Solutions to practice problems for the final

Transcription

1 Solutions to practice problems for the final Holomorphicity, Cauchy-Riemann equations, and Cauchy-Goursat theorem 1. (a) Show that there is a holomorphic function on Ω = {z z > 2} whose derivative is z (z 1)(z 2 + 1). Hint. A holomorphic function has a primitive if the integral on any triangle in the domain is zero. Solution: Call the given function f(z). Consider a triangle in Ω. Either it encloses the circle z = 2 or it doesn t. In the latter case, f(z) is holomorphic in a neighborhood of the triangle and so by Goursat s theorem the integral of f over such a triangle is zero. In case the triangle, which we denote by T, does enclose the circle z = 2, then by generalized Cauchy s theorem f(z) dz = f(z) dz, T where the circle is given positive orientation. Since the given circle encloses all poles of f(z) when thought of as a meromorphic function, again by generalized Cauchy s theorem, we see that for any R > 2, f(z) dz = f(z) dz. z =R But the latter integral goes to zero as you let R since the function grows like 1/R 2 and the length of the circle grows like R. So the original integral must have been zero. Hence in both cases integrals on triangles is zero. Hence there must exist a primitive. Remark. To show that the integral on z = 2, you could have also calculated the residue at infinity, and shown that that residue is zero. (b) Is there a holomorphic function on Ω whose derivative is z 2 (z 1)(z 2 + 1)? 1

2 Solution: No. If there was a primitive, then since the circle z = 3 lies in we would have to have f(z) =. z =3 On the other hand the integral equals Res z= f(z) = Res z= f(1/z)/z 2. But Res z= f(1/z)/z 2 = lim z 1 ( 1 ) z z 2 f 1 = lim z z (1 z)(1 + z 2 ) = 1. CIF, Cauchy-inequalities, Isolated singularities, Residue and argument theorems 2. If f is an entire function with f(z) C(1 + z ) 1/2, for all z C, then show that it is a constant. Solution: Since f(z) is entire it is given by a power series that converges on the entire complex plane to the function. We show that f n () = for all n 1. This would show that the the power series just a constant. By CIF, for any radius R > 1, and so by the triangle inequality, f (n) () n! f(z) dz 2π z n+1 C n! 2π 2Cn! z =R f (n) () = n! 2πi z =R f(z) dz, zn+1 (1 + R)1/2 R n+1 2πR. (Since on z = R, by given estimate, f(z) (1 + R) 1/2 ) R R R n, as long as n 1. This completes the proof. 3. If f : D C is defined by f(z) = 1 e w cos w 2πi w =1 (w z) 2 dw, where the circle is given the positive orientation, evaluate f (). Solution: By CIF, it is clear that f(z) = d dz ez cos z = e z cos z e z sin z. 2

3 So f (z) = 2e z sin z, and so f () =. 4. Evaluate the integrals tan z dz, where the circle is given the positive orientation. (z π/2) tan z dz, Solution: The integral can be evaluated using the residue theorem since tan z is a meromorphic function with the only poles inside z = 2 being at z = π/2 and z = π/2. But we evaluate it using the argument principle. and so (cos z) tan z = sin z/ cos z = cos z, ( ) tan z = 2πi number of zeroes of cos z inside z = 2 = 4πi. For the second integral, inside z = 2, the function has a simple pole only at z = π/2. But ) ( )( ) Res z= π/2 (z π/2 tan z = z + π/2 z π/2 tan z = π, and so the integral is 2π 2 i. lim z π/2 5. Find the residue of f(z) = cot πz at all of it s poles. Solution: The poles of cot πz are at the zeroes of sin πz, namely all the integers. The poles are all simple, and so the residue is given by cos πz Res z=n cot πz = lim(z n) z n sin πz where we used L Hospital s rule in the last equation. z n = cos πn lim z n sin πz = 1 π, 6. Find the integral of x sin x (x 2 + 1) 2 dx. Solution: The required integral is the imaginary part of I = xe ix R xe ix (x 2 dx = lim + 1) 2 R R (x 2 + 1) 2 dx, 3

