137A_HW01_solutions.pdf - Physics 137A Spring 2017 Problem...

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Physics 137A, Spring 2017 Problem Set 1 Solutions Problem 1: Complex Makes Simple a) We use Euler’s formula to get Ae = a + ib (1) A cos δ + iA sin δ = a + ib (2) We equate real and imaginary parts to get a = A cos δ (3) b = A sin δ (4) Inverting these, we find A and δ in terms of a and b A = p a 2 + b 2 (5) δ = tan - 1 b a (6) However, the principal value of the inverse tangent function is defined to be in the range - π 2 < tan - 1 ( x ) < π 2 . This gives the correct phase δ for complex numbers in the right half of the complex plane, and for those in the left half of the complex plane we must add a factor of π . Therefore, the final result is A = p a 2 + b 2 (7) δ = ( tan - 1 ( b a ) , a > 0 tan - 1 ( b a ) + π , a < 0 (8) b) We directly compute | z | 2 to show that it equals the result given | z | 2 = zz * = ( a + ib )( a - ib ) = a 2 - iab + iab - i 2 b 2 = a 2 + b 2 = A 2 (9) c) Let z = a + ib . Then, z + z * = ( a + ib ) + ( a - ib ) = 2 a = 2 Re { z } (10) z - z * = ( a + ib ) - ( a - ib ) = i 2 b = i 2 Im { z } (11) Therefore, 1 2 ( e + e - ) = 1 2 2 Re { e } = cos φ (12) 1 2 ( e - e - ) = 1 2 i i 2 Im { e } = sin φ (13) d) We can find | y s | 2 from y s e i ( kx - ωt + δ s ) = (˜ y 1 + ˜ y 2 ) e i ( kx - ωt ) (14) | y s e i ( kx - ωt + δ s ) | 2 = | y 1 + ˜ y 2 ) e i ( kx - ωt ) | 2 (15) y 2 s = | ˜ y 1 + ˜ y 2 | 2 = | ˜ y 1 | 2 + ˜ y 1 ˜ y * 2 + ˜ y * 1 ˜ y 2 + | ˜ y 2 | 2 (16) = y 2 1 + 2 Re { ˜ y 1 ˜ y * 2 } + y 2 2 = y 2 1 + 2 y 1 y 2 Re { e i ( δ 1 - δ 2 ) } + y 2 2 (17) = y 2 1 + y 2 2 + 2 y 1 y 2 cos(Δ δ ) (18) 1
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e) We can find δ s from y s e s e i ( kx - ωt ) = (˜ y 1 + ˜ y 2 ) e i ( kx - ωt ) (19) y s e s = ˜ y 1 + ˜ y 2 = y 1 e 1 + y 2 e 2 (20) = ( y
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