(a) Show that for each complex number z ‡ 1, (b) Show that (c) Show that η n Στ k=0 Text Σcos(k0) k=0 = – 1-₂n+1 1-z n Re (Σ (et) k=0 (eroja). n Σ (eig)* k=0 Hint: Factor eit/2 out of the numerator and denominator on the left-hand side. (d) Use parts (b) and (c) to show that 1 – pi(n+1)θ 1 ρίθ ei(2n+1)0/2 –10/2 e 2i sin(θ/2) 1 + cos(0) + cos(20) + ... + cos(nθ) 1 = + 2 sin((2n + 1)0/2) 2 sin(θ/2)

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.3: Zeros Of Polynomials
Problem 35E
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2. Let be a real number such that 0 < 0 < 2π. Let n be a positive integer.
(a) Show that for each complex number z ‡ 1,
(b) Show that
(c) Show that
n
k=0
=
zk
Trett
=
1 — zn+1
1-z
n
Σ cos(k0) = Re (Σ (et)
k=0
n
= Re (Σ (e²9) *).
k=0
n
Σ (eio) k
k=0
Hint: Factor e¹0/2 out of the numerator and denominator on the left-hand side.
(d) Use parts (b) and (c) to show that
1- ei(n+1)0
1- eie
ei(2n+1)0/2e-i0/2
2i sin (0/2)
1 + cos(0) + cos(20) + ... + cos(ne)
1
= +
2
sin((2n + 1)0/2)
2 sin(0/2)
Transcribed Image Text:2. Let be a real number such that 0 < 0 < 2π. Let n be a positive integer. (a) Show that for each complex number z ‡ 1, (b) Show that (c) Show that n k=0 = zk Trett = 1 — zn+1 1-z n Σ cos(k0) = Re (Σ (et) k=0 n = Re (Σ (e²9) *). k=0 n Σ (eio) k k=0 Hint: Factor e¹0/2 out of the numerator and denominator on the left-hand side. (d) Use parts (b) and (c) to show that 1- ei(n+1)0 1- eie ei(2n+1)0/2e-i0/2 2i sin (0/2) 1 + cos(0) + cos(20) + ... + cos(ne) 1 = + 2 sin((2n + 1)0/2) 2 sin(0/2)
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