4 since the degree of the denominator is at least two more than the degree of the numerator. The integral is of type III(a). So consider a contour made up of a large semi-circle of radius R traversed in the anticlockwise direction (denoted by C R ) and a straight line path from x = R to x = R along the x-axis, and the complex valued function f(z) = zeiz (z 2 + 1) 2. The only pole of the function inside the contour is at z = i, and ord z=i f(z) = 2. By the residue theorem, R R xe ix (x 2 + 1) CR 2 dx + ze iz (z 2 + 1) 2 dz = 2πiRes z=if(z). Letting R, the integral on the semicircle converges to zero, and so I = 2πiRes z=i f(z) = 2πi d (z i) 2 f(z) = πi dz z=i 2e. So the integral that we want is the imaginary part of this, that is x sin x (x 2 + 1) 2 dx = π 2e. 7. Evaluate x x dx. Solution: This is standard of Type II. Since the integrand is even, the integral we need, which we denote by I is I = 1 2 x R x x 4 dx = lim + 1 R R x dx. Again, using the same contour as above, since the degree of the denominator is two bigger than the degree of numerator, the integral on the semi-circular will approach zero as R and we obtain I = πi Res z=p f(z), where p an interior pole f(z) = 1 + z2 1 + z 4. The poles of the function are at α k = e iπ/4+iπk/2, k =, 1, 2, 3. Since the contour is in the upper half plane, the only poles in the interior of the contour are α = e iπ/4 and α 1 = e 3πi/4. 4

5 Since both poles are simple, the residues can be calculated using L Hospital s rule, namely From this one can compute that (z α k ) Res z=αk = lim z α k 1 + z 4 (1 + z2 ) = 1 + α2 k. 4α 3 k Res z=α f(z) + Res z=α1 f(z) = 1 i 2, and so x x dx = I = π 2. Open mapping, max modulus, Rouche s theorem, analytic continuation, sequences etc. 8. Let {f n } be a sequence of entire functions that converges compactly to a f with f not identically zero. If all f n have only real roots, show that all roots of f must be real. Solution: This is a direct consequence of the generalized Hurwitz s theorem from the final homework (Why?). We give a direct proof. Let p be a zero of f, whcih must be isolated. Suppose p is not real. Let r > such that D r (p) does not contain any other root and does not intersect the x-axis. Denoting the boundary circle by C, by the argument principle, 1 2πi C f (z) dz 1, f(z) since there is a root inside C, namely at p. But by compact convergence, 1 f (z) 1 f dz = lim n(z) 2πi f(z) n 2πi f n (z) dz. C But the integrals on the right are integers since they also count the number of zeroes of f n inside C. But then there must exist an N large enough so that the integrals on the right are greater than or equal to one for n > N. This is a contradiction since all roots of f n are real and the real axis does not intersect D r (p). C 9. Suppose f(z) is holomorphic in an open set containing the closure of the unit disc { z < 1}, such that f(z) < 1 whenever z = 1. Show that the equation f(z) = z 3 has exactly three solutions (counting multiplicities) inside the unit disc. Solution: Rouche s theorem! Let g(z) = z 3, h(z) = z 3 f(z). Then on z = 1, h(z) g(z) = f(z) < 1 = z 3 = g(z) 3. 5

6 Hence h and g have same number of roots in z < 1 counting multiplicities, which is three. 1. Does there exist a holomorphic function on the unit disc D such that { 1 n, n is even ( 1 f = n) 1 n+1, n is odd? Solution: No. Since f(1/2m) = 1/2m for natural numbers m, and since 1/2m as m, by the principle of analytic continuation, f(z) = z. But his contradicts f(1/(2m + 1)) = 1/(2m + 2). 11. Let f be holomorphic on a domain containing the closure of the unit disc, such that f(z) 5 for all z = 1. If f() = 3 + 4i, find f (). Solution: We claim that f is constant, and hence f () =. To see this, note that f() = 5 f(z) for all z = 1. This contradicts the maximum principle unless f(z) is a constant. Conformal maps 12. Find a fractional linear transformation that maps the circle z 1 = 1 onto the x-axis. Solution: A circle is characterized by three points on it, which in this case, we can take as z =, z = 2 and z = 1 + i. The x-axis is specified by w =, w = and w = 1. So the required fractional linear transformation is T z = z z 2 i 1 i + 1. Remark It is a good exercise in manipulating complex numbers and conjugates to show that T z is purely real if z 1 = 1. To see this note that z 1 = 1 is equivalent to z 2 = z + z (Why?). 13. Give examples of biholomorphisms between (a) {z z < 1} and {z Re(z) < }. 6

