Given $\frac{z_1}{2z_2}+\frac{2z_2}{z_1} = i$ and $0, z_1, z_2$ form two triangles with $A, B$ the least angles of each. Find $\cot A +\cot B$

Question

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?

Answer 1

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

$$\cot A + \cot B= \left|\left(\frac {z_2}{z_1}\right)_1\right|+ \left|\left(\frac {z_2}{z_1}\right)_2\right|= {\sqrt5+1}+ {\sqrt5-1}=2{\sqrt5} $$

Answer 2

Number one rule: get rid of your denominator by multiplying your equation by z1 * z2. Now, you should have just a numerator, right? Next, separate real and imaginary parts and solve the two parts as equations: f(reals) = 0 and g(imag) = 1. Got it?

Hope this helps.