Math Homework 2

Transcription

1 Math 73 Homework Due: September 8, 6 Suppose that f is holomorphic in a region Ω, ie an open connected set Prove that in any of the following cases (a) R(f) is constant; (b) I(f) is constant; (c) f is constant; (d) arg(f) is constant; we can conclude that f is a constant See also Ahlfors, p 7 If f (z) = for all z Ω, we fix a point z Ω and join it with any path γ to z Then f(z) f(z ) = f (z) dz = So f(z) is a constant We set f(z) = u+iv For (a) we note that if u is constant, then u x = and u y = But f (z) = u x + iv x = u x iu y = So the previous statement gives that f is constant For (b) we note that if v is constant, then v x = v y = But f (z) = v y + iv x = So again f is constant For (c) we argue as follows: If u + v = k, then we get by differentiation uu x + vv x =, uu y + vv y = Using the Cauchy-Riemann equations we get: uu x vu y =, uu y + vu x = We now have a homogeneous system with unknowns u x and u y The determinant is u v v u = 4(u + v ) If u + v the system has a unique solution u x = u y =, which implies that u is constant and we can use (a) If u + v = at some point, then automatically u + v = for all points, and the function f vanishes For (d) we set u = kv for fixed k Consider the analytic function g(z) = (+ki)(u+iv), which has real part u kv = Then by (a), g(z) is constant, which implies that f is constant γ

2 Show that if {a n } n= is a sequence of non-zero complex numbers such that then a n+ lim n a n = L, lim a n /n = L n This is the ratio test and it can be used for the calculation of the radius of convergence of a power series Hint: Show that lim inf a n+ a n We only need to prove the hint, since lim inf a n /n lim sup a n /n lim sup a n+ a n lim x n = L lim inf x n = lim sup x n = L Also since we work with absolute values, we might as well prove the result for the sequence b n = a n Set l = lim inf b n+ b n, L = lim inf n b n, M = lim sup n b n, m = lim sup b n+ b n Fix ɛ > Since m = lim sup b n+ /b n = inf N sup n N b n+ /b n, the number m + ɛ is not a lower bound, ie, we can find a N such that sup n N b n+ /b n < m + ɛ, ie for all n N we have b n+ b n < m + ɛ We apply this successively for n = N, N +,, n to get b N+ < (m + ɛ)b N b N+ < (m + ɛ)b N+ b n < (m + ɛ)b n We multiply together to get, after cancellations, We take n-th roots to get b n < (m + ɛ) n N b N ( ) /n n b N bn < (m + ɛ), n N (m + ɛ) N

3 We take lim sup of both sides, taking into account that this does not depend on a finite number of initial terms and that c /n for c > This gives: lim sup n b n (m + ɛ) Since this is true for all ɛ >, we get lim sup n b n = M m For the lim inf we work analogously: Fix ɛ > Find N such that inf n N b n+ /b n > l ɛ, which implies, for n N, b n+ /b n > l ɛ We apply this successively for n = N, N +,, n to get b N+ > (l ɛ)b N b N+ > (l ɛ)b N+ b n > (l ɛ)b n We multiply together to get, after cancellations, We take n-th roots to get b n > (l ɛ) n N b N ( n bn bn > (l ɛ) (l ɛ) N ) /n, n N We take lim inf of both sides, taking into account that this does not depend on a finite number of initial terms and that c /n for c > This gives: lim inf n b n (l ɛ) Since this is true for all ɛ >, we get lim inf n b n = L l 3 (a) Find the radius of convergence of the hypergeometric series F (α, β, γ; z) = + n= Here α, β C and γ,,, α(α + ) (α + n )β(β + ) (β + n ) z n n!γ(γ + ) (γ + n ) We use the ratio test We must assume that the terms are nonzero This happens as long as α and β are not negative integers (or zero) If either is a negative integer (or zero), the power series terminates and we get a polynomial with infinite radius of convergence We get in the general case a n+ a n = α(α + ) (α + n )(α + n)β(β + ) (β + n )(β + n) α(α + ) (α + n )β(β + ) (β + n ) + ) (γ + n )(γ + n) (α + n)(β + n) n! γ(γ =, n (n + )! γ(γ + ) (γ + n ) (n + )(γ + n) 3

