<s>[INST] Find modulus of $z$ given modulus of $(z-3w)/(3-z\overline{w})$ Question:

(23) If $z_1$, $z_2$ are complex numbers such that $\left|\dfrac{z_1-3z_2}{3-z_1\overline{z}_2}\right|=1$ and $|z_2|\neq 1$, then find $|z_1|$.

How would I attempt this question? I tried using values for $z_1$ and $\overline{z}_2$ but it is coming out to be extremely lengthy. There is probably a quicker solution to this. Can someone provide me with a hint to start?
 [/INST] $$\frac{z_1 - 3z_2}{3 - z_1 \overline{z_2}} = \frac{\frac{z1}{3} - z_2}{1 - \frac{z1}{3} \overline{z_2}}$$
$b = \overline{z_1} / 3, a = z_2 \Rightarrow |a - \overline{b}| = |1 - ab|$ and we should find $b$.
Let's look this problem geometrically and consider complex numbers $a, b$ as two-dimensional vectors (common approach). 
Main idea is that ($Q \angle W$ denotes angle between $Q$ and $W$)
$$a \angle \overline{b} = 1 \angle ab$$
Explanation: Let $\arg(a) = \alpha, \arg(b) = \beta$. Then $\arg(ab) = \alpha + \beta, \arg(1) = 0, \arg(\bar{b}) = -\beta$. Hence $a \angle \overline{b} = \alpha + \beta = 1 \angle ab$.
Let $A = |a|, B = |b|$. Having angles equation, we use law of cosines for corresponding triangles and get (|ab| = AB, |1| = 1)
$$(AB) ^ 2 + 1^2 - 2 AB \cos(\alpha + \beta) = A^2 + B^2 - 2 AB \cos (\alpha + \beta)$$
$$A^2 B^2 + 1 = A^2 + B^2$$
$$B^2 (A^2 - 1) = A^2 - 1$$
Given $1 \neq |z_2| = |a| = |A|$, 
$$B^2 = 1$$
Therefore, $B = 1$ because it is be positive.
But $1 = B = |\overline{z_1} / 3| = |z_1| / 3$, so answer is 3.
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