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?
 A: When dealing with absolute values, it is usually more convenient to look at the squares, since we have $\lvert a+b\rvert^2 = \lvert a\rvert^2 + a\overline{b} + \overline{a}b + \lvert b\rvert^2$ as a convenient expansion. Here, we compute
$$\begin{align}
\lvert 3 - z_1\overline{z_2}\rvert^2 - \lvert z_1 - 3z_2\rvert^2
&= (9 - 3 z_1\overline{z_2} - 3\overline{z_1}z_2 + \lvert z_1\overline{z_2}\rvert^2) - (\lvert z_1\rvert^2 - 3z_1\overline{z_2} - 3\overline{z_1}z_2 + 9\lvert z_2\rvert^2)\\
&= 9 + \lvert z_1\rvert^2\lvert z_2\rvert^2 - \lvert z_1\rvert^2 - 9\lvert z_2\rvert^2\\
&= (9 - \lvert z_1\rvert^2)(1-\lvert z_2\rvert^2),
\end{align}$$
and find that
$$\left\lvert \frac{z_1-3z_2}{3 - z_1\overline{z_2}}\right\rvert = 1 \iff \bigl( \lvert z_2\rvert = 1 \lor \lvert z_1\rvert = 3\bigr).$$
More, we can read off exactly when $\frac{z_1-3z_2}{3-z_1\overline{z_2}}$ is in the interior or in the exterior of the unit disk.
A: $$\left|\frac{z_1-3z_2}{3-z_1\overline{z_2}}\right|=1 \Rightarrow \left|z_1-3z_2\right|=\left|3-z_1\overline{z_2}\right| \Rightarrow \left|z_1-3z_2\right|^2=\left|3-z_1\overline{z_2}\right|^2 \Rightarrow (z_1-3z_2)(\overline{z_1-3z_2})=(3-z_1\overline{z_2})(\overline{3-z_1\overline{z_2}}) \\ \Rightarrow (z_1-3z_2)(\overline{z_1}-3\overline{z_2})=(3-z_1\overline{z_2})(3-\overline{z_1}z_2) \\ \Rightarrow  |z_1|^2-3z_1\overline{z_2}-3\overline{z_1}z_2+9|z_2|^2=9-3\overline{z_1}z_2-3z_1\overline{z_2}+|z_1|^2|z_2|^2 \\ \Rightarrow |z_1|^2+9|z_2|^2-|z_1|^2|z_2|^2-9=0 \\ \Rightarrow (|z_1|^2-|z_1|^2|z_2|^2)-(9-9|z_2|^2)=0 \\ \Rightarrow |z_1|^2(1-|z_2|^2)-9(1-|z_2|^2)=0 \\ \Rightarrow (|z_1|^2-9)(1-|z_2|^2)=0 \\ \Rightarrow |z_1|^2-9=0 \text{ or } 1-|z_2|^2=0 \Rightarrow |z_1|=3 \text{ or } |z_2|=1$$
Since $|z_2|=1$ we get $|z_1|=3$
A: $$
\left|\dfrac{z_1-3z_2}{3-z_1\overline{z}_2}\right|=1
$$
Let
$$
w=\dfrac{z_1-3z_2}{3-z_1\overline{z}_2}.
$$
Then
\begin{align}
w(3-z_1\bar{z}_2) & =z_1-3z_2 \\[8pt]
3w-w z_1\bar{z}_2 & = z_1 - 3z_2 \\[8pt]
3w+3z_2 & = z_1+wz_1\bar{z}_2 \\[8pt]
3w+3z_2 & = z_1(1+w\bar{z}_2) \\[8pt]
3\frac{w+z_2}{1+w\bar{z}_2} & = z_1 \tag 1
\end{align}
Finding the two fixed points of $w\mapsto \dfrac{w+z_2}{1+w\bar{z}_2}$ is solving a quadratic equation, and the solutions are $\pm z_2/\bar{z}_2$, and these are antipodal points on the unit circle centered at $0$.  Now observe that the point $iz_2/\bar{z}_2$, which is on the unit circle, is mapped by this linear fraction transformation to
$$
\frac{iz_2}{\bar{z}_2}\cdot\frac{1+\bar{z}_2}{1+z_2},
$$
which is also on th unit circle.  Since linear fractional transformations map every circle to a line or a circle, we conclude that the unit circle centered at $0$ is invariant under this l.f.t.  Consequently the number in $(1)$ has norm $3$.
A: $$\frac{z_1 - 3z_2}{3 - z_1 \overline{z_2}} = \frac{\frac{z1}{3} - z_2}{1 - \frac{z1}{3} \overline{z_2}}$$
$b = \overline{z_1} / 3, a = z_2 \Rightarrow |a - \overline{b}| = |1 - ab|$ 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.