7 Solution: First map the disc to upper half plane by z i 1 + z 1 z, and then rotate by π/2. So ϕ(z) = z + 1 z 1 (b) {z z < 1} to itself with f() = and f(1/2) = i/2. Solution: A biholomorphism of the unit disc that sends zero to zero is a rotation. That is f(z) = e iθ z. But then θ = π/2. (c) {z < arg(z) < 3π/2} and {z < arg(z) < π/2}. Solution: ϕ(z) = z 1/3 = e log z/3, where we use the branch cut as the positive real axis (that is, we restrict < argz < 2π). (d) {z z < 1, Im(z) > } and {z z < 1}. Hint. Note that you cannot simply use z 2. Find a fractional linear transformation that will map the given region conformally onto the first quadrant {z Re(z) >, Im(z) > }. Solution: The map w = 1 + z 1 z maps the upper half of the unit disc to the first quadrant. Then ξ = ( 1 + z ) 2 1 z maps the domain to the upper half plane, and ζ = ξ i ξ + i maps this to the unit disc. So the final map is Φ(z) = (1 + z)2 i(1 z) 2 (1 + z) 2 + i(1 z) 2. 7

8 Gamma and Zeta functions 14. (a) Use a change of variables to show that for all t >, ( 1 Γ = t e 2) vt (vt) 1/2 dv. (b) Show that Hint. Write ( 1 2 dv Γ =. 2) v(1 + v) Γ(1/2) 2 = Γ(1/2)Γ(1/2) = e t t 1/2 Γ(1/2) dt, use part(a) for Γ(1/2) on the right hand side, and integrate out the t. (c) Use residue calculus to compute the integral on the right and conclude that Γ(1/2) = π. (d) More generally, show that Γ(s)Γ(1 s) = π sin πs. Hint. First prove the identity for s real and in the interval (, 1), and then use analytic continuation. Solution: We will just prove the last part. Let us first assume that < s < 1. Then Γ(1 s) = e x x s dx. For t >, consider the change of variables x = tv. Then dx = tdv and so So Γ(1 s) = t Γ(s)Γ(1 s) = = = e tv t s v s dv. e t t s 1 Γ(1 s) dt e t(v+1) v s dt dv v s v + 1 dv. In class we showed (Type IV integrals; integrating around branch cuts and residue theorem) that v s 1 + v dv = π sin πs. Plugging in s = 1/2 gives and so Γ(1/2) = π. Γ(1/2) 2 = π, 8

9 Miscellaneous 15. If Ω is connected, f : Ω C is holomorphic, and e f is constant, show that f is itself constant. Solution: = (e f(z) ) = f (z)e f. Since e f is never zero, f (z) = on Ω, and hence f is constant. 16. If f(z) is an entire function such that f(z) as z, show that there exists a constant c > and R > such that f(z) > c z for all z > R. Solution: This directly follows from the fact that any entire function with a pole at infinity is a non-constant polynomial. But we can give a direct proof. Since f(z) has a pole at infinity, g(w) = f(1/w) has a pole at w = of order M 1. In particular, there exists c > such that g(w) c w m for all w < r for some r >. We can choose r < 1 so that, w m < w, and so g(w) > c w on w < r. Or equivalently ( 1 ) f > c w w, for w < r. Putting z = 1/w, this is the same as saying f(z) > c z for z > 1/r. THe proof is complete by setting R = 1/r. 9