let give p(x)= (((1+ix)/(1−ix)))^n − ((1+itanα )/(1−itanα)) factorize p(x) inside C[x].


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Question Number 27384 by abdo imad last updated on 05/Jan/18

$l e t g i v e p (x) = {(\frac{1 + i x}{1 - i x})}^{n} - \frac{1 + i t a n α}{1 - i t a n α} f a c t o r i z e p (x) i n s i d e$ $C [x] .$
Commented by abdo imad last updated on 07/Jan/18

$r o o t s o f p (x)$ $p (x) = 0 \Leftrightarrow {(\frac{1 + i x}{1 - i x})}^{n} = \frac{1 + i t a n α}{1 - i t a n α} = \frac{c o s α + i s i n α}{c o s α - i s i n α} = e^{i (2 α)}$ $l e t p u t \frac{1 + i x}{1 - i x} = t s o p (x) = 0 \Leftrightarrow t^{n} = e^{i (2 α)} w i c h h a v e f o r s o l u t i o n s$ $t_{k} = e^{i 2 (α + k π) \frac{1}{n}} a n d k \in [[0, n - 1]] b u t \frac{1 + i x}{1 - i x} = t_{k}$ $\Leftrightarrow 1 + i x = (1 - i x) t_{k} \Leftrightarrow i x (1 + t_{k}) = t_{k} - 1$ $i x = \frac{t_{k} - 1}{1 + t_{k}} \Leftrightarrow x = - i \frac{t_{k} - 1}{1 + t k} = i \frac{1 - t_{k}}{1 + t_{k}} s o t h e c o m p l e x w i c h$ $v e r i f y p (x) = 0 a r e x_{k} = i \frac{1 - t_{k}}{1 + t_{k}} a n d k \in [[0, n - 1]]$ $b u t x_{k} = i \frac{1 - e^{i 2 \frac{α + k π}{n}}}{1 + e^{i 2 \frac{α + k π}{n}}} = i \frac{1 - c o s (2 \frac{α + k π}{n}) - i s i n (2 \frac{α + k π}{n})}{1 + c o s (2 \frac{α + k π}{n}) + i s i n (2 \frac{α + k π}{n})}$ $x_{k} = t a n (\frac{α + k π}{n}) w i t h k f r o m [[0, n - 1]]$ $s o p (x) = λ \prod_{k = 0}^{n - 1} (x - t a n (\frac{α + k π}{n})) .$
	Answered by sma3l2996 last updated on 06/Jan/18

	$p (x) = {(\frac{i (- i + x)}{- i (i + x)})}^{n} - \frac{c o s α + i s i n α}{c o s α - i s i n α} = {(\frac{- (- 2 i + i + x)}{i + x})}^{n} - \frac{e^{i α}}{e^{- i α}}$ $p (x) = {(- 1)}^{n} {(1 - \frac{2}{i + x})}^{n} - e^{2 i α}$ $p (x) = {(- 1)}^{n} \underset{k = 0}{\sum^{n}} {(\frac{2}{i + x})}^{k}^{n} C_{k} - e^{2 i α}$
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