4 This implies that R = (b) Find the radius of convergence of the Bessel function of order r: We use the ratio test a n+ a n = ( z r ( ) J r (z) = ) n ( z ) n, r N n!(n + r)! n= n n!(n + r)! n+ (n + )!(n + + r)! = 4(n + )(n + + r) So the radius of convergence is 4 Prove that, although all the following power series have R =, (a) nz n does not converge on any point of the unit circle, (b) z n /n converges at every point of the unit circle,, n (c) z n /n converges at every point of the unit circle except z = summation by parts) (Hint: Use Since lim n n = it is easy to see that R = For (a) we notice that on z =, nz n = n, so the series cannot converge, as the general term does not go to For (b) we notice that we can apply the comparison test with b n = n, where bn < We have on z = the bound z n /n = b n The comparison test implies that the series converges For (c) and z = we notice that we get the harmonic series n, which diverges by the integral test: compare with the integral dx = x For z and z = we get by summation by parts, setting s N ( z N+ )/( z), s = M n= z n n = M n= s n s n n = M = s M M M + ( s n n ) = s M n + M s n n M M + s n n + s n n(n + ) = N n= zn = Since s M / z (note that it is exactly in this geometric series that we use z ), the first term goes to zero, while the series can be compared with z n(n + ) = z 4

5 The Fibonacci numbers are defined by c =, c =, Define their generating function as c n = c n + c n, n =, 3, F (z) = c n z n n= (a) Find a quadratic polynomial Az + Bz + C such that We get Ac n z n+ + n= Ac n z n + n= (Az + Bz + C)F (z) = Bc n z n+ + n= Bc n z n + n= Cc n z n =, n= Cc n z n =, n= (Ac n + Bc n + Cc n )z n + (Bc + Cc )z + Cc = n= By the uniqueness of the coefficients of the power series, we get Ac n + Bc n + Cc n =, Bc + Cc =, Cc = Using c = c = we get C =, B =, while the first equation becomes Ac n c n + c n = The recurrence formula c n = c n + c n implies A = So the quadratic polynomial is z z + (b) Use partial fractions to determine the following closed expression for c n c n = (Qualifying exam Sept 6) The roots of the quadratic polynomial are ( + ) n+ ( ) n+ with ρ ρ = We set z z + = ρ, = ± A z ρ + B z ρ

6 to get = A(z ρ ) + B(z ρ ) = A = ρ ρ, B = ρ ρ = A This gives A = /, B = / We expand the fractions into geometric series valid for z/ρ < and z/ρ < respectively to get and A = z ρ ρ ( z/ρ ) = B z ρ = ρ ( z/ρ ) = This gives, by the uniqueness of the power series using ρ ρ = c n = ( ρ n+ ρ n+ ) c n = ( ( ρ ) n+ ( ρ ) n+) 6 Expand ( z) m in powers of z, for m N Let then a n ( z) m = a n z n, n= (m )! nm, n, z n ρ n+ z n ρ n+ where means that the quotient of the expressions to the left and the right of it tends to The geometric series is ( z) = We differentialte m -times to get (m )! ( z) = m n=m z n, z < n= n(n )(n ) (n m+)z n m+ =, (n+m )(n+m ) (n+)z n n= So a n = (n + m ) (n + ) (m )! 6

7 Since m is fixed and n we easily see that as a n has m factors in the numerator 7 Show that for z < we have a n n m /(m )!, n z z + z z z n z n+ + = z z, () and z + z + z + z + + k z k + = z + z k z () Justify any change in the order of summation Hint: Use the dyadic expansion of an integer and the sum k = k+ We first prove the Lemma: Every positive integer n has a unique expression of the t + k t t+, with t, k t We use for this the dyadic expansion of n If n = a + a + a + + a s s, a i {, } we pick t to be the first nonzero a i : ie a t =, while a = a = a t = Then n = t + a t+ t+ + + a s s = t + t+ (a t+ + + a s s t ) = t + t+ k t This expression is unique as t + k t t+ = j + j+ k j, j t implies that either j > t or j < t By symmetry we assume j > t, ie, j t + Look at the remainder when we divide by t+ The left-hand side gives t, while the right-hand side gives We expand the left-side of () using the geometric series to get z k+ + k = z 4k+ + + k = z m+ k m+ m + k m= while the right-hand side is just n= zn The lemma proves the identity, provided that we have absolute convergence of the double series m= k m= z m+ k m+ m, 7

