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Unformatted text preview: We apply the Mathematica command Solve[(z+1)^5==z^5,z] to ﬁnd the four solutions of the equation to be z 1 =1 2i 10 q 5(52 √ 5) z 2 =1 2 + i 10 q 5(52 √ 5) z 3 =1 2i 10 q 5(5 + 2 √ 5) z 4 =1 2 + i 10 q 5(5 + 2 √ 5) which veriﬁes our geometric argument. 1 We also note in the absence of a symbolic manipulator, we may solve the above equation in the following manner where we apply the identity sin( θ ) = e iθeiθ 2 i We start by dividing both sides by z 5 ( z + 1) 5 = z 5 ⇐⇒ (1 + z1 ) 5 = 1 ⇐⇒ 1 + z1 = 1 1 / 5 ⇐⇒ 1 + z1 = e i 2 πk 5 Solving for z , we ﬁnd z = 1 e i 2 πk 51 = 1 e i πk 5 ( e i πk 5ei πk 5 ) = ei πk 5 2 i sin( πk 5 ) = cos( πk 5 )i sin( πk 5 ) 2 i sin( πk 5 ) which reduces to z =1 2i 2 cot ± 2 πk 5 ² Letting k run from 1 to 4 we obtain the above roots. ± 2...
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 Three '09
 Cong
 Math, Geometry, Degree of a polynomial, Konrad Zuse, Stephen Taylor

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