<s>[INST] Complex sets: factoring into circle How must $|z|=3|z-1|$ be factored so I end up with a circle, plugging in $z=x+iy$ seems to just up with square roots everywhere. Detailed steps is much appreciated, thanks!
 [/INST] \begin{align*}
 z \overline{z} = 9 (z - 1)(\overline{z}-1) = 9 z \overline{z} - 9 z - 9 \overline{z} + 9 \\
8 z \overline{z} - 9 z - 9 \overline{z} + 9 = 0
\end{align*}
Now a circle is supposed to look like $$(z - a)(\overline{z} - \overline{a}) = r^2= z \overline{z} - a \overline{z} - \overline{a} z + |a|^2$$
So taking our old relation and diving by 8, we get
\begin{align*}
z \overline{z} - \frac{9}{8} z - \frac{9}{8} \overline{z} + \frac{9}{8} = 0 \\
z \overline{z} - \frac{9}{8} z - \frac{9}{8} \overline{z} + \frac{81}{64} = \frac{81}{64} - \frac{9}{8} = \frac{9}{64} 
\end{align*}
So with some algebra we have a circle centered at $\frac{9}{8}$ with radius $\frac{3}{8}$.
For fun: you can view the locus of solutions $S$ as the inverse image of the unit circle of the fractional linear transformation $z \mapsto \frac{z}{3z-3}$. Conjugate points are sent to conjugate points and the center of $S$'s conjugate is $\infty$. Infinity maps to $1/3$ and the conjugate about the unit circle is $3$, so the inverse of that is $8/9$. Similarly I can find what is sent to 1 and that'll give me a radius.
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