1. DO NOT LIFT THIS COVER PAGE UNTIL INSTRUCTED TO DO SO. Write your student number and name at the top of this page. This test has SIX pages.
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1 Student Number Name (Printed in INK Mathematics 54 July th, 007 SIMON FRASER UNIVERSITY Department of Mathematics Faculty of Science Midterm Instructor: S. Pimentel 1. DO NOT LIFT THIS COVER PAGE UNTIL INSTRUCTED TO DO SO. Write your student number and name at the top of this page. This test has SIX pages.. ONLY SCIENTIFIC CALCULATORS ARE PERMITTED: Calculators with alphanumeric storage, graphics tools or communication capacity are not allowed.. Full marks will be given for correct, complete, well-organized solutions. Full justification is required, and requested methods must be used. 4. Total marks for each question are shown in boxes at the beginning. Part marks are shown at the right of each part. The total is 40 marks You have 50 minutes to complete the test. Check that you have pages including this cover page. FOR EXAMINER S USE ONLY Question Mark available Mark P1 5 P 10 P 8 P4 8 P5 5 P 4 Total 40
2 Mathematics 54 July th, 007 of P1. 5 (a State the generalized cauchy integral formula and the conditions under which it applies. Let f(z be analytic on and inside a positively oriented simple closed contour C and z be any point inside C. Then f (n (z = n! f(ξ πi (ξ z n+1dξ, n N C (b Evaluate z = e z dz (z iπ The singularity is at z = iπ, this lies outside the circle z =, therefore e z dz (z iπ is analytic on and inside C and therefore by Cauchy-Goursat Theorem z = e z dz (z iπ = 0
3 Mathematics 54 July th, 007 of P. 8 (a State the Cauchy-Goursat Theorem. Given a simple closed contour C and let f(z be analytic on and inside C, then f(zdz = 0 C (b Determine the value of the integral 8 (z 4iz + z πi cos z e z Re(zdz = (z 4iz + z dz (z 4iz + z dz = 0 πi cos(zdz e z Re(zdz by Cauchy-Goursat as, z = π, thus z 4iz + z is a polynomial and is therefore entire. πi cos zdz = 0 by Cauchy-Goursat as, cos z is entire. ez Re(zdz = ez 1 (z+zdz = ze z dz = 0 by the Cauchy-Goursat Theorem as ze z is analytic. e z z dz = e z π z dz = π ie 0 = π i using the Cauchy integral formula, as e z is entire. ( ze z + z e z z dz
4 Mathematics 54 July th, of P. 8 Find the values of z for which z 4 = ( + i. + ( ( i = + = = 1 4 = arg ( + i = arctan ( ( / = arctan / = π So + i = (cos( π + kπ (π + kπ, k Z ( 1/4 [ ( ( π(1 + 1k π(1 + 1k z = + i = ( 1/ cos ( ( ] π(1 + 1k π(1 + 1k = [cos 1/8 4 4 ( If k = 0, then z 1 = [cos 1/8 π4 If k = 1, then z 1 = 1/8 [cos ( 1π 4 If k =, then z 1 = 1/8 [cos ( 5π 4 If k =, then z 1 = 1/8 [cos ( 7π 4 ( ] π4 ( 1π 4 ( 5π 4 ( 7π 4 ] ] ] ] 1/4
5 Mathematics 54 July th, of P4. 8 Verify that u(x, y = e y sin(x is harmonic and find its harmonic conjugate v(x, y. If the nd order partial derivatives are continuous and u xx + u yy = 0 then u is harmonic. e y sin(x is analytic and so its partial derivatives of all orders exist and are continuous. u x = e y cos(x u y = e y sin(x u xx = 4e y sin(x u yy = 4e y sin(x Therefore u xx + u yy = 0 and u is harmonic. If u is harmonic and the Cauchy-Riemann Equations are satisfied (for f = u + iv then v is the harmonic conjugate. u x = v y and u y = v x So and Integrating (1 gives v y = e y cos(x (1 v x = e y sin(x ( v = cos(x e y dy ( = cos(xe y + g(x (4 Differentiating (4 gives v x = sin(xe y + g (x comparing with ( shows that g (x = 0 and therefore g(x = C, where C is a constant. Thus v(x, y = e y cos(x + C
6 Mathematics 54 July th, 007 of P5. 5 Write (i 7 in the form re iθ and in the form x + iy. r = + i = ( + (1 = 4 = ( θ = arctan 1 + π = π π = 5π (i 7 = ( e 5πi 7 = 7 e 5πi re iθ = r(cos θ θ So (i ( ( ] 5π 5π 7 = [cos 7 [ ( ] [ ( ] 5π 5π = 7 cos + i 7 sin P. 4 Find the general solution of e z = 1 + i z = ln(1 + i = ln 1 + i + i arg(1 + i = ln + i(arctan(1 + kπ, k Z = 1 ln + i ( π 4 + kπ
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