5 - we have n cos(j θ) = j =0 n j =0 = = = 1 eiθ/2...

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Unformatted text preview: we have n cos(j θ) = j =0 n j =0 = = = 1 eiθ/2 e−iθ/2 − ei(n+ 2 )θ eiθ/2 e−iθ/2 − eiθ/2 1 − ei(n+1)θ 1 − eiθ eij θ = e−iθ/2 − ei(n+ 2 )θ e−iθ/2 − eiθ/2 1 i e−iθ/2 − ei(n+ 2 )θ 1 i e−iθ/2 − eiθ/2 Now we know eiθ − e−iθ = sin(θ), 2i so i e−iθ − eiθ = 2 sin(θ) is purely real. Thus i e−iθ/2 − ei(n+ 2 )θ 1 i e−iθ/2 − eiθ/2 = = ie−iθ/2 − iei(n+ 2 )θ 1 θ 2 sin 2 sin n + 1 θ + sin θ/2 2 2 sin θ/2 1 1 sin n + 2 θ =+ . 2 2 sin θ/2 . 6. (a) Recall that nonzero polynomial of degree n has at most n roots. Now consider the polynomial p(z ) = (z − ω0 ) · · · (z − ωn−1 ) − z n + 1. We know that p(z ) has degree at most n − 1 since the z n terms cancel. Yet p(ωi ) = 0 for each i with 0 ≤ i < n. Thus p(z ) has n 5 ...
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