Proof | Heptagon Construction by Neusis let \(f(x)\) be \(7^{th}\) roots of unity $$f^x=f(x)$$ for all \(x\in\mathbb{Z}\), \(f(x)\) forms a regular heptagon on unit circle in complex plane and satisfies $$f(x)^7=1$$ which factors into $$\Big(f(x) - 1\Big)\Big(f(x)^6 + f(x)^5 + f(x)^4 + f(x)^3 + f(x)^2 + f(x)^1 + 1\Big)=0$$ which we can obtain from (where \(f(x)\neq1\)) $$f(6x) + f(5x) + f(4x) + f(3x) + f(2x) + f(x) = -1$$ let $$p(x)=f(x) + f(2x) + f(4x)$$ we can find that $$\begin{array}{llll}p(x)&=p(2x)&=p(4x)\\ \\p(3x)&=p(6x)&=p(5x)\\ \\p(7x)&=3\end{array}$$ $$\begin{array}{ll}\Big(p(x)\Big)^*&=f(x)^*+f(2x)^*+f(4x)^*\\ \\&=f(6x)+f(5x)+f(3x)\\ \\&=p(3x)\end{array}$$ $$p(x) + p(3x) = -1$$ which shows that \(p(x), p(3x)\) lies on line \(z+z^*=-1\), and $$\begin{array}{llllllll}p(x)\cdot p(3x) &=& &f(3x) &f(6x) &(5x)&=p(x)+p(3x)+3&=2\\ \\ &&f(x)&f(4x)&1&f(6x)\\ \\ &&f(2x)&f(5x)&f(x...
Proof | Hendecagon construction by Neusis for any complex number \(z\) : $$z + z^*\in\mathbb{R}$$ $$z z^*\in\mathbb{R}^{+}$$ $$\dfrac{\sqrt{\overset{\phantom{0}}{zz^*}}}{z^*}\in\mathbb{S}^{1}$$ $$z - z^*\in\Big\{ix\mid x\in\mathbb{R}\Big\}$$ let \(f(x)\) be \(11^{th}\) roots of unity $$f^x=f(x)$$ for all \(x\in\mathbb{Z}\), \(f(x)\) forms a regular heptagon on unit circle in complex plane and satisfies $$f(x)^{11}=1$$ which factors into $$\Big(f(x) - 1\Big)\Big(f(x)^{10} + f(x)^9 + \cdots + f(x)^2 + f(x)^1 + 1\Big)=0$$ which we can obtain from (where \(f(x)\neq1\) $$f(10x) + f(9x) + f(8x) + f(7x) + f(6x) + f(5x) + f(4x) + f(3x) + f(2x) + f(x) = -1$$ let $$p(x)=f(x) + f(4x) + f(5x) + f(9x)+f(3x)$$ we have $$\begin{array}{lllll}p(x)&=p(4x)&=p(5x)&=p(9x)&=p(3x)\\ \\p(2x)&=p(8x)&=p(10x)&=p(7x)&=p(6x)\\ \\p(11x)&=5\end{array}$$ $$\Big(p(x)\Big)^*=p(2x)$$ $$p(x) + p(2x) = -1$$ which shows that \(p(x),p(2x)\) lies on line \(z+z^*=-1\) ...
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