<s>[INST] Identify the plane defined by $|z-2i| = 2|z+3|$ I tried:
$$|z-2i| = 2|z+3| \Leftrightarrow \\
|x+yi-2i|=2|x+yi+2|\Leftrightarrow \\
\sqrt{x^2+(y-2)^2}=\sqrt{4((x+2)^2+y^2)} \Leftrightarrow \\
\sqrt{x^2+y^2-4y+4} = \sqrt{4x^2+24x+36+4y^2} \Leftrightarrow \\
x^2+y^2-4y+4 = 4x^2+24x+36+4y^2 \Leftrightarrow \\
y^2-4y-4y^2=4x^2+24x+36+x^2 \Leftrightarrow \\
-3y^2-4y=5x^2+24x+26 \Leftrightarrow \\
???$$
What do I do next?
 [/INST] You can square without adding spurious solutions, because $|w|\ge0$ by definition. So $|z-2i|^2=4|z+3|^2$ and, using $|w|^2=w\bar{w}$,
$$
(z-2i)(\bar{z}+2i)=4(z+3)(\bar{z}+3)
$$
This simplifies to
$$
3z\bar{z}+12(z+\bar{z})-2i(z-\bar{z})+32=0
$$
Now recall that, for $z=x+yi$, $z\bar{z}=x^2+y^2$, $z+\bar{z}=2x$ and $z-\bar{z}=2yi$, to get
$$
3(x^2+y^2)+24x+4y+32=0
$$
that can also be written as
$$
x^2+y^2+8x+\frac{4}{3}y+\frac{32}{3}=0
$$
or, completing the squares,
$$
(x+4)^2+\left(y+\frac{2}{3}\right)^{\!2}=\frac{52}{9}
$$
which represents a circle.
Your computation is also good up to
$$
x^2+y^2-4y+4 = 4x^2+24x+36+4y^2
$$
Then you do wrong simplifications: taking everything to the right hand side you get
$$
3x^2+3y^2+24x+4y+32=0
$$
exactly the same I got.
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