let z_1 and z_2 the roots of x^2 −2x+2=0 1) calculate z_1 ^3 +z_2 ^3 then (1/z_1 ^3 ) +(1/z_2 ^3 ) 2) calculate z_1 ^4 +z_2 ^4 then (1/z_1 ^4 ) +(1/z_2 ^4 ) 3) let n from N simplify A_n = z_1 ^n +z_2 ^n and B_n = z_1 ^n −z_2 ^n 4) simplify S_n =Σ_(k=0) ^(n−1) (z_1 ^k +z


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Question Number 41622 by math khazana by abdo last updated on 10/Aug/18

$l e t z_{1} a n d z_{2} t h e r o o t s o f x^{2} - 2 x + 2 = 0$ $1) c a l c u l a t e z_{1}^{3} + z_{2}^{3} t h e n \frac{1}{z_{1}^{3}} + \frac{1}{z_{2}^{3}}$ $2) c a l c u l a t e z_{1}^{4} + z_{2}^{4} t h e n \frac{1}{z_{1}^{4}} + \frac{1}{z_{2}^{4}}$ $3) l e t n f r o m N s i m p l i f y$ $A_{n} = z_{1}^{n} + z_{2}^{n} a n d B_{n} = z_{1}^{n} - z_{2}^{n}$ $4) s i m p l i f y S_{n} = \sum_{k = 0}^{n - 1} (z_{1}^{k} + z_{2}^{k})$
Commented by rahul 19 last updated on 10/Aug/18

$Wow, i like the change from calculus$ $to algebra Q . now - a - days : -)$
Commented by alex041103 last updated on 10/Aug/18

$: D$
Commented by tanmay.chaudhury50@gmail.com last updated on 10/Aug/18

$w h a t i s t h e m e a n i n g o f D$
Commented by alex041103 last updated on 10/Aug/18

$I t s l i k e b i g s m i l e i f y o u r o t a t e D 90 °$
Commented by tanmay.chaudhury50@gmail.com last updated on 10/Aug/18

$h a h a h a . . .$
Commented by math khazana by abdo last updated on 12/Aug/18

$1) l e t f i n d z_{1} a n d z_{2}$ $Δ^{^{'}} = 1 - 2 = - 1 = i^{2} \Rightarrow z_{1} = 1 + i a n d z_{2} = 1 - i$ $z_{1} = \sqrt{2} e^{\frac{i π}{4}} a n d z_{2} = \sqrt{2} e^{- \frac{i π}{4}} \Rightarrow$ $z_{1}^{3} + z_{2}^{3} = z_{1}^{3} + {\overset{-^{3}}{z}}_{1} = 2 R e (z_{1}^{3}) = 2 R e (2 \sqrt{2} e^{\frac{i 3 π}{4}})$ $= 2 R e (2 \sqrt{2} (- \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}})) = 2 (- 2) = - 4 a l s o$ $\frac{1}{z_{1}^{3}} + \frac{1}{z_{2}^{3}} = \frac{z_{1}^{3} + {\overset{-^{3}}{z}}_{1}}{{(z_{1} z_{2})}^{3}} = \frac{- 4}{8} = - \frac{1}{2}$
Commented by math khazana by abdo last updated on 12/Aug/18

$2) z_{1}^{4} + z_{2}^{4} = z_{1}^{4} + {\overset{-^{4}}{z}}_{1} = 2 R e (z_{1}^{4}) = 2 R e ({(\sqrt{2})}^{4} e^{i π})$ $= 2 R e (- 4) = - 8$ $\frac{1}{z_{1}^{4}} + \frac{1}{z_{2}^{4}} = \frac{z_{1}^{4} + {\overset{-^{4}}{z}}_{1}}{{(z_{1} . z_{2})}^{4}} = \frac{- 8}{2^{4}} = - \frac{8}{16} = - \frac{1}{2}$
Commented by math khazana by abdo last updated on 12/Aug/18

