we are in C. solve z^5 =1. show that the sum of its solutions is null the deduct that cos(((2π)/5))+cos(((4π)/5))=−(1/2)


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Question Number 125669 by mathocean1 last updated on 12/Dec/20

$w e a r e i n C .$ $s o l v e z^{5} = 1 .$ $s h o w t h a t t h e s u m o f i t s s o l u t i o n s i s$ $n u l l t h e d e d u c t t h a t c o s (\frac{2 π}{5}) + c o s (\frac{4 π}{5}) = - \frac{1}{2}$
	Answered by mr W last updated on 13/Dec/20

	$z^{5} + 0 z^{4} - 1 = 0$ $\underset{k = 0}{\sum^{4}} z_{k} = 0 .$ $z_{k} = \cos \frac{2 k π}{5} + i \sin \frac{2 k π}{5}$ $Σ z_{k} = (1 + \cos \frac{2 π}{5} + \cos \frac{4 π}{5} + \cos \frac{6 π}{5} + \cos \frac{8 π}{5}) + i (9 + \sin \frac{2 π}{5} + \sin \frac{4 π}{5} + \sin \frac{6 π}{5} + \sin \frac{8 π}{5}) = 0$ $1 + \cos \frac{2 π}{5} + \cos \frac{4 π}{5} + \cos \frac{6 π}{5} + \cos \frac{8 π}{5} = 0$ $1 + \cos \frac{2 π}{5} + \cos \frac{4 π}{5} + \cos (2 π - \frac{4 π}{5}) + \cos (2 π - \frac{2 π}{5}) = 0$ $1 + \cos \frac{2 π}{5} + \cos \frac{4 π}{5} + \cos \frac{4 π}{5} + \cos \frac{2 π}{5} = 0$ $1 + 2 (\cos \frac{2 π}{5} + \cos \frac{4 π}{5}) = 0$ $\Rightarrow \cos \frac{2 π}{5} + \cos \frac{4 π}{5} = - \frac{1}{2}$
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