Math 220A Homework 4 Solutions

Transcription

1 Math 220A Homework 4 Solutions Jim Agler 26. (# pg. 73 Conway). Prove the assertion made in Proposition 2. (pg. 68) that g is continuous. Solution. We wish to show that if g : [a, b] [c, d] C is a continuous function and g : [c, d] C is defined by g(t) = b a φ(s, t) ds, then g is a continuous function. Fix ɛ > 0. Since f is continuous and [a, b] [c, d] is compact, f is uniformly continuous on [a, b] [c, d]. Hence, there exists a δ > 0 such that s,s 2 [a,b] t,t 2 [c,d] s s 2 < δ and t t 2 < δ = f(s, t ) f(s 2, t 2 ) < Consequently, if t t 2 < δ, then g(t ) g(t 2 ) = = = ɛ. b a b a b a b This proves that g is continuous. a φ(s, t ) ds b a φ(s, t 2 ) ds (φ(s, t ) φ(s, t 2 )) ds φ(s, t ) φ(s, t 2 ) ds ɛ b a ds ɛ b a.

2 27. (#4 pg. 74 Conway) (a) Prove Abel s Theorem: Let a n (z a) n have radius of convergence and suppose that a n converges to A. Prove that lim an r n = A. (0.) r Note: At least spend some time trying to prove the assertion stated above. If you have never seen a proof of Abel s Theorem, then I would consider this to be a very challenging (or at least time consuming) problem. Therefore, it would be reasonable for you to adapt a proof from another text or from the excellent discussion on Wikipedia of a slightly juiced up version of the Abel s Theorem stated above. (b) Use Abel s Theorem to prove that log 2 = Solution to (a). First consider the special case where A = 0. Fix ɛ > 0. If for n 0 we set s n = a 0 + a a n, then s n 0 and a 0 + a r +... a n r n = ( r)(s 0 + s r +... s n r n ) + s n r n for all n 0 and r R. Consequently, if r (, ), then n s nr n converges and a n r n = ( r) s n r n. If we choose N so that n > N = s n < ɛ, 2

3 then for r (0, ) we have that a n r n = ( r) s n r n ( r) ( r) = ( r) < ( r) N s n r n + ( r) N s n + ( r) ɛ N n=n+ n=n+ s n + ( r) ɛ rn+ r N s n + ɛ Hence, for r sufficiently close to, a n r n < ɛ. r n s n r n This proves that (0.) holds in the special case when A = 0. The general case follows by replacing a 0 with a 0 A and then invoking the special case just proved. Solution to (b). Since B(; ) is a subset of the domain of the principal branch of the logarithm, log( + z) is an analytic function on B(0; ), and consequently, by Theorem 2.8 (pg. 72 Conway) has a power series representation on B(0; ) which is easily computed to be Since the series n ( ) n+ n= n ( ) n+ n log( + z) = = lim r ( ) n+ z n. n n= converges, it follows by Abel s Theorem that ( ) n+ n= n 3 r n = lim log( + r) = log 2. r

4 28. In Homework Problem #4 (cf. #7 pg. 33 Conway) it was shown that the power series ( ) n n= converges when z = i, i.e., that the series n zn(n+) converges. Evaluate the sum of this series. Solution. We group the series and define a function by the formula f(z) = , z2 2 z z6 5 6 z Since the power series this formula has radius of convergence, it follows from Abel s Theorem that = lim f(x). (0.2) x To compute f notice that Proposition 2.5 (pg. 35 Conway) implies that f (z) = z 2 + z 4 z for all z D. Since this formula for f is a geometric series that sums to ( + x 2 ), it follows that the function f(x) is the unique twice differentiable function that solves the 2 nd order differential equation f(0) = f (0) = 0 and f (x) = + x 2 x (, ). By elementary calculus, f (x) = arctan(x) and f(x) = x arctan x 2 log( + x2 ). 4

