P-a-z-z-2n-2z-n-cosa-1-montrer-que-p-a-z-k-0-n-1-z-2-2zcos-a-pi-2kpi-n-1-

Question Number 147302 by puissant last updated on 19/Jul/21

P_{a} (z) = z^{2 n} - 2 z^{n} c o s a + 1

m o n t r e r q u e p_{a} (z) = \underset{k = 0}{\prod^{n - 1}} (z^{2} - 2 z c o s (\frac{a}{π} + \frac{2 k π}{n}) + 1)

Answered by mathmax by abdo last updated on 20/Jul/21

p_{a} (z) = z^{2 n} - {2 z}^{n} cosa + 1 let z^{n} = x \Rightarrow

p_{a} (z) = x^{2} - 2 (cosa) x + 1

Δ^{^{'}} = \cos^{2} a - 1 = - \sin^{2} a \Rightarrow x_{1} = cosa + isina = e^{ia} and x_{2} = e^{- ia}

\Rightarrow p_{a} (z) = (z^{n} - e^{ia}) (z^{n} - e^{- ia})

z^{n} = e^{ia} for z = e^{i θ} \Rightarrow ni θ = i (a + 2 k π) \Rightarrow θ_{k} = \frac{a + 2 k π}{2 n} k \in [[0, n - 1]]

\Rightarrow z_{k} = e^{i (\frac{a + 2 k π}{n})} the roots of z^{n} - e^{ia} = 0 are α_{k} = e^{- i (\frac{a + 2 k π}{n})} \Rightarrow

p_{a} (z) = \prod_{k = 0}^{n - 1} (z - e^{i (\frac{a + 2 k π}{n})}) (z - e^{- i (\frac{a + 2 k π}{n})})

= \prod_{k = 0}^{n - 1} (z - z_{k}) (z - {\bar{z}}_{k}) = \prod_{k = 0}^{n - 1} (z^{2} - 2 Re (z_{k}) z + ∣ z_{k} ∣^{2})

= \prod_{k = 0}^{n - 1} (z^{2} - 2 \cos (\frac{a}{n} + \frac{2 k π}{n}) z + 1)

Commented by puissant last updated on 20/Jul/21

thanks sir