<s>[INST] Equation of the complex locus: $|z-1|=2|z +1|$ This question requires finding the Cartesian equation for the locus:

$|z-1| = 2|z+1|$

that is, where the modulus of $z -1$ is twice the modulus of $z+1$
I've solved this problem algebraically (by letting $z=x+iy$) as follows:

$\sqrt{(x-1)^2 + y^2} = 2\sqrt{(x+1)^2 + y^2}$
$(x-1)^2 + y^2 = 4\big((x+1)^2 + y^2\big)$
$x^2 - 2x + 1 + y^2 = 4x^2 + 8x + 4 + 4y^2$
$3x^2 + 10x + 3y^2 = -3$
$x^2 + \frac{10}{3}x + y^2 = -1$
$(x + \frac{5}{3})^2 +y^2 = -1 + \frac{25}{9}$

therefore, $(x+\frac{5}{3})^2 + y^2 = \frac{16}{9}$, which is a circle.
However, I was wondering if there is a method, simply by inspection, of immediately concluding that the locus is a circle, based on some relation between the distance from $z$ to $(1,0)$ on the plane being twice the distance from $z$ to $(-1,0)$?
 [/INST] A perhaps easier way to see that it is a circle is to rewrite as
$\displaystyle |z'| = 2|z' + 2|, \text{where } z' = z -1$ 
And so
$\displaystyle 1/4 = |1/2 + 1/z'|$
Thus $\displaystyle \frac{1}{z'}$ gives a circle centred at $(-1/2,0)$ and radius $1/4$.
As we can see from the answers and comments to this recent question: If $0$, $z_1$, $z_2$ and $z_3$ are concyclic, then $\frac{1}{z_1}$,$\frac{1}{z_2}$,$\frac{1}{z_3}$ are collinear that, the map $\displaystyle z \to \frac{1}{z}$ maps circles to circles, if the circle does not pass through the origin.
In this case, it does not pass through the origin, and so the original curve you seek would be a circle too.
Of course the fact that $\displaystyle z \to \frac{1}{\bar{z}}$ is inversion with respect to the unit circle probably uses the Appollonius theorem, but it looks like a useful fact to keep in mind.
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