TECHNISCHE UNIVERSITEIT EINDHOVEN Faculteit Wiskunde en Informatica Re-exam Complex Analysis 2DN40 + 2DE15 + 2Y480, Wednesday 29 January 2014, 14:00 – 17:00 hour. The exam consists of 6 questions on 1 page. For each question the following points can be obtained: qst. pnt. qst. pnt. qst. pnt. qst. pnt. qst. pnt. qst. pnt. qst. pnt. 1 : 4 2 : 4 3 : 6 4a : 3 4b : 3 5 : 4 6 : 6 The final mark is obtained by dividing the total number of points by 3, and then rounding off the result to an integer between 1 and 10. The answers have to be motivated, formulated clearly, and written down orderly. Any computer (notebook, laptop, smartphone, tablet, iPad), books or sources of information are not allowed. One is not allowed to have a mobile phone. 1. For which holomorphic functions f (z) is Re( f (x + iy)) = x 4 − 6x 2 y 2 + y 4 + x y for all x + iy ∈ C ? Express all possible functions f (z) explicitly in z. n 2. The function f ∶ C → C, analytic in C/{0}, has the Laurent series ∑∞ n=−∞ cn z in 0 < ∣z∣ < ∞. Let γ be a positively oriented Jordan curve enclosing the origin. Determine (in terms of cn ) 1 ∫ f (2z) dz, 2π i γ 1 2 ∫ f (z ) dz, 2π i γ 1 1 ∫ z f ( z ) dz, 2π i γ f (z 2 ) 1 dz. ∫ 2π i γ z 3 3. Let a > b > 0 be real constants. Evaluate the integral 2π ∫ 0 1 dθ. a + b cos θ Hint: the algebra simplifies if you write a = b cosh(ϕ) for some ϕ > 0. 4. a. Find the solutions of z 3 − i = 0. b. Determine the integral ∫ ∞ −∞ x 6 by using the identity (for x ∈ R) x6 1 dx +1 1 1 = Im [ 3 ] . +1 x −i √ 5. Let z denote the principal value square root of z (i.e. with branch cut along the negative real √ axis and 1 = 1). Let a be positive real. Find and sketch the branch cuts of the functions √ √ √ √ √ a2 − z 2, z 2 − a2, z 2 + a2 , z −a ⋅ z+a. 6. Determine the integral ∫ ∞ −∞ sin x dx. x(1 + x 2 ) 3 ANSWERS 1. Write f (z) = u(x, y) + iv(x, y), such that u = x 4 − 6x 2 y 2 + y 4 + x y. For analytic f , u and v satsify the Cauchy-Riemann equations, so v y = u x = 4x 3 − 12x y 2 + y, so v = 4x 3 y − 4x y 3 + 12 y 2 + C(x). Furthermore, also vx = 12x 2 y − 4y 3 + C ′ (x) = −u y = −(−12x 2 y + 4y 3 + x) = 12x 2 y − 4y 3 − x, so C ′ = −x, and hence C = − 12 x 2 + C0 , with C0 a real constant. Finally f (z) = z 4 − 12 iz 2 + iC0 . 2. Expand c−1 + ⋯, 2z c2 z f ( 1z ) = ⋯ + c1 + + ⋯, z f (2z) = ⋯ + c−1 + c0 + ⋯, z2 f (z 2 ) c1 = ⋯ + c0 + + ⋯ z3 z f (z 2 ) = ⋯ + such that the integrals are respectively equal to 12 c−1 , 0, c2 and c1 . 3. Consider the integral as a contour integral along the unit circle. Replace eiθ by z with cos θ = 21 (z + z −1 ) and dθ = −idz/z. Then ∫ −i ∣z∣=1 z(a + 1 b(z 2 + z −1 )) dz = ∫ ∣z∣=1 dz = + 1) 2 1 1 dz dz = ∫ z 2 + 2 cosh(ϕ)z + 1 ib ∣z∣=1 (z + eϕ )(z + e−ϕ ) −i az + 2 ∫ ib ∣z∣=1 1 b(z 2 2 Since only the pole z = − e−ϕ is inside the contour, we have finally 4π i 2π 2π = =√ . −ϕ − e ) b sinh(ϕ) (a 2 − b2 ) ib(eϕ 4. a. z 3 − i = 0 can writing z and i in polar coordinates. We find z 1 = −i, √ be solved by √ z 2 = 12 (i + 3), z 3 = 12 (i − 3). b. Since the integral of 1/(z 3 − i) converges, the integral of the imaginary part of the integrand is equal to the imaginary part of the integral. If we close the contour downward we find ∫ ∞ −∞ z 3 2π i 1 dz = −2π i Res(z 1 ) = − = 2 π i. −i (z 1 − z 2 )(z 1 − z 3 ) 3 So the required integral is 23 π 5. We write z = x + iy and λ ⩾ 0. Then we need respectively: a 2 − z 2 = a 2 − (x 2 − y 2 + 2ix y) = −λ leading to (∣x∣ ⩾ a ∧ y = 0). z 2 − a 2 = (x 2 − y 2 + 2ix y) − a 2 = −λ leading to (y ∈ R ∧ x = 0) ∨ (∣x∣ ⩽ a ∧ y = 0). z 2 + a 2 = (x 2 − y 2 + 2ix y) + a 2 = −λ leading to (∣y∣ ⩾ a ∧ x = 0). For the last function we observe that the branch cuts (−∞, −a] and (−∞, a] partly cancel each other, such that we have in the end a branch cut along [−a, a]. 4 6. The integrand as a function of z ∈ C has a removable singularity in z = 0 and is analytic except for the points z = ±i. So if we stay away from z = ±i, we can deform the integration contour such, that it doesn’t pass through z = 0 but with the integral remaining the same. Choose a deformation below z = 0. Now we can split the integrand in complex exponentials: ∫ ∞ −∞ ∞ eiz − e−iz ∞ ∞ eiz e−iz 1 1 sin x dx = dz = dz − dz. ⨚ ⨚ ⨚ −∞ 2iz(1 + z 2 ) x(1 + x 2 ) 2i −∞ z(1 + z 2 ) 2i −∞ z(1 + z 2 ) Note: without contour deformation this split is meaningless, because the separate integrals would not exist. The contour of the first integral can be closed upwards (because eiz is bounded and the integrand ∼ 1/z 3 for ∣z∣ → ∞). This integral is equal to the sum of the residues in z = 0 and z = i. The contour of the second integral can be closed downwards (because e−iz is bounded and the integrand ∼ 1/z 3 for ∣z∣ → ∞). This integral is equal to the residue in z = −i. Altogether we find for the total integral π (1 − 12 e−1 ) − π 21 e−1 = π − π e−1 .