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
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 $0$:

Now:

Hence:

and

We can now simplify the sum of the real parts of the roots of $x^5 - 1 = 0$:

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:

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 $2 \cos 36 \degrees, 2 \cos 72 \degrees, 2 \cos 216 \degrees, 2 \cos 252 \degrees$