8 which would mean that we can rearrange the terms in whatever way we would like But the lemma shows that the series z m+ k m+ m m= k m= is exactly the geometric series z n, z <, which converges For () we expand the left-hand side to get z ( ) n z n + z ( ) n z n + + k z k ( ) n z kn + = ( ) m+ z m + ( ) m+ z m + + k ( ) m+ z km + Looking at the dyadic expansion of n we pick the first a t =, such that a = a = = a t = Then i n for i t, while t+ does not divide n and the same is true for higher powers of For i t, n = i l with l even, so the coefficient it picks from the i-series in the sum is i ( ) l+ = i On the other hand for i = t, n = t l with l odd, so that it picks a coefficient t ( ) l+ = t For i > t it picks nothing So the coefficient of z n is + ( ) + ( ) + + ( t ) + t = by the identity k = k+ Example: n = 4+higher powers of We write the first 4 series in the sum: z z + z 3 z 4 + (z z 4 + z 6 z 8 + ) 4(z 4 z 8 + z z 6 + ) 8(z 8 z 6 + z 4 z 3 + ) The first three give to z 4 coefficients,, 4, while the last and the subsequent ones give We need to explain the interchange of summations We look at the series with absolute values on the terms This is k z km k= m= If this converges, then the original double sum converges absolutely and we can sum in whatever order we like In this series the coefficient of z n is equal to k = t+ < t+ n, n = t + a t+ t+ + k m k = k t So we compare with the series n z n, which has radius of convergence This suffices 8

9 8 Find the holomorphic function of z that vanishes at z = and has real part u(x, y) = x( + x + y ) + x y + (x + y ) We substitute z + z = x, x y = R(z ) = (z + z )/ and z z = x + y to get u(x, y) = (z + z)( + z z) + z + z + z z = z + z + z z + z z ( + z )( + z ) = ( z + z + z ) + z We set f(z) = z/( + z ), which is holomorphic, to get u(x, y) = (f(z) + f(z)) ie, u(x, y) is the real part of f(z), and f(z) =, this gives the answer to be f(z) Second method: (Look at p 7 in Ahlfors) Since f(z) is holomorphic, the conjugate function g(x, y) = f(z) depends only on z, so we write g( z) = f(z) to get u(x, y) = (f(z) + g( z)) = (f(x + iy) + g(x iy)) This is true for x and y real Formally we can plug x = z and y = z/(i), so that u(z/, z/(i)) = (f(z) + g()) This determines f(z) up to a constant If u(, ) = and we ask for f() =, then g() = In our case (z/)( + (z /4) + ( z /4)) f(z) = u(z/, z/(i)) = + (z /4 + z /4) + (z /4 z /4) = z + z 9 (i) Show that the Laplace operator can be calculated as = 4 z z = 4 z z (ii) Show that for any analytic function f(z) we have f(z) = 4 f (z), and log( + f(z) ) = 4 f (z) ( + f(z) ) (i) We have for a function f with continuous second partial derivatives f z = ( ) f x i f f, y z = ( ) f x + i f y 9

