Question #f312d

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Question #f312d

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Answer 1 · 2017-04-21T17:24:37

$\frac{4}{1 - \sqrt{5}} + \frac{4}{1 + \sqrt{5}} = - 2$ $4/(1 - sqrt[5]) + 4/(1 + sqrt[5])=-2$

Explanation:

Given $z_{1}$ $z_1$ we can calculate $z_{2}$ $z_2$ according to

$4 z_{2}^{2} + z_{1}^{2} - 2 i z_{1} z_{2} = 0$ $4z_2^2+z_1^2-2iz_1z_2=0$ or

$z_{2} = \frac{i}{4} (1 \pm \sqrt{5}) z_{1}$ $z_2 =i/4(1pm sqrt(5)) z_1$

which means that given $z_{1}$ $z_1$ to obtain $z_{2}$ $z_2$ we need two transformations
1) Scaling by $\frac{1}{4} (1 \pm \sqrt{5})$ $1/4(1pm sqrt5)$
2) Rotating $\frac{π}{2}$ $pi/2$ counterclockwise as associated to the product by $i$ $i$

so with $z_{0} = 0$ $z_0 = 0$ we can build two triangles

$[z_{0}, z_{1}, z_{2}^{a}]$ $[z_0,z_1,z_2^a]$ and $[z_{0}, z_{1}, z_{2}^{b}]$ $[z_0,z_1,z_2^b]$

with

$z_{2}^{a} = \frac{i}{4} (1 + \sqrt{5}) z_{1}$ $z_2^a=i/4(1+ sqrt(5)) z_1$
$z_{2}^{b} = \frac{i}{4} (1 - \sqrt{5}) z_{1}$ $z_2^b=i/4(1- sqrt(5)) z_1$

so $z_{2}^{a}, z_{0}, z_{2}^{b}$ $z_2^a,z_0,z_2^b$ are aligned and

$[z_{2}^{a}, z_{2}^{b}]$ $[z_2^a,z_2^b]$ is perpendicular to $[z_{0}, z_{1}]$ $[z_0,z_1]$

and given ${(z_{1})}_{k}$ $(z_1)_k$ all triangles ${[z_{2}^{a}, z_{1}, z_{2}^{b}]}_{k}$ $[z_2^a,z_1,z_2^b]_k$ are similar.

The least angles at $[z_{2}^{a}, z_{0}, z_{1}]$ $[z_2^a,z_0,z_1]$ and $[z_{2}^{b}, z_{0}, z_{1}]$ $[z_2^b,z_0,z_1]$ are respectively

$α = arctan (\frac{1 - \sqrt{5}}{4}), β = arctan (\frac{1 + \sqrt{5}}{4})$ $alpha = arctan((1-sqrt(5))/4), beta = arctan((1+sqrt(5))/4)$

and

$cot (α) + cot (β) = \frac{4}{1 - \sqrt{5}} + \frac{4}{1 + \sqrt{5}} = - 2$ $cot(alpha)+cot(beta) = 4/(1 - sqrt[5]) + 4/(1 + sqrt[5])=-2$

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Question #f312d

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