Describe the set of points z=x+iy satisfying |z-2|<|z|
Describe the set of points $z=x+iy$ satisfying $\ |z-2|<|z|$.
Explanation of Method
Both sides are nonnegative, so squaring preserves the inequality. Writing $z=x+iy$ turns it into an inequality in $x,y$ that collapses to a simple condition on $x$ alone.
Step 1 · Square both sides
Since $|z-2|\ge0$ and $|z|\ge0$, squaring is reversible:
$$ |z-2|^2<|z|^2 $$
Step 2 · Substitute $z=x+iy$
$$ (x-2)^2+y^2<x^2+y^2 $$
Step 3 · Expand and simplify
$$ x^2-4x+4+y^2<x^2+y^2 \quad\Longrightarrow\quad -4x+4<0 \quad\Longrightarrow\quad x>1 $$
Step 4 · Conclusion
$$ \boxed{|z-2|<|z| \ \Longleftrightarrow\ \text{Re}(z)>1} $$
Geometrically: the set of points closer to $2$ than to $0$ is exactly the open half-plane strictly to the right of the perpendicular bisector of the segment from $0$ to $2$, which is the vertical line $x=1$.
The open half-plane $\text{Re}(z)>1$, bounded by the perpendicular bisector of $0$ and $2$.
Result
An inequality comparing distances to two points always defines a half-plane bounded by their perpendicular bisector — here the bisector of $0$ and $2$ is the line $x=1$.