# Question #f312d

Apr 21, 2017

$\frac{4}{1 - \sqrt{5}} + \frac{4}{1 + \sqrt{5}} = - 2$

#### Explanation:

Given ${z}_{1}$ we can calculate ${z}_{2}$ according to

$4 {z}_{2}^{2} + {z}_{1}^{2} - 2 i {z}_{1} {z}_{2} = 0$ or

${z}_{2} = \frac{i}{4} \left(1 \pm \sqrt{5}\right) {z}_{1}$

which means that given ${z}_{1}$ to obtain ${z}_{2}$ we need two transformations
1) Scaling by $\frac{1}{4} \left(1 \pm \sqrt{5}\right)$
2) Rotating $\frac{\pi}{2}$ counterclockwise as associated to the product by $i$

so with ${z}_{0} = 0$ we can build two triangles

$\left[{z}_{0} , {z}_{1} , {z}_{2}^{a}\right]$ and $\left[{z}_{0} , {z}_{1} , {z}_{2}^{b}\right]$

with

${z}_{2}^{a} = \frac{i}{4} \left(1 + \sqrt{5}\right) {z}_{1}$
${z}_{2}^{b} = \frac{i}{4} \left(1 - \sqrt{5}\right) {z}_{1}$

so ${z}_{2}^{a} , {z}_{0} , {z}_{2}^{b}$ are aligned and

$\left[{z}_{2}^{a} , {z}_{2}^{b}\right]$ is perpendicular to $\left[{z}_{0} , {z}_{1}\right]$

and given ${\left({z}_{1}\right)}_{k}$ all triangles ${\left[{z}_{2}^{a} , {z}_{1} , {z}_{2}^{b}\right]}_{k}$ are similar.

The least angles at $\left[{z}_{2}^{a} , {z}_{0} , {z}_{1}\right]$ and $\left[{z}_{2}^{b} , {z}_{0} , {z}_{1}\right]$ are respectively

$\alpha = \arctan \left(\frac{1 - \sqrt{5}}{4}\right) , \beta = \arctan \left(\frac{1 + \sqrt{5}}{4}\right)$

and

$\cot \left(\alpha\right) + \cot \left(\beta\right) = \frac{4}{1 - \sqrt{5}} + \frac{4}{1 + \sqrt{5}} = - 2$