Complex Algebra/Examples/z^4 - 3z^2 + 1 = 0
== Example of Complex Algebra ==  The roots of the equation: :$z^4 - 3z^2 + 1 = 0$ are: :$2 \cos 36 \degrees, 2 \cos 72 \degrees, 2 \cos 216 \degrees, 2 \cos 252 \degrees$  == Proof == {{begin-eqn}} {{eqn | l = z^4 - 3z^2 + 1       | r = z^4 - 2z^2 + 1 - z^2       | c = Separating the $z^2$ term }} {{eqn | r = (z^2 - 1)^2 - z^2       | c = Completing the Square }} {{eqn | r = (z^2 - 1 - z)(z^2 - 1 + z)       | c = Difference of Two Squares }} {{end-eqn}} From the Quadratic Formula applied to each of the above quadratic factors we can easily see that the four roots are: :$\dfrac {\pm 1 \pm \sqrt 5} 2$ $360 \degrees = 2 \pi \radians$, so $72 \degrees = 2 \pi / 5 \radians$ From De Moivre's Formula, the roots of $x^5 - 1 = 0$ are: :$(\cos(2 n \pi) + i \sin(2 n \pi))^{1/5}= \cos \dfrac {2 n \pi} 5 + i \sin \dfrac {2 n \pi} 5$ However, the coefficient of $x^4$ is $0$ and therefore, by Viète's Formulas, the sum of the roots of $x^5 - 1 = 0$ are also $0$ which means that the sum of the real parts of the roots are also $0$: {{begin-eqn}} {{eqn | l = 0       | r = \cos 0 + \cos \dfrac {2 \pi} 5 + \cos \dfrac {4 \pi} 5 + \cos \dfrac {6 \pi} 5 + \cos \dfrac {8 \pi} 5       | c }} {{eqn | r = 1 + \cos \dfrac {2 \pi} 5 + \cos \dfrac {4 \pi} 5 + \cos \dfrac {6 \pi} 5 + \cos \dfrac {8 \pi} 5       | c }} {{end-eqn}} Now: {{begin-eqn}} {{eqn | l = \cos(\pi + x)       | r = -\cos(x)       | c = Cosine of Angle plus Straight Angle }} {{eqn | r = \cos(-x)       | c = Cosine Function is Even }} {{eqn | r = -\cos(\pi - x)       | c = Cosine of Angle plus Straight Angle }} {{eqn | r = \cos(\pi - x)       | c = Cosine Function is Even }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \cos \dfrac {6 \pi} 5       | r = \map {\cos} {\pi + \dfrac \pi 5}       | c = }} {{eqn | r = \map {\cos} {\pi - \dfrac \pi 5}       | c = }} {{eqn | r = \cos \dfrac {4 \pi} 5       | c = }} {{end-eqn}} and {{begin-eqn}} {{eqn | l = \cos \dfrac {8 \pi} 5       | r = \map {\cos} {\pi + \dfrac {3 \pi} 5}       | c = }} {{eqn | r = \map {\cos} {\pi - \dfrac {3 \pi} 5}       | c = }} {{eqn | r = \cos \dfrac {2 \pi} 5       | c = }} {{end-eqn}} We can now simplify the sum of the real parts of the roots of $x^5 - 1 = 0$: {{begin-eqn}} {{eqn | l = 0       | r = 1 + \cos \dfrac {2 \pi} 5 + \cos \dfrac {4 \pi} 5 + \cos \dfrac {6 \pi} 5 + \cos \dfrac {8 \pi} 5       | c = }} {{eqn | r = 1 + 2 \cos \dfrac {2 \pi} 5 + 2 \cos \dfrac {4 \pi} 5       | c = $\cos(\pi + x) = \cos(\pi - x)$ }} {{eqn | r = 1 + 2 \cos \dfrac {2 \pi} 5 + 4 \cos^2 \dfrac {2 \pi} 5 - 2       | c = Double Angle Formula for Cosine, $\cos 2 \theta = 2 \cos^2 \theta - 1$ }} {{eqn | r = 4 \cos^2 \dfrac {2 \pi} 5 + 2 \cos \dfrac {2 \pi} 5 - 1       | c = simplifying }} {{end-eqn}} From the Quadratic Formula we then have two '''potential''' values: :$\cos \dfrac {2 \pi} 5 = \dfrac {-1 \pm \sqrt 5} 4$ $0 < 2 \pi / 5 < \pi / 2$, so we know that $\cos \dfrac {2 \pi} 5 > 0$, hence: :$2 \cos 72 \degrees = 2 \cos \dfrac {2 \pi} 5 = \dfrac {-1 + \sqrt 5} 2$ From Cosine of Angle plus Straight Angle, $\map \cos {x + 180 \degrees} = -\cos x$, hence: :$2 \cos 252 \degrees = \dfrac {1 - \sqrt 5} 2$ Now: {{begin-eqn}} {{eqn | l = 2 \cos 36 \degrees       | r = 2 \cos \dfrac \pi 5       | c = }} {{eqn | r = 2 \map {\cos} {\dfrac {- \pi} 5}       | c = Cosine Function is Even }} {{eqn | r = - 2 \map {\cos} {\pi - \dfrac \pi 5}       | c = Cosine of Angle plus Straight Angle }} {{eqn | r = - 2 \map {\cos} {\dfrac {4 \pi} 5}       | c = simplifying }} {{eqn | r = 1 + 2 \cos \dfrac {2 \pi} 5       | c = Sum of the real parts, $1 + 2 \cos \dfrac {2 \pi} 5 + 2 \cos \dfrac {4 \pi} 5 = 0$ }} {{eqn | r = 1 + \dfrac {-1 + \sqrt 5} 2       | c =  }} {{eqn | r = \dfrac {1 + \sqrt 5} 2       | c =  }} {{end-eqn}} From Cosine of Angle plus Straight Angle, $\map \cos {x + 180 \degrees} = -\cos x$, hence: :$2 \cos 216 \degrees = \dfrac {- 1 - \sqrt 5} 2$ We have therefore shown that the four roots of $z^4 - 3z^2 + 1 = 0$ are $\dfrac {\pm 1 \pm \sqrt 5} 2$ and that these four values are also equal to $2 \cos 36 \degrees, 2 \cos 72 \degrees, 2 \cos 216 \degrees, 2 \cos 252 \degrees$  == Sources == * {{BookReference|Theory and Problems of Complex Variables|1981|Murray R. Spiegel|ed = SI|edpage = SI (Metric) Edition|prev = Triple Angle Formula for Cosine|next = Cosine of 36 Degrees/Proof 2}}: $1$: Complex Numbers: Supplementary Problems: De Moivre's Theorem: $91$