5 Therefore, by (0.2), = π 4 log (#5 pg. 74 Conway). Give the power series expansion of log z about z = i and find its radius of convergence. Solution. If f(z) = log z, then f(i) = π i and for n, 2 f (n) (z) = ( ) n (n )!z n, so that Hence, f (n) (i) = ( ) n (n )! i n = (n )! i n. f(z) = π 2 i (n )! i n (z i) n = π n! 2 i i n n (z i)n. (0.3) n= Since lim sup n /n =, Theorem.3 (pg. 3 Conway) implies that the radius of convergence is. Remark. Let R be the radius of convergence of the power series in (0.3). In the solution we concluded that R = by computing that lim sup n /n = and then invoking the theorem which asserts that R = lim sup a n /n. Alternately, in this instance, we can see that R = without any computation whatsoever. As there is no branch of the logarithm defined on a neighborhood of 0 (HW # 8), that the radius of convergence is at most follows from Corollary 2.9 (pg. 37 Conway). That R is at least follows from Theorem 2.8 (pg. 72 Conway). n= 30. (#7 pg. 74 Conway) Use the results of this section to evaluate the 5

6 following integrals: e iz (a) z dz, (t) = 2 eit, 0 t 2π; dz (b) z a, (t) = a + reit, 0 t 2π; sin z (c) z dz, (t) = 3 eit, 0 t 2π; log z (d) z dz, (t) = + n 2 eit, 0 t 2π and n 0. Note: If these did not all immediately come to you, then you could get more practice by working through Exercise #9 pg. 75 Conway. Solution. Each integral can be evaluated using Corollary 2.3 which implies that if f : G C is analytic and B(a; r) G, then f(z) 2πif(n) (z a) dz = (a), n+ n! where is the path defined by (t) = a + re it, 0 t 2π. (a) Letting f = e iz, n =, a = 0 and r = gives e iz z dz = ) (0) 2 2πi(eiz = 2π.! (b) Letting f =, n = 0, a = a and r = r gives z a dz = 2πi 0! = 2πi. (c) Letting f = sin z, n = 2, a = 0 and r = gives sin z z dz = 2πi(sin z) (0) = ! (d) Letting f = (z ) log z, n = 0, a = and r = /2 gives z n log z z n dz = (z ) log z z n z dz = 2πi((z ) log z z n )() 0! 6 = 0.

7 3. (#0 pg. 75 Conway) Evaluate z 2 + z(z 2 + 4) dz where (t) = re it, 0 t 2π, for all possible values of r, 0 < r < 2 and 2 < r <. (Hint: use partial fractions). Solution. The method of partial fractions gives the identity Also, for 0 < r < 2, dz = 2πi and z and for 2 < r <, z dz = z 2 + z(z 2 + 4) = /4 z + 3/8 z + 2i + 3/8 z 2i. z + 2i dz = z + 2i dz = dz = 0, z 2i dz = 2πi. z 2i Therefore, for 0 < r < 2, z 2 + z(z 2 + 4) dz = /4 z dz + 3/8 z + 2i dz + 3/8 z 2i dz = 4 2πi = π 2 i, 7

8 and for 2 < r <, z 2 + z(z 2 + 4) dz = /4 z dz + 3/8 z + 2i dz + 3/8 z 2i dz = 4 2πi πi πi = 2πi. 32. (# pg. 80 Conway) Let f be an entire function and suppose there is a constant M, an R > 0, and an integer n such that f(z) M z n for z > R. Show that f is a polynomial of degree n. Solution. If r > R and k > n, then using Corollary 2.3 (pg. 73 Conway) with (t) = re it gives that f (k) (0) = k! f(w) 2πi w dw k+ = k! 2π f(re it ) 2π 0 (re it ) k+ ireit dt k! 2π f(re it dt 2πr k 0 k! 2π max 2πrk 0 t 2π f(reit ) k! r k Mrn = Mk! r. k n Since k > n, it follows by letting r that f (k) (0) = 0. Summarizing, we have shown that if k > n, then f (k) (0) = 0. This implies that the power series representation of f at 0 is in fact a polynomial with degree n. 33. (#6 pg. 80 Conway) Let G be a region and suppose that f : G C is analytic and a G such that f(a) f(z) for all z G. Show that either f(a) = 0 or f is constant. 8

