Empma u3 heptagono regular

𝑧7 = 1 2𝜋 2𝜋/7

𝑧7 = 1

𝑧 7 − 1 = 0.

𝑧 7 − 1 = (𝑧 − 1)(𝑧 6 + 𝑧 5 + 𝑧 4 + 𝑧 3 + 𝑧 2 + 𝑧 + 1) = 0

𝑧=1 𝑧6 + 𝑧5 + 𝑧4 + 𝑧3 + 𝑧2 + 𝑧 + 1 = 0 𝜃 = 2𝜋/7

2𝜋

𝑧𝑘 = 𝑒 𝑘 7 𝑖

𝑝𝑎𝑟𝑎

2𝜋

𝑧𝑘 = 𝑒 𝑘 7 𝑖 = cos (𝑘

𝑘 = 0, 1, … ,6

2𝜋 )+ 7

𝑖 ∙ 𝑠𝑒𝑛 (𝑘

2𝜋 ) 7

𝑘=1

2𝜋 2𝜋 2𝜋 𝑧1 = 𝑒 7 𝑖 = cos ( ) + 𝑖 ∙ 𝑠𝑒𝑛 ( ) 7 7

2𝜋

cos ( 7 )

𝑧1

2𝜋

𝑧̅1 = 𝑒 −𝑖𝑘 7 = cos (

2𝜋 2𝜋 ) − 𝑖 ∙ 𝑠𝑒𝑛 ( ) 7 7

𝑧6 + 𝑧5 + 𝑧4 + 𝑧3 + 𝑧2 + 𝑧 + 1 = 0

2𝜋

2𝜋

𝑧1 + 𝑧̅1 = 𝑒 𝑖𝑘 7 + 𝑒 −𝑖𝑘 7 = 𝑧1 +

1 𝑧1

2𝜋 7

= 2 cos ( )

𝑧6 + 𝑧5 + 𝑧4 + 𝑧3 + 𝑧2 + 𝑧 + 1 = 0

𝑧3

1 𝑧

𝑧3 + 𝑧2 + 𝑧 + 1 + +

𝑧3 +

1 𝑧2

+

1 𝑧3

=0

1 1 1 + 𝑧2 + 2 + 𝑧 + +1=0 3 𝑧 𝑧 𝑧

1 3 3 1 2 1 (𝑧 + ) − 3𝑧 − + (𝑧 + ) − 2 + ( 𝑧 + ) + 1 = 𝑧 𝑧 𝑧 𝑧 1 3 1 1 2 1 (𝑧 + ) − 3 (𝑧 + ) + (𝑧 + ) + ( 𝑧 + ) − 1 = 0 𝑧 𝑧 𝑧 𝑧 1 3

1 2

1

(𝑧 + 𝑧) − 2 (𝑧 + 𝑧) + (𝑧 + 𝑧) − 1 = 0

1

𝑥 = 𝑧+𝑧

𝑥 3 + 𝑥 2 − 2𝑥 − 1 = 0

2𝜋

2 cos ( 7 )

2𝜋 7

2cos( ) 2𝜋 7

cos( )

𝜃=

2𝜋 9