Assignment 2 - Complex Analysis MATH 440/508 M.P. Lamoureux Sketch of solutions. Γ z dz = Γ (x iy)(dx + idy) = (xdx + ydy) + i Γ Γ ( ydx + xdy) = (/2)(x 2 + y 2 ) endpoints + i [( / y) y ( / x)x]dxdy interiorγ = 0 2i dxdy = 2i Area( interior of curve Γ. interiorγ by Green s theorem for the second term, exact differential on the first term (evaluated at a common end point, so cancels to zero). 2. a) Does there exist a holomorphic function f on the unit disk with f( n ) = f( n ) =, for n =, 2, 3,...? n2 Is it unique? Yes, and Yes. f(z) = z 2 works, and any other function g(z) that works would agree with f on the sequence z n = /n, which has a limit point in the range of analyticity. Therefore f and g are identical. b) Does there exist a holomorphic function f on the unit disk with f( n ) = f( n ) =, for n =, 2, 3,...? n3 Is it unique? No. On the first sequence z n = /n, the function f (z) = z 3 works, while on the other sequence z n = /n, the function f 2 (z) = z 3 works. A function f that satisfies both sequences would have to agree with both f and f 2 on their domain, by the limit point argument as above. Which is impossible, since one function is the negative of the other.
3. a) z = zn sin(z) dz = 0 for n since the integrand is holomorphic, or has at most a removable singularity at z = 0. For other values, we use the Cauchy integral formula for derivatives, noting that the k-th derivative of sin(z) at z = 0 is ± or zero, depending on the value of k (modulo 4). So z = zn sin(z) dz = 0 when n is negative and odd, and z = z n sin(z) dz = when n is negative and even. z = sin(z) 2πi dz = z n ( n )! ( )n/2, 3. b) zn ( z) m dz = 0 when n, m 0, as the integrand is holomorphic. It is also zero for n, m < 0 because we can replace the circle of radius 2 with a circle of radius R > 2 (by the fact the integrand is holomorphic on the region R > ). Letting R goes to infinity, we see the integral grow no faster than R/R n + m, which tends to zero. So the integral on the original curve is zero. The only interesting cases are when n 0, m < 0 or m 0, n < 0. Either way, we have a polynomial divided by a power of z or z, in which case we apply the Cauchy integral formula to a simple polynomial. For n 0, m < 0, z n ( z) m dz = ( ) m z n 2πi( )m dz = (z ) m ( m )! (d/dz) m [z n ] z= = 2πi( )m n(n )(n 2) (n + m + 2) = 2πi( )m ( m )! where the binomial coefficient is defined to be zero when the bottom term is negative. (You might want to check that the signs are correct. Try n = 0, m = to see that you sometimes need negatives!) Similarly, for n < 0, m 0, z n ( z) m dz = = 2πi( ) n ( n )! (z ) m z n dz = (You might want to check that the signs are correct. ) 2πi ( n )! (d/dz) n [( z) m ] z=0 ( ) n, m ( ) m m(m )(m 2) (m + n + 2) = 2πi( ) n+, n 3. c) For z =ρ z z 0 4 dz z 0 = ρ, note the integrand and dz are all positive, so we expect a positive answer. Use the parameterization 2
z = ρe it, dz = iρe it so dz = ρdt = ρdz/iz. Then z z 0 4 ρdz dz = z =ρ z =ρ (z z 0 ) 2 (z z 0 ) 2 iz = ρ z i z =ρ (z z 0 ) 2 (zz zz 0 ) 2 dz = ρ z i z =ρ (z z 0 ) 2 (ρ 2 zz 0 ) 2 dz = ρ i 2πif z (z 0 ), where f(x) = (ρ 2 zz 0 ) 2 by the Cauchy integral formula for derivatives, with z 0 < ρ. First derivatives of quotients are easy to compute, so f (z) = Thus (ρ 2 zz 0 ) 2 + 2zz 0 (ρ 2 zz 0 ) 3, f (z 0 ) = z =ρ z z 0 4 dz = 2πρ ρ2 + z 0 2 (ρ 2 z 0 ) 3 when z 0 < ρ. A similar calculation shows that z =ρ z z 0 4 dz = 2πρ ρ2 + z 0 2 ( z 0 ρ 2 ) 3 when z 0 > ρ. Notice both answers are positive, as expected. 4. a) Use the residue theorem to compute + x 4 dx. (ρ 2 z 0 ) 2 + 2 z 0 (ρ 2 z 0 ) 3 = ρ2 + z 0 2 (ρ 2 z 0 ) 3. The function f(z) = +z 4 has simple poles at the four points z k = (± ± i)/ 2 and the residues are computed using L Hopital s rule: z z k Res zk = lim z z k + z 4 = lim z z k 4z 3 = 4(z k ) 3 = z k 4(z k ) 4 = z k/4, since zk 4 =. Using a semicircular arc on the upper half plane, and letting the radius go to infinite (standard argument), we are left with picking up the residues on the upper half plane, so + x 4 dx = 2πi(Res z 0 + Res z ) = 2πi( + i + + i)/4 2 = π/ 2. 3
b) For any even integer 2N, how would you compute dx. + x2n A similar calculation of residues for f(z) = +z 2N shows Res zk = lim z z k z z k = lim + z2n z z k 2Nz 2N = 2N(z k ) 2N = z k 2N(z k ) 2N = z k/2n, since (z k ) 2N =. There are N poles in the upper half plane, z k = e πi(2k+)/2n, k = 0,,..., N, so we add those residues to get N + x 2N dx = 2πi Res zk = 2πi 2N k=0 N k=0 e πi(2k+)/2n. This last sum we recognize as a geometric series, summing to Thus z N z 0 2eπi/2N = eπi/n e πi/n = 2 e πi/2n = i csc(π/2n). e πi/2n + x 2N dx = π N csc(π/2n). 5. Prove, that if f is a non-constant, entire function satisfying f(z) < z N for some fixed N and all z sufficiently large, then f has at most N zeros in the complex plane. Proof: use the Cauchy integral formula for derivatives to observe that the N-th derivative f (N) is bounded on the plane, and therefore a constant by Liouville s theorem. Computing N antiderivatives, we find f is a polynomial of at most degree N, hence it has no more than N zeros (or roots). 6. a) Show that the real part u(x, y) = Real{f(x + iy)} of an analytic function f in the open unit disk D is harmonic. That is, u is twice continuously differentiable (in the real variables sense) and satisfies the PDE 2 u x 2 + 2 u y 2 = 0. Proof: With f = u + i v, the fact that f is holomorphic tells us it is infinitely differentiable in the complex sense, and tis by computing difference 4
