Solve complex number inequality $|(z-2)^{2}|\leq|(z-1)^{2}|$ $$
|(z-2)^{2}|\leq|(z-1)^{2}|
$$
What I did was to let $z=(x+iy)$ and obtain the following inequality:
$$
\sqrt{[(x-2)^{2}-y^{2}]^{2}+[2(x-2)y]^{2}}\leq\sqrt{[(x-1)^{2}-y^{2}]^{2}+[2(x-1)y]^{2}}
$$
But I don't know how to continue from here, and it seems not to be the most appropriate method for this question.
Is there any way to visualise their relationship in the complex plane?
Many thanks in advance!
 A: You can apply the property $|z^{n}| = |z|^{n}$ which holds for any natural number $n$.
At the present case, one has that
\begin{align*}
|(z - 2)^{2}| \leq |(z - 1)^{2}| & \Longleftrightarrow |z - 2|^{2} \leq |z - 1|^{2}\\\\
& \Longleftrightarrow (z - 2)(\overline{z} - 2) \leq (z - 1)(\overline{z} - 1)\\\\
& \Longleftrightarrow z\overline{z} - 2(z + \overline{z}) + 4 \leq z\overline{z} - (z + \overline{z}) + 1\\\\
& \Longleftrightarrow z + \overline{z} \geq 3\\\\
& \Longleftrightarrow 2\text{Re}(z) \geq 3\\\\
& \Longleftrightarrow \text{Re}(z) \geq \frac{3}{2}
\end{align*}
and we are done.
Hopefully this helps!
A: 
$|(z-2)^{2}|\leq|(z-1)^{2}|$

Alternative approach:
The problem can be solved without (much) Math.
First, consider the alternative (simpler) problem:
$$|(z-2)^{2}|= |z-2|^2 = |z-1|^2 = |(z-1)^{2}|. \tag1 $$
(1) above will be true if and only if
$$|(z-2)| = |(z-1)|. \tag2 $$
Geometrically, (2) above is equivalent to the constraint that $z$ is equidistant between $[2 + i(0)]$ and $[1 + i(0)]$, which will be true if and only if $z$ is on the perpendicular bisector between $[2 + i(0)]$ and $[1 + i(0)].$
Since the locus of this perpendicular bisector is Re$(z) = (3/2)$, this is the locus of the solution to (2) above.

It remains only to consider the distinction between the original problem, and the problem posed in (2).
Similar to the distinction between (1) and (2) above, the original problem will be satisfied if and only if
$$|z - 2| \leq |z - 1|. \tag3 $$
Geometrically, you can visualize that the line given by Re$(z) = (3/2)$ divides the Complex plane into two half-planes, and that (3) above will be satisfied if and only if $z$ is on the right half-plane, or the border between the two half-planes.  This right half-plane (plus the corresponding border) is represented by the constraint
$$\text{Re}(z) \geq (3/2).$$
