Question: If $z_1$ and $z_2$ are two complex numbers satisfying
$\frac{z_1}{2z_2}+\frac{2z_2}{z_1} = i$
and $0, z_1, z_2$ form two non-similar triangles. $A, B$ are the least angles in the two triangles, then $\cot A +\cot B$ equals:
I tried solving the first equation by trying to complete the square but to no avail. Then I tried taking $\frac{z_1}{z_2}$ as another variable $z$, hoping to use the rotation method but I couldn't figure out what to do with it. I think that I'm missing something, but even if I calculate $z_1$ and $z_2$ then would it not be insufficient to find a condition for minimum values of the cotangents of the angles?
Best Answer
Write the equation $\frac{z_1}{2z_2}+\frac{2z_2}{z_1} = i$ as $(\frac{z_1}{2z_2})^2 -i \frac{z_1}{2z_2} +1 =0 $ and solve to obtain
$$\left(\frac{z_1}{z_2}\right)_{1,2}=(\sqrt5+1)e^{i\frac\pi2},\> (\sqrt5-1)e^{i\frac{3\pi}2}$$
Thus, the two are right triangles with $|z_1|>|z_2|$ and
$$\cot A + \cot B= \left|\left(\frac {z_2}{z_1}\right)_1\right|+ \left|\left(\frac {z_2}{z_1}\right)_2\right|= {\sqrt5+1}+ {\sqrt5-1}=2{\sqrt5} $$