9 Solution. We prove that if f(a) 0, then f is constant. If f(a) 0, then by the continuity of f, there exists r > 0 such that B(a; r) G and f(z) 0 for all z B(a; r). Since f(z) 0 for all z B(a; r), the function g defined on B(a; r) by the formula g(z) =, z B(a; r) f(z) is a well defined analytic function on B(a; r). Furthermore, since f(a) f(z) for all z G, g(z) g(a) for all z B(a; r). Therefore, the Maximum Modulus Theorem (pg. 79 Conway) implies that g is constant on B(a; r). Therefore, f is constant on B(a; r) as well. Since G is a region (and hence connected), it follows that f is constant on all of G. 34. Use the previous exercise to give a proof of the Fundamental Theorem of Algebra that does not use Liouville s Theorem. Solution. Fix a nonconstant polynomial p. Since there exists R > 0 such that lim p(z) =, z z > R = p(z) > p(0) +. By the Extreme Value Theorem for continuous functions, there exists a B(0; R) such that p(a) p(z) for all z B(0; R). Note that since by continuity, p(z) p(0) + for all z B(0; R), we have that necessarily a B(0; R). Therefore, p is nonconstant and attains its minimum on B(0; R) at a B(0; R). It follows from the previous exercise that p(a) = (#8 pg. 67 Conway). Let G be a region and let f and g be analytic functions on G such that f(z)g(z) = 0 for all z G. Show that either 9

10 f 0 or g 0. Solution. We show that if f is not identically 0 on G, then g is identically 0 on G. If f is not identically 0 on G, then there exists a G such that f(a) 0. By continuity, there exists δ > 0 such that f(z) 0 for all z B(a; δ). But f(z)g(z) = 0 for all z G. Therefore, g(z) = 0 for all z B(a; δ). Since G is connected, it follows that g is identically 0 on G. 36. (#3 pg. 83 Conway) Let p(z) be a polynomial of degree n and let R > 0 be sufficiently large so that p never vanishes in {z z R} (fixed a typo here). If (t) = Re it, 0 t 2π, show that p (z) dz = 2πin. p(z) Solution. By Corollary 3.6 (pg. 77 Conway) we may write p(z) = c(z a )(z a 2 )... (z a n ) where for each k =,..., m, a k < R. Consequently, using the generalized product rule, (f f 2... f n ) = f f 2... f n + f f 2... f n f f 2... f n, we have that p (z) = c [ (z a 2 )... (z a n ) ] +c [ (z a )(z a 3 )... (z a n ) ] +...+c [ (z a )... (z a n ) ]. It follows that if z R, then Hence, p (z) p(z) dz = p (z) p(z) = z a + z a z a n. dz + z a dz z a 2 z a n dz = 2πi + 2πi πi = 2πin. 0

11 Solution 2. This solution is not quite rigorous but gives a picture of what is going on. If you believe that for large R, t p(re it ), 0 t 2π is a curve that winds around the origin n times, then But then, 2πi p def dz = n(p ; 0) = n. z p (z) 2π p(z) dz = p ((t)) 0 p((t)) (t)dt = = 2π 0 p p((t)) (p (t))dt z dz = 2πin. 37. (#4 pg. 83 Conway) Fix w = re iθ 0 and let be a rectifiable path in C \ {0} from to w. Show that there is an integer k such that z dz = log r + iθ + 2πik. Solution. Suppose that : [a, b] C \ {0} where (0) = and (b) = w. Define two paths ρ and ω in C \ {0} by and ρ(t) = + t(r ), 0 0 ω(t) = re itθ, 0 t. Evidently, ω ρ is a closed rectifiable path in C \ {0}. Therefore, by Proposition 4. (pg.8 Conway) there exists an integer k such that dz 2πi z = k, ω ρ

12 or equivalently, But dz = log r and ρ z ω was to be proved. dz z = dz ρ z + dz ω z + 2πik. dz z = iθ. Therefore, z dz = log r + iθ + 2πik, as 2