# 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

 $\displaystyle z^{10} - 1$ $=$ $\displaystyle \paren {z - 1} \paren {z + 1} \displaystyle \prod_{k \mathop = 1}^4 \paren {z^2 - 2 z \cos \dfrac {k \pi} 5 + 1}$ Factors of Difference of Two Even Powers $\displaystyle \leadsto \ \$ $\displaystyle \dfrac {z^{10} - 1} {z^2 - 1}$ $=$ $\displaystyle \paren {z^2 - 2 z \cos \dfrac \pi 5 + 1} \paren {z^2 - 2 z \cos \dfrac {2 \pi} 5 + 1} \paren {z^2 - 2 z \cos \dfrac {3 \pi} 5 + 1} \paren {z^2 - 2 z \cos \dfrac {4 \pi} 5 + 1}$

Then we have:

 $\displaystyle \paren {z^2 - 2 z \cos \dfrac \pi 5 + 1} \paren {z^2 - 2 z \cos \dfrac {4 \pi} 5 + 1}$ $=$ $\displaystyle z^4 - 2 z \paren {z^2 + 1} \paren {\cos \dfrac \pi 5 + \cos \dfrac {4 \pi} 5} + z^2 \paren {2 + 4 \cos \dfrac \pi 5 \cos \dfrac {4 \pi} 5} + 1$ $\displaystyle$ $=$ $\displaystyle z^4 + z^2 \paren {2 + 4 \cos \dfrac \pi 5 \cos \dfrac {4 \pi} 5} + 1$ as $\cos \dfrac \pi 5 + \cos \dfrac {4 \pi} 5 = 0$ $\displaystyle$ $=$ $\displaystyle z^4 + z^2 \paren {2 + 4 \paren {\dfrac {\map \cos {\dfrac {4 \pi} 5 - \dfrac \pi 5} + \map \cos {\dfrac {4 \pi} 5 + \dfrac \pi 5} } 2} } + 1$ Simpson's Formula for Cosine by Cosine $\displaystyle$ $=$ $\displaystyle z^4 + z^2 \paren {2 + 2 \paren {\cos \dfrac {3 \pi} 5 + \cos \pi} } + 1$ simplifying $\displaystyle$ $=$ $\displaystyle z^4 + z^2 \paren {2 + 2 \paren {\cos \dfrac {3 \pi} 5 - 1} } + 1$ Cosine of Full Angle $\displaystyle$ $=$ $\displaystyle z^4 + 2 z^2 \cos \dfrac {3 \pi} 5 + 1$ simplifying

and similarly:

 $\displaystyle \paren {z^2 - 2 z \cos \dfrac {2 \pi} 5 + 1} \paren {z^2 - 2 z \cos \dfrac {3 \pi} 5 + 1}$ $=$ $\displaystyle z^4 - 2 z \paren {z^2 + 1} \paren {\cos \dfrac {2 \pi} 5 + \cos \dfrac {3 \pi} 5} + z^2 \paren {2 + 4 \cos \dfrac {2 \pi} 5 \cos \dfrac {3 \pi} 5} + 1$ $\displaystyle$ $=$ $\displaystyle z^4 + z^2 \paren {2 + 4 \cos \dfrac {2 \pi} 5 \cos \dfrac {3 \pi} 5} + 1$ as $\cos \dfrac {2 \pi} 5 + \cos \dfrac {3 \pi} 5 = 0$ $\displaystyle$ $=$ $\displaystyle z^4 + z^2 \paren {2 + 4 \paren {\dfrac {\map \cos {\dfrac {3 \pi} 5 - \dfrac {2 \pi} 5} + \map \cos {\dfrac {3 \pi} 5 + \dfrac {2 \pi} 5} } 2} } + 1$ Simpson's Formula for Cosine by Cosine $\displaystyle$ $=$ $\displaystyle z^4 + z^2 \paren {2 + 2 \paren {\cos \dfrac \pi 5 + \cos \pi} } + 1$ simplifying $\displaystyle$ $=$ $\displaystyle z^4 + z^2 \paren {2 + 2 \paren {\cos \dfrac \pi 5 - 1} } + 1$ Cosine of Full Angle $\displaystyle$ $=$ $\displaystyle z^4 + 2 z^2 \cos \dfrac \pi 5 + 1$ simplifying

Hence:

 $\displaystyle \dfrac {z^{10} - 1} {z^2 - 1}$ $=$ $\displaystyle \paren {z^4 + 2 z^2 \cos \dfrac {3 \pi} 5 + 1} \paren {z^4 + 2 z^2 \cos \dfrac \pi 5 + 1}$

 $\displaystyle z^{10} - 1$ $=$ $\displaystyle \paren {z^5 + 1} \paren {z^5 - 1}$ Difference of Two Squares $\displaystyle \leadsto \ \$ $\displaystyle \dfrac {z^{10} - 1} {z^2 - 1}$ $=$ $\displaystyle \paren {\dfrac {z^5 + 1} {z + 1} } \paren {\dfrac {z^5 - 1} {z - 1} }$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac {z^{10} - 1} {z^2 - 1}$ $=$ $\displaystyle \paren {z^4 - z^3 + z^2 - z + 1} \paren {z^4 + z^3 + z^2 + z + 1}$

which doesn't seem to help much.

What about binomial theorem on $\paren {z - 1}^{10}$ or $\paren {z + 1}^{10}$? See how that goes:

 $\displaystyle \paren {z + 1}^{10}$ $=$ $\displaystyle \sum_{k \mathop = 0}^{10} \binom {10} k z^k$ Binomial Theorem $\displaystyle \paren {z - 1}^{10}$ $=$ $\displaystyle \sum_{k \mathop = 0}^{10} \paren {-1}^k \binom {10} k z^k$ Binomial Theorem

Or we could use the Quintuple Angle Formula for Cosine:

$-1 = \cos \pi = 16 \cos^5 \dfrac \pi 5 - 20 \cos^3 \dfrac \pi 5 + 5 \cos \dfrac \pi 5$

Let $z = \cos \dfrac \pi 5$, which leads to:

$16 z^5 - 20 z^3 + 5 z + 1 = 0$

Next thing: try to factor by $z^4 - 3z^2 + 1$ -- but the day job calls.