let Z = cos(((2π)/7)) +isin(((2π)/7)) and A= Z+Z^2 +Z^4 B=Z^3 +Z^5 +Z^6 1) prove that A^− =B 2) prove that A+B =−1 and A.B =2 3) find A and B.


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Question Number 41520 by maxmathsup by imad last updated on 08/Aug/18

$l e t Z = c o s (\frac{2 π}{7}) + i s i n (\frac{2 π}{7}) a n d A = Z + Z^{2} + Z^{4}$ $B = Z^{3} + Z^{5} + Z^{6}$ $1) p r o v e t h a t \bar{A} = B$ $2) p r o v e t h a t A + B = - 1 a n d A . B = 2$ $3) f i n d A a n d B .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 09/Aug/18

	$A + B = z + z^{2} + z^{3} + z^{4} + z^{5} + z^{6}$ $= \frac{z (z^{6} - 1)}{z - 1} = \frac{z^{7} - z}{z - 1} = \frac{c o s 2 Π + i s i n 2 Π - z}{z - 1}$ $= \frac{1 - z}{z - 1} = - 1$
	Answered by tanmay.chaudhury50@gmail.com last updated on 09/Aug/18
	$let t=((2Π)/7) z=e^(it) A=e^(it) +e^(i2t) +e^(i4t) B=e^(i3t) +e^(i5t) +e^(i6t) A.B=e^(it) (1+e^(it) +e^(i3t) ).e^(i3t) (1+e^(i2t) +e^(i3t) ) =e^(i4t) (1+e^(i2t) +e^(i3t) +e^(it) +e^(i3t) +e^(i4t) +e^(i3t) +e^(i5t) +e^(i6t) ) =e^(i4t) (1+e^(it) +e^(i2t) +3e^(i3t) +e^(i4t) +e^(i5t) +e^(i6t) ) =e^(i4t) {(((1−e^(i7t) )/(1−e^(it) ))+2e^(i3t) )} e^(i7t) =(e^(it) )^7 =(e^(i×((2Π)/7)) )^7 =cos2Π+isin2Π=1 =e^(14t) {(((1−1)/(1−e^(((2Π)/7)i) ))+2e^(i3t) } =e^(i4t) 2.e^(i3t) =2e^(i7t) =2(cos2Π+isin2Π)=2×1=2$
	$l e t t = \frac{2 Π}{7}$ $z = e^{i t}$ $A = e^{i t} + e^{i 2 t} + e^{i 4 t} B = e^{i 3 t} + e^{i 5 t} + e^{i 6 t}$ $A . B = e^{i t} (1 + e^{i t} + e^{i 3 t}) . e^{i 3 t} (1 + e^{i 2 t} + e^{i 3 t})$ $= e^{i 4 t} (1 + e^{i 2 t} + e^{i 3 t} + e^{i t} + e^{i 3 t} + e^{i 4 t} + e^{i 3 t} + e^{i 5 t} + e^{i 6 t})$ $= e^{i 4 t} (1 + e^{i t} + e^{i 2 t} + 3 e^{i 3 t} + e^{i 4 t} + e^{i 5 t} + e^{i 6 t})$ $= e^{i 4 t} {(\frac{1 - e^{i 7 t}}{1 - e^{i t}} + 2 e^{i 3 t})}$ $e^{i 7 t} = {(e^{i t})}^{7} = {(e^{i \times \frac{2 Π}{7}})}^{7} = c o s 2 Π + i s i n 2 Π = 1$ $= e^{14 t} {(\frac{1 - 1}{1 - e^{\frac{2 Π}{7} i}} + 2 e^{i 3 t}}$ $= e^{i 4 t} 2 . e^{i 3 t} = 2 e^{i 7 t} = 2 (c o s 2 Π + i s i n 2 Π) = 2 \times 1 = 2$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Aug/18

	$A = e^{i t} + e^{i 2 t} + e^{i 4 t} t = \frac{2 Π}{7} 7 t = 2 Π$ $B = z^{3} + z^{5} + z^{6}$ $= e^{i 3 t} + e^{i 5 t} + e^{i 6 t}$ $= e^{i (7 t - 4 t)} + e^{i (7 t - 2 t)} + e^{i (7 t - t)}$ $= e^{i 2 Π} . e^{- i 4 t} + e^{i 2 Π} . e^{- i 2 t} + e^{i 2 Π} . e^{- i t}$ $e^{i 2 Π} = c o s 2 Π + i s i n 2 Π = 1$ $= e^{- i 4 t} + e^{- i 2 t} + e^{- i t}$ $= \bar{A}$
	Commented by rahul 19 last updated on 10/Aug/18

	$nice!$
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