# 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

 $\displaystyle z^4 - 3z^2 + 1$ $=$ $\displaystyle z^4 - 2z^2 + 1 - z^2$ Separating the $z^2$ term $\displaystyle$ $=$ $\displaystyle (z^2 - 1)^2 - z^2$ Completing the Square $\displaystyle$ $=$ $\displaystyle (z^2 - 1 - z)(z^2 - 1 + z)$ Difference of Two Squares

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 $0$:

 $\displaystyle 0$ $=$ $\displaystyle \cos 0 + \cos \dfrac {2 \pi} 5 + \cos \dfrac {4 \pi} 5 + \cos \dfrac {6 \pi} 5 + \cos \dfrac {8 \pi} 5$ $\displaystyle$ $=$ $\displaystyle 1 + \cos \dfrac {2 \pi} 5 + \cos \dfrac {4 \pi} 5 + \cos \dfrac {6 \pi} 5 + \cos \dfrac {8 \pi} 5$

Now:

 $\displaystyle \cos(\pi + x)$ $=$ $\displaystyle -\cos(x)$ Cosine of Angle plus Straight Angle $\displaystyle$ $=$ $\displaystyle \cos(-x)$ Cosine Function is Even $\displaystyle$ $=$ $\displaystyle -\cos(\pi - x)$ Cosine of Angle plus Straight Angle $\displaystyle$ $=$ $\displaystyle \cos(\pi - x)$ Cosine Function is Even

Hence:

 $\displaystyle \cos \dfrac {6 \pi} 5$ $=$ $\displaystyle \map {\cos} {\pi + \dfrac \pi 5}$ $\displaystyle$ $=$ $\displaystyle \map {\cos} {\pi - \dfrac \pi 5}$ $\displaystyle$ $=$ $\displaystyle \cos \dfrac {4 \pi} 5$

and

 $\displaystyle \cos \dfrac {8 \pi} 5$ $=$ $\displaystyle \map {\cos} {\pi + \dfrac {3 \pi} 5}$ $\displaystyle$ $=$ $\displaystyle \map {\cos} {\pi - \dfrac {3 \pi} 5}$ $\displaystyle$ $=$ $\displaystyle \cos \dfrac {2 \pi} 5$

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

 $\displaystyle 0$ $=$ $\displaystyle 1 + \cos \dfrac {2 \pi} 5 + \cos \dfrac {4 \pi} 5 + \cos \dfrac {6 \pi} 5 + \cos \dfrac {8 \pi} 5$ $\displaystyle$ $=$ $\displaystyle 1 + 2 \cos \dfrac {2 \pi} 5 + 2 \cos \dfrac {4 \pi} 5$ $\cos(\pi + x) = \cos(\pi - x)$ $\displaystyle$ $=$ $\displaystyle 1 + 2 \cos \dfrac {2 \pi} 5 + 4 \cos^2 \dfrac {2 \pi} 5 - 2$ Double Angle Formula for Cosine, $\cos 2 \theta = 2 \cos^2 \theta - 1$ $\displaystyle$ $=$ $\displaystyle 4 \cos^2 \dfrac {2 \pi} 5 + 2 \cos \dfrac {2 \pi} 5 - 1$ simplifying

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:

 $\displaystyle 2 \cos 36 \degrees$ $=$ $\displaystyle 2 \cos \dfrac \pi 5$ $\displaystyle$ $=$ $\displaystyle 2 \map {\cos} {\dfrac {- \pi} 5}$ Cosine Function is Even $\displaystyle$ $=$ $\displaystyle - 2 \map {\cos} {\pi - \dfrac \pi 5}$ Cosine of Angle plus Straight Angle $\displaystyle$ $=$ $\displaystyle - 2 \map {\cos} {\dfrac {4 \pi} 5}$ simplifying $\displaystyle$ $=$ $\displaystyle 1 + 2 \cos \dfrac {2 \pi} 5$ Sum of the real parts, $1 + 2 \cos \dfrac {2 \pi} 5 + 2 \cos \dfrac {4 \pi} 5 = 0$ $\displaystyle$ $=$ $\displaystyle 1 + \dfrac {-1 + \sqrt 5} 2$ $\displaystyle$ $=$ $\displaystyle \dfrac {1 + \sqrt 5} 2$

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

$\blacksquare$