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

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?

Quanto · 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
$$left(frac{z_1}{z_2}right)_{1,2}=(sqrt5+1)e^{ifracpi2},> (sqrt5-1)e^{ifrac{3pi}2}$$
Thus, the two are right triangles with $|z_1|>|z_2|$ and
$$cot A + cot B= left|left(frac {z_2}{z_1}right)_1right|+ left|left(frac {z_2}{z_1}right)_2right|= 
{sqrt5+1}+ {sqrt5-1}=2{sqrt5}
$$

Bru · Answer

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.

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 $cot A +cot B$

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