2 Write down the range of values of α (real) or β (complex) for which the following integrals converge. (i) e z2 dz where {γ : z = se iα, < s < }

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1 Mathematical Tripos Part II Michaelmas term 2007 Further Complex Methods, Examples sheet Dr S.T.C. Siklos Comments and corrections: to Sheet with commentary available for supervisors. All handouts and examples sheets are available on Show that: dx (i) (x 2 + ) 2 (x 2 + 4) = π 8 ; (ii) (iii) 0 cos x dx x 2 + a 2 = π a e a where a > 0 ; x sin x x 3 dx = π 2 ; What is the value of the integral in part (ii) when a < 0? These are just revision of the Part IB course. For part (i), they can use l Hôpital for the single pole and the derivative formula for the double pole. The answer to part (ii) is obviously πe a / a. You don t need small circles for part (iii): just deform the contour to avoid the origin, then split the sin into exponentials. You need the deformation so as not to have singularities on the contour (unless you use the Cauchy PV). 2 Write down the range of values of α (real) or β (complex) for which the following integrals converge. (i) e z2 dz where { : z = se iα, < s < } (ii) (iii) (iv) 0 e /z dz where { : z = se iα, 0 s } x β dx + x ( + tanh z) dz, where { : z = se iα, 0 s < } Answers: (i) π/4 α 3π/4 or 5π/4 α 7π/4; (ii) π/2 α 3π/2; (iii) < Reβ < 0; (iv) π/2 < α < 3π/2.

2 3 Let f(t) be analytic at t = 0 with f(0) = 0 and f (0) 0. Let C be a circle centred on the origin, with interior D, such that f is analytic in D and the inverse of f exists on f(d). For a fixed point z within C, let w = f(z). Assuming that w is small, show (using the residue theorem) that and hence that z = b n w n, where b n = 2πi n= C z = 2πi C tf (t) dt = (f(t)) n+ 2πin C tf (t) f(t) w dt, (f(t)) n dt = n! lim d n ( ) t n t 0 dt n. f(t) Show that the equation w = ze z has a solution, for sufficiently small w (how small?), z = n= n n Find also one solution of the equation w = 2z z 2. This give a rather remarkable formula for the power series for the inverse of a function, known for example to Lagrange long before the theory of complex variable. The existence of a disc D such that f exists on f(d) is guaranteed by the conditions that f is analytic at t = 0 and f (0) 0. (This first result shows that the inverse is also analytic.) The existence of the inverse means of course that the equation w = f(t) has only one solution (t = z) for w f(d). The uniqueness of the solution to the equation w = f(t) guarantees that the only singularity of the integrand is at t = f (w). The first result comes easily from l Hôpital, and the second from the series expansion of the denominator of the integrand which converges for w/f(z) <. This latter condition can be satisfied if w is taken to be small enough, since f (0) = 0. It is a nice exercise, using the ratio test and lim n ( + /n) n e, to show that the radius of convergence of the final series is e. The solution of the quadratic is the one closest to the origin, and the D must have radius less than (the inverse function has a singularity at w = ). 4 Let φ(x, y) be a harmonic function. Show that φ is the real part of any analytic function f(z) of the form f(z) = 2φ ( (z + )/2, (z )/2i ) φ(, 0) + ic where c is a real constant (provided φ is such that the right hand side exists). Use this formula to find analytic functions whose real parts are (i) x/(x 2 + y 2 ) and (ii) tan y/x. If you set = 0 wherever occurs in the above formula you get the more straightforward case that I did in the lectures which doesn t work for functions that are bad at the origin, as in the last part of the question. For the proof, you consider g(z, z) φ( 2 (z + z), 2i (z z)) + iψ( 2 (z + z), 2i (z z)). f(z) = g(z, z) so g(z, z) is independent of z. We take z =. Then note that φ and ψ are real functions. Taking the complex conjugate of the constant function g(, z) shows that the two halves of g(z, z) are equal. The obvious function with real part equal to tan y/x is i log z, so this may give them pause for thought (it requires a little work to establish the equivalence) or maybe not. n! w n. 2

