furliinotes

  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...

Notebook post  this article expresses complex conjugate of \(c\) with asterisk notation \(c^*\) Implicit | Clines the set of all real numbers \(\mathbb{R}\) forms a horizontal axis on complex plane. Its element satisfy $$r=r^* \quad \quad \dfrac{r}{r^*}=1$$ the set of all unit complex number (denoted \(\mathbb{T}\) or \(\mathbb{S}^1\) forms a unit circle on complex plane. Its elements satisfy $$s s^* = 1$$ complex number \(z\) that lies on a perpendicular bisector between \(0\) and \(1\) satisfy $$z + z^* = 1$$ complex number \(z\) that lies on orthogonal axis satisfy $$z + z^* = 0$$ complex number \(z\) that lies on a line across \(a_0\) and \(a_1\) satisfy $$(z - a_0)(a_1 - a_0)^* = (z - a_0)^*(a_1 - a_0)$$ $$z(a_1 - a_0)^* -z^*(a_1 - a_0) = a_0 a_1 ^* - a_0 ^* a_1$$ complex number \(z\) that lies on a circle centered at \(a_0\) across \(a_1\) satisfy $$(z - a_0)(z - a_0)^* = (a_1 - a_0)(a_1 - a_0)^*$$ $$zz^* - a_0 ^* z - a_0 z^* = a_1 a_1 ^* - a_1 a_0^* - a_1 ^* a_0 $$...

this page uses \(z^*\) as an expression for complex conjugate of \(z\) Mechanism | Cline If \(l_1\) and \(l_2\) are (defined as) straight lines across two same points $$\begin{array}{lllll}\text{line } l_1\ni&a_0 & ,a_1 & ,a_2 \\ \\ \text{line } l_2 \ni&a_0 & ,a_1 & ,a_3\end{array}$$ then $$l_1 = l_2$$ If \(c_1\) and \(c_2\) are (defined as) two circles across three same points $$\begin{array} {lllll} \text{circle } c_1 \ni&a_0 & ,a_1 & ,a_2 & ,a_3 \\ \\ \text{circle } c_2 \ni&a_0 & ,a_1 & ,a_2 & ,a_4 \end{array}$$ then $$c_1 = c_2$$ ________________________________________________ addition by \(z\) corresponds to translation preserve both orientation, angle, and distance multiplication by \( z \neq 0\) corresponds to rotation centered at \( 0\) and scaling centered at \( 0\), preserves angles if \( z=z^*\) then the multiplication corresponds to pure scaling centered at \( 0\) without rotation, preserving orientation if \( zz^* =...

 \(3x +1\) $$\dfrac12$$ $$\begin{array}{llll}a&aa&aaa&aaaa\\ \\ bbbb&bbb&bb&b \end{array}$$ 225225 \(\phi \varphi\)