Plot $2|z|=|z-2|$ over the complex plain I was given a home work to Sketch $2|z|=|z-2|$ on the complex plane.I tried to simplify and found out that the last expression was $3x^2+3y^2+4x-4$. However I don't know how to sketch this on the complex plane and how I can describe the set of points
 A: The equation $2|z| = |z - 2|$ represents the locus of points in the complex plane such that the distance from $(2,0) = 2 + 0i$ is twice the distance from the origin.  So one obvious example of such a point would be $z = 2/3$, since the distance from the origin is $2/3$ and the distance from $2$ is $2 - 2/3 = 4/3$.
Another (slightly less) obvious example would be $z = -2$, since its distance to the origin is $2$, but the distance to $2$ is $4$.
Since you found an equation $3x^2 + 3y^2 + 4x - 4$, where $x$ is the real part of $z$ and $y$ the imaginary part (i.e., $z = x + iy$), what happens when you plug in $(x,y) = (2/3, 0)$?  What happens when you plug in $(x,y) = (-2, 0)$?  Why do both of these choices give you $0$ for this expression?
Now, what does the equation $$3x^2 + 3y^2 + 4x - 4 = 0$$ correspond to in the Cartesian coordinate plane?  What curve does it look like?
A: Hint
$$2|z|=|z-2| \Leftrightarrow \\
4|z|^2 = |z-2|^2 \Leftrightarrow \\
4z \bar{z}= (z-2)(\bar{z}-2) = z \bar{z}-2z -2 \bar{z}+4 \Leftrightarrow \\
3z \bar{z} +2z +2 \bar{z}-4 =0 \Leftrightarrow \\
9z \bar{z} +6z +6 \bar{z}-12 =0\Leftrightarrow \\
(3z+2)(3 \bar{z}+2)=8\Leftrightarrow \\
|3z+2|^2=8
$$
A: Note that $\phi(z) = {z-2 \over 2z}$ is a Möbius transformation and you are looking for $\{ z | |\phi(z)|=1 \} = \phi^{-1}(\partial B(0,1))$.
Since the inverse of a Möbius transformation is a Möbius transformation, and Möbius transformations transform circles into circles, we see that the set in question is a circle.
It is simple to verify that $-2, {2 \over 3}$ and ${2 \over \sqrt{3}}i$ are in this locus and from these three points we can easily plot the set.
