1) simplify W_n (z)=(1+z)(1+z^2 )....(1+z^2^n ) (z from C) 2) simplify P_n (θ) =(1+e^(iθ) )(1+e^(2iθ) ).....(1+e^(i2^n θ) ) and sove P


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Question Number 63893 by mathmax by abdo last updated on 10/Jul/19

$1) s i m p l i f y W_{n} (z) = (1 + z) (1 + z^{2}) . . . . (1 + z^{2^{n}}) (z f r o m C)$ $2) s i m p l i f y P_{n} (θ) = (1 + e^{i θ}) (1 + e^{2 i θ}) . . . . . (1 + e^{i 2^{n} θ}) a n d s o v e$ $P_{n} (θ) = 0$
Commented by mathmax by abdo last updated on 05/Nov/19

$1) w e h a v e W_{n} (z) = (1 + z) (1 + z^{2}) . . . . (1 + z^{2^{n}})$ $l e t p r o v e b y r e c u r r e n c e t h a t W_{n} (z) = \frac{1 - z^{2^{n + 1}}}{1 - z}$ $n = 0 \to W_{n} (z) = 1 + z = \frac{1 - z^{2}}{1 - z} (t r u e) l e t s u p p o s e t h e r e l a t i o n t r u e$ $W_{n + 1} = (1 + z) (1 + z^{2}) . . . . (1 + z^{2^{n + 1}}) = (1 + z) (1 + z^{2}) . . . (1 + z^{2^{n}}) (1 + z^{2^{n + 1}})$ $= \frac{1 - z^{2^{n + 1}}}{1 - z} \times (1 + z^{2^{n + 1}}) = \frac{1 + z^{2^{n + 1}} - z^{2^{n + 1}} - z^{2^{n + 2}}}{1 - z} = \frac{1 - z^{2^{n + 2}}}{1 - z}$ $c / c W_{n} (z) = \frac{1 - z^{2^{n + 1}}}{1 - z} w i t h z \neq 1$ $2) P_{n} (θ) = (1 + e^{i θ}) (1 + e^{2 i θ}) . . . . . (1 + e^{i 2^{n} θ})$ $= (1 + e^{i θ}) (1 + {(e^{i θ})}^{2}) . . . . . (1 + {(e^{i θ})}^{2^{n}}) = W_{n} (e^{i θ})$ $= \frac{1 - {(e^{i θ})}^{2^{n + 1}}}{1 - e^{i θ}} = \frac{1 - e^{i 2^{n + 1} θ}}{1 - e^{i θ}} (i f θ \neq 2 k π)$ $P_{n} (θ) = 0 \Leftrightarrow 1 - e^{i 2^{n + 1} θ} = 0 \Rightarrow e^{i 2^{n + 1} θ} = e^{i 2 k π} \Rightarrow$ $θ_{k} = \frac{2 k π}{2^{n + 1}} = \frac{k π}{2^{n}} w i t h k \in [[1, 2^{n + 1} - 1]]$
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