Complex Algebra/Examples/z^4 - 3z^2 + 1 = 0

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

$\blacksquare$


Sources