3 5 Let F (z) = e uz + e u du. For what region of the z-plane does F (z) define an analytic function? Show by closing the contour (use a rectangle) in the upper half plane that F (z) = π cosec πz. Explain how this result provides the analytic continuation of F (z). F (z) is analytic if 0 < Rz <. I don t expect a serious justification of this they might say that nothing goes wrong with the integrand (it is continuous as a function of t and z, and analytic as a function of z for each t, and converges exponentially). The exponential behaviour for large u means that the integral converges uniformly i.e. given ɛ > 0, one can find U 0 (indep. of z) such that U < ɛ for all U > U 0. This is certainly true if Rz k for any small positive k, so F is analytic on any strip of the form m Rz k, i.e. on 0 < Rz <. You can also do this with the Weierstrass comparison test (comparing the itengrand with a function that is independent of z), but the more intuitive approach is appropriate for this course. Obviously, the analytic continuation of F (z) has the value of π cosec πz, for z n: both functions are analytic on the strip where they agree, but π cosec πz is analytic on the wider domain. 3

4 6 Let F n (z) = ( + iz ) n ( iz ) n, n n where n is an odd integer. By setting z = n tan θ, or otherwise, show that F n (z) has simple zeros at z = ±n tan(rπ/n) for r =, 2,..., m, where 2m + = n. Deduce that F n (z) = 2iz m r= Hence (by taking the limit n ) show that sin z = z ( z 2 ) n 2 tan 2. (rπ/n) r= ) ( z2 r 2 π 2. By comparing this with the Taylor expansion for sin z, show that ζ(2) = π 2 /6 and find an expression for ζ(4). [Note: ζ(k) n k.] n= This is essentially due to Euler rather clever, I think. It was one of his 5 (!) solutions to the previously baffling Basel problem of evaluating ζ(2). It seems that he realised that this method lacked rigour; he gave a more elementary method which is rigorous. The final step requires a condition along the lines of uniform convergence (which of course Euler could not have supplied). Tannery s theorem says that if F (z, n) = and w r (z) = lim n v r (z, n) (fixed r) then n v r (z, n) r=0 lim F (z, n) = w r (z) n provided that v r (z, n) M r (z), where M r (z) is independent of n and M r (z) converges. I don t expect them to know this or even think about it. ζ(4) = π 4 /90 since the coefficient of x 5 ( in the product is 2π /n 2 ) 2 4 2π /n Let f (z) be the branch of (z 2 ) 2 defined by branch cuts in the z-plane along the real axis from to and from to, with f (z) real and positive just above the latter cut. Let f 2 (z) be the branch of (z 2 ) 2 defined by a cut along the real axis from to +, with f 2 (x) real and positive for (x ) real and positive. Show that f (z) = f ( z) but f 2 (z) = f 2 ( z). Set z = + re iθ and z = + se iφ, where θ = φ = 0 just above the right hand branch cut, so f (z) = rse i(θ+φ)/2. Take z in the uhp, where (say) both θ and φ are acute. Then see how θ and φ vary along the straight line path to z. A little geometry shows that θ π + φ, φ θ π, so f ( z) = f (z). r=0 4

5 8 Show, without making substitutions or manipulating integrands, that (i) du ( u 2 ) 2 ( + u 2 ) = π 2 ; (ii) du u(u 2 ) 2 = π 2. State how the residues in (i) would differ if the integrand were u 2 ( u 2 ) 2 ( + u 2 ), and explain why the value of the integral would then be π π/ 2. The extra term in (i) comes from a non-zero contribrution from the large semi-circle - essentially a pole at infinity. You could also do the integral by using the residue theorem on this pole, moving it to the origin by z = t. You have to worry about the sense of the curve. 5