use the ratio test to show that the series kk k=1 Solution: Applying the ratio test, we find that L = lim k→∞  k k  (k + 1)! kk 1 −k = lim · = lim = e−1 . 1 + k→∞ k + 1 k→∞ (k + 1)k+1 k! k Since e−1 < 1, the series converges by the ratio test. Exercise 5.1.10: For each positive integer n, let Fn (z) = and that {Fn (z)} converges to 1 for |z| > 3. zn . Show that {Fn (z)} converges to 0 for |z| < 3 z n − 3n Solution: Suppose first that |z| < 3. Since |z/3| < 1, the sequence {(z/3)n } converges to 0. Since Fn (z) = zn zn (z/3)n = , n −3 (z/3)n − 1 the sequence {Fn (z)} converges to 0. Now suppose that |z| > 3. In this case, we see that |3/z| < 1 and thus the sequence {(3/z)n } converges to 0. Since Fn (z) = zn zn 1 = , n −3 1 − (3/z)n the sequence {Fn (z)} converges to 1. 13 Exercise 4.6.17: Find the maximum modulus for the analytic function f (z) = (z − 1)(z + 21 ) on the unit disk. Solution: By the maximum modulus principle, we know that the maximum modulus of f (z) occurs when |z| = 1. Since |f (z)| and |f (z)|2 attain their maximum value at the same point, we work with the latter since it avoids square roots. Writing z = x + iy, we find that |f (z)|2 = |z − 1|2 · z + 1 2 2
1
= (x − 1)2 + y 2 · (2x + 1)2 + (2y)2 . 4 On the boundary |z| = 1, we know that x2 + y 2 = 1 and thus 1 1 |f (z)|2 = (2 − 2x) · (5 + 4x) = (−4x2 − x + 5). 4 2 The problem has now been reduced to the following calculus problem: Find the maximum value of the function g(x) = −4x2 − x + 5 on the closed interval [−1, 1]. It is easy to see that the only critical point for g is x = − 81 . Evaluating g at the critical point and the endpoints of the interval, we find that g(−1) = 2, g(− 18 ) = − 1 81 1 + +5= , 16 8 16 g(1) = 0. The maximum value of g on [−1, 1] is thus 81/16. It follows that the maximum modulus for f on the unit disk is r 81 9 9√ = √ = 2. 32 8 4 2 Although it is not necessary, we can locate the points on the unit circle where this maximum value occurs. Given √ that x2 + y 2 = 1, when x = − 18 , we find that y = ± 63/8. The maximum modulus thus occurs at the points 1 3√ − + 7i 8 8 and − 1 3√ − 7 i. 8 8 For the record, it is possible to solve this problem using trigonometry (and many of you did), but the details are a little bit messier. It does make it easy to see that the maximum occurs at eiθ , where θ = π ± arccos( 18 ). 14 Solutions for Special Assignment 4 Problem 1: Present a different solution to Example 4 in Section 4.2. Solution: Since z 2 is a continuous function that has an antiderivative throughout C, we can use Corollary 2 (see page 175) to conclude that Z Z z 2 dz =
z 2 − 4xy i dz = Z Γ Γ z 2 dz − 4i Γ Z Z xy dz = −4i Γ xy dz. Γ Using the three parametrizations from the textbook, we obtain Z 2 Z Z 2ti dt + 0 dt + xy dz = 0 0 Γ 2 Z 0 8 t2 (−1 − i) dt = 0 + 4i − (1 + i). 3 −2 It then follows that Z   32 16 32 8 + i. z 2 dz = −4i 4i − (1 + i) = 16 + (i − 1) = 3 3 3 3 Γ Z 1 Problem 2: Use the fact that dz = 2πi to evaluate C z−1 traversed once in the positive direction. Z 2π 0 2 − cos t dt, where C is the circle |z| = 2 5 − 4 cos t Solution: Representing the curve C by z = 2eit for 0 ≤ t ≤ 2π, we find that Z 2πi = C 1 dz = z−1 Z 0 2π 2ieit dt = 2i 2eit − 1 and thus Z π= 0 2π Z 0 2π eit (2e−it − 1) dt = 2i |2eit − 1)|2 Z 0 2π 2 − eit dt 5 − 4 cos t 2 − cos t − i sin t dt. 5 − 4 cos t Equating real parts shows that the value of the requested trig integral is π. We now present a solution using calculus techniques. As a start, we evaluate the following integral: Z π Z π Z π Z π 1 1 1 1 dt = dt = dt = dt 2 2 2 5 − 4 cos t 1 + 4(1 − cos t) 0 0 0 1 + 8 sin (t/2) 0 cos (t/2) + 9 sin (t/2) Z π  π π 2 2π sec2 (t/2) = dt = arctan(3 tan(t/2)) = −0 = . 2 3 3 2 3 0 0 1 + 9 tan (t/2) We then note that (using the substitution s = 2π − t) Z 2π π 1 dt = 5 − 4 cos t Z 0 π and conclude that Z 0 2π −1 ds = 5 − 4 cos(2π − s) Z 0 π 1 dt 5 − 4 cos t 1 2π dt = . 5 − 4 cos t 3 It then follows that Z 2π Z Z Z 2 − cos t 1 2π 3 + (5 − 4 cos t) 3 2π 1 1 2π π π dt = dt = dt + 1 dt = + = π. 5 − 4 cos t 4 0 5 − 4 cos t 4 0 5 − 4 cos t 4 0 2 2 0 15 Z Problem 3: Use partial fractions and the result of Exercise 4.3.6 to evaluate √ the circle |z − i| = 3 traversed once in the positive direction. C 3z + 1 dz, where C is (z − 1)(z 2 + 1) Solution: Factoring the denominator over the complex numbers, we find that 3z + 1 3z + 1 A B D = = + + . (z − 1)(z 2 + 1) (z − 1)(z − i)(z + i) z−1 z−i z+i Note that the distance from 1 to i is √ 2 and that the distance from −i to i is 2. Given that 1 and i lie within the circle C and that −i lies outside of C, the result of Exercise 4.3.6 shows that Z 3z + 1 dz = (A + B)2πi. 2 C (z − 1)(z + 1) Using our methods for finding the coefficients of partial fractions, we obtain 3z + 1 = 2; z 2 + 1 z=1 3z + 1 3i + 1 (3i + 1)(−2 + 2i) −8 − 4i i 3i + 1 B= = = = = −1 − . = (z − 1)(z + i) z=i (i − 1)2i −2 − 2i 8 8 2 A= It follows that Z C  3z + 1 i dz = 1 − 2πi = π + 2πi. (z − 1)(z 2 + 1) 2 Although the value of D is not needed, we note that the coefficient of z 2 when the partial fractions are added is A + B + D. Since this value needs to be 0 (the coefficient of z 2 for 3z + 1), we find that i D = −(A + B) = −1 + . 2 Of course, we could find D using the formula D= 3z + 1 (z − 1)(z − i) = z=−i −3i + 1 1 − 3i (1 − 3i)(−2 − 2i) −8 + 4i i = = = = −1 + , (−i − 1)(−2i) −2 + 2i 8 8 2 but this requires many more calculations. 16