1) If ω is an imaginary fifth root of unity, then find value of log _2 ∣1+ω+ω^2 +ω^3 −(1/ω)∣ ? 2) Find value of : (i+(√3))^(100) +(i−(√3))^(100) +2^(100) ?


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Question Number 49202 by rahul 19 last updated on 04/Dec/18

$1) I f ω i s a n i m a g i n a r y f i f t h r o o t o f$ $u n i t y, t h e n f i n d v a l u e o f$ $\log_{2} ∣ 1 + ω + ω^{2} + ω^{3} - \frac{1}{ω} ∣ ?$ $2) F i n d v a l u e o f :$ ${(i + \sqrt{3})}^{100} + {(i - \sqrt{3})}^{100} + 2^{100} ?$
Commented by Abdo msup. last updated on 04/Dec/18

$1) w^{5} = 1 \Rightarrow w^{5} - 1 = 0 \Rightarrow (w - 1) (1 + w + w^{2} + w^{3} + w^{4}) = 0 b u t w \in i R$ $\Rightarrow 1 + w + w^{2} + w^{3} + w^{4} = 0 \Rightarrow 1 + w + w^{2} + w^{3} = - w^{4} \Rightarrow$ $1 + w + w^{2} + w^{3} - \frac{1}{w} = - (w^{4} + \frac{1}{w}) = - \frac{w^{5} + 1}{w} = - \frac{2}{w} \Rightarrow$ $∣ 1 + w + w^{2} + w^{3} - \frac{1}{w} ∣= \frac{2}{∣ w ∣} = \frac{2}{1} = 2 \Rightarrow$ ${l o g}_{2} ∣ 1 + w + w^{2} + w^{3} - \frac{1}{w} ∣= {l o g}_{2} (2) = 1 .$
Commented by Abdo msup. last updated on 04/Dec/18

$2) w e h a v e i + \sqrt{3} = 2 (\frac{\sqrt{3}}{2} + \frac{i}{2}) = 2 e^{i \frac{π}{6}} a n d$ $i - \sqrt{3} = - (\sqrt{3} - i) = - 2 e^{- \frac{i π}{6}} \Rightarrow$ ${(\sqrt{3} + i)}^{100} + {(- \sqrt{3} + i)}^{100} + 2^{100}$ $= 2^{100} e^{i \frac{100 π}{6}} + 2^{100} e^{- i \frac{100 π}{6}} + 2^{100}$ $= 2^{101} c o s (\frac{100 π}{6}) + 2^{100} b u t$ $c o s (\frac{100 π}{6}) = c o s (\frac{50 π}{3}) = c o s (\frac{48 π + 2 π}{3}) = c o s (16 π + \frac{2 π}{3})$ $= c o s (\frac{2 π}{3}) = - \frac{1}{2} \Rightarrow {(\sqrt{3} + i)}^{100} + {(- \sqrt{3} + i)}^{100} + 2^{100}$ $= 2^{101} . (- \frac{1}{2}) + 2^{100} = - 2^{100} + 2^{100} = 0 .$
Commented by rahul 19 last updated on 05/Dec/18
thank you prof Abdo ☺️
Commented by maxmathsup by imad last updated on 05/Dec/18

$y o u a r e w e l c o m e$
	Answered by tanmay.chaudhury50@gmail.com last updated on 04/Dec/18
	$2)(√3) +i =2(((√3)/2)+i(1/2))=2(cos(π/6)+isin(π/6))=2e^(i(π/6)) i−(√3) =−2(((√3)/2)−i(1/2))=−2e^(−i(π/6)) {(2e^(i(π/6)) )}^(100) +{(−2e^(−i(π/6)) )}^(100) +2^(100) =2^(100) {e^(i((100π)/6)) +e^(−i((100π)/6)) }+2^(100) =2^(100) (2cos((100π)/6))+2^(100) cos(3000^o )=cos(8×360^o +120^o )=cos(120^o ) cos(120^o )=((−1)/2) so ans is 2^(100) ×2×((−1)/2)+2^(100) =0$
	$2) \sqrt{3} + i$ $= 2 (\frac{\sqrt{3}}{2} + i \frac{1}{2}) = 2 (c o s \frac{π}{6} + i s i n \frac{π}{6}) = 2 e^{i \frac{π}{6}}$ $i - \sqrt{3}$ $= - 2 (\frac{\sqrt{3}}{2} - i \frac{1}{2}) = - 2 e^{- i \frac{π}{6}}$ ${(2 e^{i \frac{π}{6}})}^{100} + {(- 2 e^{- i \frac{π}{6}})}^{100} + 2^{100}$ $= 2^{100} {e^{i \frac{100 π}{6}} + e^{- i \frac{100 π}{6}}} + 2^{100}$ $= 2^{100} (2 c o s \frac{100 π}{6}) + 2^{100}$ $c o s (3000^{o}) = c o s (8 \times 360^{o} + 120^{o}) = c o s (120^{o})$ $c o s (120^{o}) = \frac{- 1}{2}$ $s o a n s i s$ $2^{100} \times 2 \times \frac{- 1}{2} + 2^{100} = 0$
	Commented by rahul 19 last updated on 05/Dec/18
	thank you sir! ��
	Answered by tanmay.chaudhury50@gmail.com last updated on 04/Dec/18

	$1) x^{5} = 1$ $(x - 1) (x^{4} + x^{3} + x^{2} + x + 1) = 0$ $w^{4} + w^{3} + w^{2} + w + 1 = 0$ $1 + w + w^{2} + w^{3} - \frac{1}{w}$ $= - w^{4} - \frac{1}{w}$ $= \frac{- w^{5} - 1}{w} = \frac{- 1 - 1}{1} = - 2$ $s o ∣ 1 + w + w^{2} + w^{3} - \frac{1}{w} ∣=∣ - 2 ∣= 2$ $h e n c e {l n}_{2} 2 = 1$
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