Let z1 and z2 be any two non-zero complex numbers such that 3|z1| = 4|z2|. If z = (3z1)/(2z2) + (2z2)/(3z1), then?

Let z1 and z2 be two non-zero complex numbers such that 3|z1| = 4|z2|.
Let z = (3z1)/(2z2) + (2z2)/(3z1).
We have
z = (3z1)/(2z2) + (2z2)/(3z1) = (9|z1|^2 + 4|z2|^2)/(6z1z2)
Let w = (3z1)/(2z2). Then z = w + 1/w
|w| = (3|z1|)/(2|z2|) = (3/2)(4/3) = 2
Let w = 2e^(iθ). Then 1/w = (1/2)e^(-iθ)
z = w + 1/w = 2e^(iθ) + (1/2)e^(-iθ)
= 2(cosθ + isinθ) + (1/2)(cosθ - isinθ)
= (5/2)cosθ + (3/2)isinθ
|z|^2 = (25/4)cos^2θ + (9/4)sin^2θ
If θ = 0, then z = 5/2 and |z| = 5/2
If θ = π/2, then z = (3/2)i and |z| = 3/2
|z|^2 = (25/4)cos^2θ + (9/4)sin^2θ = (25/4)cos^2θ + (9/4)(1 - cos^2θ) = (16/4)cos^2θ + 9/4 = 4cos^2θ + 9/4
Let x = (3z1)/(2z2). Then 1/x = (2z2)/(3z1)
z = x + 1/x
|x| = (3|z1|)/(2|z2|) = (3/2)(4/3) = 2
Let x = a + ib
|x|^2 = a^2 + b^2 = 4
Then z = x + 1/x = (x^2 + 1)/x
|z| = |x^2 + 1|/|x| = |x^2 + 1|/2
|x^2 + 1|^2 = (a^2 + 1 - b^2)^2 + (2ab)^2
= (a^2 - b^2 + 1)^2 + 4a^2b^2
Since |x| = 2, x = 2(cosθ + isinθ)
x^2 = 4(cos2θ + isin2θ)
x^2 + 1 = 4cos2θ + 1 + i(4sin2θ)
|x^2 + 1|^2 = (4cos2θ + 1)^2 + 16sin^22θ = 16cos^22θ + 8cos2θ + 1 + 16sin^22θ = 17 + 8cos2θ
|z|^2 = (17 + 8cos2θ)/4
If cos2θ = 1, |z|^2 = 25/4, |z| = 5/2
If cos2θ = -1, |z|^2 = 9/4, |z| = 3/2
Let u = 3z1/(2z2). Then |u| = 2. z = u + 1/u. Then |z|^2 = u^2 + 1/u^2 + 2 = |u|^2 + 1/|u|^2 = 4 + 1/4 = 17/4. Thus |z| = √17/2. This is incorrect.
Let's use the fact that 3|z1| = 4|z2|. Then |3z1/(2z2)| = 2. Let w = 3z1/(2z2). Then |w| = 2. z = w + 1/w. |z|^2 = w^2 + 1/w^2 + 2 = (w + 1/w)^2 - 2 = |z|^2. |z|^2 = (w + 1/w)(w* + 1/w*) = 17/4. |z| = sqrt(17)/2. This is wrong again.
Consider z = (3z1)/(2z2) + (2z2)/(3z1) = (9|z1|^2 + 4|z2|^2)/(6z1z2)
|z|^2 = (9|z1|^2 + 4|z2|^2)^2/(36|z1|^2|z2|^2) = (9(16/9)|z2|^2 + 4|z2|^2)^2/(36(16/9)|z2|^4) = (20|z2|^2)^2/(64|z2|^4) = 400/64 = 25/4
|z| = 5/2