Adv Alegbra HW Solutions 16

Adv Alegbra HW Solutions 16 - 16 1.35 Let 1 , . . . , n be...

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16 1.35 Let ε 1 ,...,ε n be complex numbers with | ε j |= 1 for all j , where n 2. (i) Prove that ± ± ± n X j = 1 ε j ± ± ± n X j = 1 ± ± ε j ± ± = n . Solution. The triangle inequality gives | u + v |≤| u |+| v | for all complex numbers u and v , with no restriction on their norms. The inductive proof is routine. (ii) Prove that there is equality, ± ± ± n X j = 1 ε j ± ± ± = n , if and only if all the ε j are equal. Solution. The proof is by induction on n 2. For the base step, suppose that | ε 1 + ε 2 |= 2. Therefore, 4 =| ε 1 + ε 2 | 2 = 1 + ε 2 ) · 1 + ε 2 ) =| ε 1 | 2 + 2 ε 1 · ε 2 +| ε 2 | 2 = 2 + 2 ε 1 · ε 2 . Therefore, 2 = 1 + ε 1 · ε 2 , so that 1 = ε 1 · ε 2 =| ε 1 || ε 2 | cos θ = cos θ, where θ is the angle between ε 1 and ε 2 (for | ε 1 |= 1 =| ε 2 | ). Therefore, θ = 0or θ = π , so that ε 2 ε
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