quotients along horizontal and vertical directions, we conclude both u and v are infinitely differentiable in the real sense (i.e. with respect to variables x and y). The Cauchy-Riemann equations tell us x u = y v and y u = x v, so x x u + y y u = x y v y x v = 0, since the order of differentiation does not matter for infinitely differentiable function v. Thus u is harmonic (as is v). b) Show that given a (real-valued) harmonic function u(x, y) on the open unit disk, there exists another harmonic function v(x, y) such that the function f(z) = f(x + iy) = u(x, y) + iv(x, y) is analytic in the unit disk. Proof: our motivation is that if such an f exists, its derivative, computed along the x direction would be f (z) = x u + i x v = x u i y u, by the Cauchy-Riemann equation. So, although we don t know (yet) what f is, or its derivative, we can define a function g(z) = x u(x, y) i y u(x, y) and verify that the real and imaginary parts of g satisfy the Cauchy-Riemann equation. To check this: x Real(g) = x x u, y Imag(g) = y y u, and since u xx + u yy = 0, these two guys are equal. i.e. x Real(g) = y Imag(g). A similar calculation, noting that u xy = u yx since u is twice continuously differentiable, shows. x Imag(g) = y Real(g). Thus g satisfies the C-R equation, hence is holomorphic, hence has an antiderivative f in the open disk, and we can add a constant to ensure that f(0) = u(0, 0). With f = U + iv, define the difference function D(x, y) = U(x, y) u(x, u). Since f = g we compute that x D = y D = 0, with D(0, 0) = 0, (the difference of f(0, 0) and u(0, 0).) The only solution to this linear PDE is D 0, so U u, confirming that we have found an analytic function f on the open disk whose real part is 5
u. Of course, V is also harmonic, by part a), since it is the real part of the analytic function if(z). c) Assume u(x, y) is harmonic in the open unit disk, and continuous on its closure. Show we can compute its values inside the disk by the real integral u(re iθ ) = 2π r 2 2π 0 2r cos(θ φ) + r 2 u(eiφ ) dφ. That is, we are integrating the values of u on the circle with the Poisson kernel. (Hint. Use the Cauchy integral formula on f = u + iv, applied to a disk of radius slightly less than. Then take a limit.) Proof: Rather than using magic to get the right form for the answer, let s try to figure this out step by step. We know the solution has something to do with two points, one inside the unit disk z = re iθ and one on the unit circle, w = e iφ. The distance (squared) between them is w z 2 = (e iφ re iθ )(e iφ re iθ ) = 2r cos(θ φ) + r 2, so we can write the Poisson kernel as r 2 2r cos(θ φ) + r 2 = r 2 (w z)(w z) = w w z + z w z, where we get these last two terms by computing the partial fractions. Parametrizing the circle by w = e iφ, dw = ie iφ dφ = iw dφ, we can write r 2 [ 2r cos(θ φ) + r 2 dφ = w w z + z ] dw w z iw = [ i w z + ] z dw. wz Now we are in good shape, because the first term has the /(w z) which we need for the Cauchy integral formula, and the second term has a pole only at w = (/r)e iθ, which is outside the unit circle, so it is holomorphic inside the disk. Thus if f is an analytic function on, and inside, the closed unit disk, we have 2π 2π 0 r 2 2r cos(θ φ) + r 2 f(eiφ )dφ = [ 2πi w = w z + ] z f(w)dw = f(z), wz 6
by the Cauchy integral formula (first term) and analyticity (second term). Taking real parts of this equation, we have 2π r 2 2π 0 2r cos(θ φ) + r 2 Real(f(eiφ ))dφ = Real(f(z)). ( ) Now, we are not quite done, because the original u we started with was only harmonic inside the unit disk, not on its boundary. So when we construct analytic f from u, it is only analytic inside the unit disk, not on its boundary. So, for a fixed parameter 0 < ρ <, define f ρ (z) = f(ρz) = u(ρz) + iv(ρz). This function is analytic on the open disk of radius /ρ > and in particular, on an open set containing the unit disk and its boundary. Thus we can apply the above formula (**) to f ρ to obtain 2π r 2 2π 0 2r cos(θ φ) + r 2 u(ρeiφ )dφ = u(ρz). Holding z fixed, and let ρ, by uniform continuity of u on the closed unit disk (which is compact), we can exchange limits with integrals to find 2π r 2 2π 0 2r cos(θ φ) + r 2 u(eiφ )dφ = u(z) = u(re iθ ), as desired. (Notice for fixed z, the Poisson kernel is a nice continuous, bounded function of φ, so it is only the uniform convergence of u(ρe iφ ) u(e iφ ) that needs to concern us.) That was a bit longer than I would expect the students to do for this question, but I decided to fill in some details as to why the proof should proceed this way. 7