u = 0; thus v = 0 (and v = 0). Consequently,

Transcription

1 MAT40 - MANDATORY ASSIGNMENT #, FALL 00; FASIT REMINDER: The assignment must be handed in before 4:30 on Thursday October 8 at the Department of Mathematics, in the 7th floor of Niels Henrik Abels hus, Blindern. Be careful to give reasons for your answers. To have a passing grade you must have correct answers to at least 50% of the questions. Exercise. a. Consider the function f(z) = (x 3 + 3xy 3x) + i (y 3 + 3x y 3y). Determine the points z = x + iy at which f(z) is differentiable. Answer. Set u(x, y) = x 3 + 3xy 3x and v(x, y) = y 3 + 3x y 3y. We compute x u = 3x + 3y 3 = y v, but y u = 6xy = x v, and thus u = v if and only if x = 0 or y = 0, i.e., by the Cauchy-Riemann y x equations f(z) = u(x, y) + i v(x, y) is differentiable at the coordinate axes. b. Let f(z) = u(x, y) + i v(x, y) be a function that is analytic in a domain D C, and suppose the real part of f, Re(f), is constant in D. Show that then f must be constant in D. Answer. The function f(z) = u(x, y) + i v(x, y) is differentiable at a point z implies that the Cauchy-Riemann equations hold at that point: u x = v y, u y = v x. By assumption, u = 0 and u = 0; thus v = 0 (and v = 0). Consequently, x y x y f (z) = u x + i v = 0 z D, x and by Theorem 6 (p. 76) this imples f = constant in D. Date: November 4, 00.

2 MAT40 - MANDATORY ASSIGNMENT #, FALL 00; FASIT c. Let f(z) be a function such that both f(z) and f(z) are analytic in a domain D C. Show that then f must be constant in D. Answer. Re(f) = f+f is (real-valued) analytic if and only if f and f are analytic. If f = u + i v, then f = u i v. These functions are differentiable at a point z if the Cauchy-Riemann equations hold: u x = v y, u y = v (for f); x u x = v y, u y = v (for f). x Adding these equations yields u x = u y = 0, i.e., the real part of f is constant. The result follows form b). d. Determine the poles of the function z 4 + z 3 z 5 + ( + i)z 4 + 4iz 3. Answer. First, write the polynomial in the denominator z 5 + ( + i)z 4 + 4iz 3 in factored form. Real roots z = and z = 0 (multiplicity 3). Complex root z = i. Factored form is z 3 (z + )(z + i). Next, write z 4 + z 3 = z 3 (z + ). Hence z 4 + z 3 z 5 + ( + i)z 4 + 4iz = z 3 (z + ) 3 z 3 (z + )(z + i) = z + i. i.e, the (single) pole is z = i.

3 Exercise. MAT40 - MANDATORY ASSIGNMENT #, FALL 00; FASIT 3 a. Consider the function f(z) = cos (z) + sin (z). Prove that this function is entire and f (z) = 0 for all z. Use this to conclude that sin (z) + cos (z) = for each z. Answer. The functions sin(z) and cos(z) are entire, and the chain rule implies that cos (z) and sin (z) are entire functions. Moreover, f (z) = cos(z) sin(z) + sin(z) cos(z) = 0. Hence, f(z) = C for some constant C. We determine C by f(0) = cos (0) + sin (0) = + 0 =, so C =. b. Determine the domain D on which f(z) = Log(4+i z) is analytic. Draw a picture of D in the complex plane. Compute f (z). What is the domain of analyticity of L π (z), where L π refers to the branch of log z based on the cut along the positive imaginary axis (draw a picture of the domain). Compute d L dz 0(z). Answer. Domain of analyticity of Log(4 + i z) is and d dz C \ {z = x + iy : x 4, y = }, Log(4 + i z) =. 4+i z Domain of analyticity of L π (z) is C \ {z = x + iy : x = 0, y 0}, and d dz L 0(z) = z. D D Figure. Left. Domain of analyticity of Log(4 + i z). Right. Domain of analyticity of L π (z). c. Consider the set (annulus) D = {z C : < z < 4}. Prove that there does not exist a function f(z) that is analytic on D such that f (z) = for all z D. [Hint: z Assume that such a function exists and then show that this leads to a contradiction.]

4 4 MAT40 - MANDATORY ASSIGNMENT #, FALL 00; FASIT Answer. If such a function exists, set h(z) = f(z) Log(z). Then h (z) = 0 for all z D\[ 3, ]. Consequently, h is constant in D\[ 3, ] and therefore f(z) = Log(z) + C for some constant C, in D \ [ 3, ]. This contradicts the assumption that f analytic in D. d. Consider the function sec(z) =. Derive the expression cos(z) ( ( ) ) sec (z) = i log z + z for the inverse of sec(z). Compute the principal value Sec ( ). Answer. Set w = sec (z). Then z = sec(w) = e iw +e iw ze iw + ze iw = or ze iw + z e iw = 0. Setting W = e iw, we obtain the quadratic equation zw W + z = 0 = W = z + ( z ). or In other words, e iw = z + ( z ) and so iw = log ( ( ) ) z + z, which, after multiplying both sides by i, yields ( sec (z) = i log z + ( z Next, we compute the principal value Sec ( ): ) ) Sec ( ) = i Log( ) = i (Log + iarg( )) = Arg( ) = π..

5 MAT40 - MANDATORY ASSIGNMENT #, FALL 00; FASIT 5 Exercise 3. a. Determine an admissible parametrization z(t), t [0, 3], of the contour Γ shown in Figure, which has initial point 0 and terminal point. Figure. Contour Γ for Exercise 3. Answer. A parameterization of Γ reads { t, t [0, ], z(t) = + e iπt, t [, 3]. b. Compute the length of the contour shown in Figure, using the parameterization from a) and formula () on page 59 in the book. Answer. The length l(γ) = 0 z (t) dt + 3 z (t) dt is + π, since d dt t dt = and c. Compute the integral Γ 3 0 d ( ) + e iπt dt = π. dt dz, where Γ is the contour shown in Figure. z+

6 6 MAT40 - MANDATORY ASSIGNMENT #, FALL 00; FASIT Answer. The contour Γ consists of two parts (γ, γ ), where z (t) = t, t [0, ], and z (t) = + e iπt, t [, 3]. Hence z + dz = First, z+ Γ γ z + dz + γ z + dz. has an antiderivative, namely the principal value of log(z + ): Log(z + ) = Log z + + i Arg(z + ). Since γ is a closed contour, it thus follows that (consult Corollary, page 75 in the book) Finally, Therefore, γ z + dz = [ ] Log(z + ) 0 Γ γ z + dz = 0. = Log() Log() = (Log + i Arg()) (Log + i Arg()) = ( i 0) (0 + i 0) = dz = z +