What are the solutions to 1 + z + z^2 + z^3 + z^4 + z^5 = 0
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What are the solutions to 1 + z + z^2 + z^3 + z^4 + z^5 = 0

[From: ] [author: ] [Date: 11-12-15] [Hit: ]
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Where z is in the complex plane

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1 + z + z^2 + z^3 + z^4 + z^5 = 0
(1 + z) + z^2(1 + z) + z^4(1 + z) = 0
(1 + z)[1 + z^2 + z^4] = 0
z = - 1 OR
1 + z^2 + z^4 = 0
(1+z^2)^2 - z^2 = 0
(1+z^2- z)(1+z^2+ z) = 0
Solve this two quadratic you get
z = (1 ± i√3)/2
z = (- 1 ± i√3)/2
You have five roots
z = - 1
z = (1 ± i√3)/2
z = (- 1 ± i√3)/2

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Geometric series formula :

1+z+z²+z³+z^4+z^5 = (1 - z^6)/(1 - z) = 0

so we need to find the 6th-roots of 1, except for the value z=1 selff

=> 1 = cos(2pi) + i sin(2pi)
=> 1^(1/6) = cos(n*pi/3) + i sin(n*pi/3) (formula of De Moivre) (n=1->5)
so we have if we write these roots out

1/2 + i sqrt(3)/2
-1/2 + i sqrt(3) /2
-1
-1/2 - i sqrt(3)/2
1/2 - i sqrt(3)/2

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-1, +- i cuberoot of 1, +- i(1)^2/3
1
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