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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

This note was uploaded on 07/27/2010 for the course MATH Math2009 taught by Professor Koskesh during the Spring '09 term at SUNY Empire State.

Ask a homework question - tutors are online