10 which gives 4 f z z = ( ) ( ) ( ) x i f (f x + if y ) = y x i f f x + i y x i f f y y = f xx if yx + i(f xy if yy ) = f xx if yx + if xy + f yy = f xx + f yy = f, since the mixed partial f xy, f yx are equal for functions with continuous second partial derivatives Reversing the order of the calculation gives 4 z z f = f (ii) For an analytic function z f(z) = f (z), while z f(z) = f (z), since f does not depend on z, so the result from applying z on its conjugate can be calculated from f (z): eg f(z) = z, f(z) = z, z f(z) = z = z = z f(z) This gives: f(z) = 4 z z (f(z)f(z)) = 4 z (f(z)f (z)) = 4f (z) z f(z) = 4f (z)f (z) = 4 f (z), since f(z) is a constant when we differentiate in z, and f (z) is a constant when we differentiate in z, as it does not depend on z For the second part we will use the chain rule as expressed in z and z coordinates (exercise 3 in homework ) We use g(w) = log( + w), which is holomorphic in w for Rw > (there is no difficulty in defining the argument of + w in this domain For f we have f(z) We have z log( + f(z) ) = We now apply the quotient rule g w =, and g w = + w + f(z) ( + f(z) ) z = f(z)f (z) + f(z) 4 z z log( + f(z) f(z)f (z) ) = 4 z + f(z) = f(z)f (z) 4 z + f(z)f(z) = 4 z(f(z)f (z))( + f(z) ) f(z)f (z) z ( + f(z)f(z)) ( + f(z) ) = 4 f (z)f (z)( + f(z) ) f(z)f (z)f (z)f(z) ( + f(z) ) = 4f (z)f (z) ( + f(z) ) Let f(z) be holomorphic and one-to-one on a set containing the unit disc D Let D = f(d) Then the area A(D ) of D is given by A(D ) = π n a n, where f(z) = n= a nz n (Qualifying exam September 6) n=

11 The Jacobian of the mapping w = f(z) as a mapping from D C is f (z) So by the change of variables formula A(D ) = du dv = f (z) dx dy D D If f(z) = n= a nz n, then f (z) = na n z n = f (z) = n= n,m= nma n ā m z n z m We switch to polar coordinates z = re iθ to get (dxdy = rdrdθ) A(D ) = π n,m= nma n ā m r n+m e iθ(n m) r dr dθ If n m, the integral π e iθ(n m) dθ =, while for n = m we get π This gives A(D ) = n a n r n π = π n= (a) Compute the integral z =r [ r n a n n n n= x dz ] = π a n n for the positive sense of the circle, in two ways: first by using a parametrization, and second, by observing that x = (/)(z + z) = (/)(z + r /z) on the circle First Method: Parametrization of the circle as z = re iθ, x = r cos θ, dz = rie iθ We get z =r x dz = π r cos θrie iθ dθ = π r i cos θ(cos θ+i sin θ)dθ = π = r + cos(θ) i + i [ θ sin(θ)dθ = sin θ r i n= π r i(cos θ+i cos θ sin θ)dθ ] π i cos θ = r iπ 4 Second method: On the circle z = r, we have z z = r = z = r /z, so that x = (z + z)/ = (/)(z + r /z) We plug into the integral to get z x dz = + r r dz = z πi = r iπ, z =r z =r since we know that z =r zn dz = for n and for n = we get πi

12 (b) Compute the integral z = dz z for the positive sense of the circle Hint: Find a primitive function of the integrand We use partial fractions to find that z = ( z ) z + We would like to integrate to get (log(z ) log(z + ))/ However, this cannot make sense in the region we are working: The log(z ) requires a cut from, say to and log(z + ) requires a cut from to These cuts meet our domain, since the circle z = contains both points and Make sense can be thought of being single-valued holomorphic functions that give us the primitive of the terms we consider However, if we write the expected answer as F (z) = ( ) z log z + things are much better: First of all, (z )/(z + ) is real only for real z This is seen as follows: z z + R z z + = z z + z z+z z = z z+ z z (z z) = I(z) = Moreover, it is a negative number exactly between and So on the complement of [, ] the function takes nonnegative values If we define the principal branch of the logarithm to be log w = log w + i arg w, π < arg w < π, then F (z) as the composition of holomorphic maps is holomorphic and is the primitive of /(z ) in a region that contains our contour Consequently dz/(z ) is exact and the integral is Remark: One can solve this problem with the residue theorem, which will be seen later in the course