$3) w e h a v e z_{1} = \sqrt{2} e^{\frac{i π}{4}} a n d z_{2} = \sqrt{2} e^{- \frac{i π}{4}} \Rightarrow$ $A_{n} = {(\sqrt{2})}^{n} e^{\frac{i n π}{4}} + {(\sqrt{2})}^{n} e^{- \frac{i n π}{4}}$ $= {(\sqrt{2})}^{n} 2 R e (e^{\frac{i n π}{4}}) = 2 {(\sqrt{2})}^{n} c o s (\frac{n π}{4})$ $B_{n} = {(\sqrt{2})}^{n} e^{\frac{i n π}{4}} - {(\sqrt{2})}^{n} e^{- \frac{i n π}{4}} = 2 i {(\sqrt{2})}^{n} I m (e^{\frac{i n π}{4}})$ $= 2 i {(\sqrt{2})}^{n} s i n (\frac{n π}{4})$
Commented by math khazana by abdo last updated on 12/Aug/18
$4) we have S_n =Σ_(k=0) ^(n−1) z_1 ^k +Σ_(k=0) ^(n−1) z_2 ^k but Σ_(k=0) ^(n−1) z_1 ^(k ) =((1−z_1 ^n )/(1−z_1 )) = ((1− ((√2))^n e^((inπ)/4) )/(1−(1+i))) =i( 1−((√2))^n e^((inπ)/4) ) Σ_(k=0) ^(n−1) z_2 ^k =((1−z_2 ^n )/(1−z_2 )) =((1−((√2))^n e^(−((iπ)/4)) )/(1−(1−i))) =−i(1−((√2))^n e^(−((iπ)/4)) ) ⇒ S_n =i −i((√2))^n e^((inπ)/4) −i +i((√2))^n e^(−((inπ)/4)) =−i((√2))^n { e^((inπ)/4) − e^(−((inπ)/4)) } =−i((√2))^n (2isin(((nπ)/4))) = 2 ((√2))^n sin(((nπ)/4)) .$
$4) w e h a v e S_{n} = \sum_{k = 0}^{n - 1} z_{1}^{k} + \sum_{k = 0}^{n - 1} z_{2}^{k} b u t$ $\sum_{k = 0}^{n - 1} z_{1}^{k} = \frac{1 - z_{1}^{n}}{1 - z_{1}} = \frac{1 - {(\sqrt{2})}^{n} e^{\frac{i n π}{4}}}{1 - (1 + i)}$ $= i (1 - {(\sqrt{2})}^{n} e^{\frac{i n π}{4}})$ $\sum_{k = 0}^{n - 1} z_{2}^{k} = \frac{1 - z_{2}^{n}}{1 - z_{2}} = \frac{1 - {(\sqrt{2})}^{n} e^{- \frac{i π}{4}}}{1 - (1 - i)}$ $= - i (1 - {(\sqrt{2})}^{n} e^{- \frac{i π}{4}}) \Rightarrow$ $S_{n} = i - i {(\sqrt{2})}^{n} e^{\frac{i n π}{4}} - i + i {(\sqrt{2})}^{n} e^{- \frac{i n π}{4}}$ $= - i {(\sqrt{2})}^{n} {e^{\frac{i n π}{4}} - e^{- \frac{i n π}{4}}}$ $= - i {(\sqrt{2})}^{n} (2 i s i n (\frac{n π}{4})) = 2 {(\sqrt{2})}^{n} s i n (\frac{n π}{4}) .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Aug/18

	$z_{1} = \sqrt{2} e^{i \frac{Π}{4}} z_{2} = \sqrt{2} e^{- i \frac{Π}{4}}$ $\frac{1}{z_{1}^{4}} + \frac{1}{z_{2}^{4}} = \frac{1}{2^{2}} (\frac{1}{e^{i Π}} + \frac{1}{e^{- i Π}}) = \frac{1}{4} (\frac{1}{- 1} + \frac{1}{- 1}) = \frac{- 1}{2}$ $A_{n} = {(\sqrt{2} e^{i \frac{Π}{4}})}^{n} + {(\sqrt{2} e^{- i \frac{Π}{4}})}^{n} = 2^{\frac{n}{2}} (2 c o s \frac{n Π}{4})$ $= 2^{\frac{n}{2} + 1} (c o s \frac{n Π}{4})$ $B_{n} = {(\sqrt{2} e^{i \frac{Π}{4}})}^{n} - {(\sqrt{2} e^{- i \frac{Π}{4}})}^{n}$ $= 2^{\frac{n}{2}} (e^{i \frac{n Π}{4} -}) - 2^{\frac{n}{2}} (e^{- i \frac{n Π}{4}})$ $= 2^{\frac{n}{2}} (2 i s i n \frac{n Π}{4})$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Aug/18

	$x^{2} - 2 x + 2 = 0$ ${(x - 1)}^{2} + 1 = 0$ $x - 1 = \pm i$ $x = 1 \pm i$ $z_{1} = 1 + i z_{2} = 1 - i$ $z_{1}^{3} + z_{2}^{3} = {(z_{1} + z_{2})}^{3} - 3 z_{1} z_{2} (z_{1} + z_{2})$ $= 8 - 3 (1 + 1) (2) = - 4$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Aug/18

	$\frac{1}{z_{1}^{3}} + \frac{1}{z_{2}^{3}} = \frac{z_{1}^{3} + z_{2}^{3}}{{(z_{1} z_{2})}^{3}} = \frac{{(z_{1} + z_{2})}^{3} - 3 z_{1} z_{2} (z_{1} + z_{2})}{{(z_{} z_{2})}^{3}} \frac{}{}$ $= \frac{2^{3} - 3.2 (2)}{2^{3}} = \frac{- 4}{8} = \frac{- 1}{2}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Aug/18

	$2) z_{1} = 1 + i z_{2} = 1 - i$ $z_{1} = \sqrt{2} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i) = \sqrt{2} (c o s \frac{Π}{4} + i s i n \frac{Π}{4}) = \sqrt{2} e^{i \frac{Π}{4}}$ $z_{2} = \sqrt{2} e^{- i \frac{Π}{4}}$ $s o z_{1}^{4} + z_{2}^{4} = 2^{2} (e^{i Π} + e^{- i Π}) = 4 (2 c o s Π) = - 8$
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