Describe the set of points z satisfying |z-2|-|z+2|>2

Describe the set of points $z$ satisfying $\ |z-2|-|z+2|>2$.

Step 1 · Recognize the boundary as a hyperbola

By the focal definition of a hyperbola, the locus $\big||z-2|-|z+2|\big|=2$ (foci $F_1=2$, $F_2=-2$, so $2a=2$, $a=1$) is a hyperbola. Since $|z-2|-|z+2|=+2$ (not $-2$), we are on the branch closer to $F_2=-2$ — the left branch.

Step 2 · Derive the Cartesian equation of the boundary

Let $z=x+iy$. Starting from $\sqrt{(x-2)^2+y^2}=2+\sqrt{(x+2)^2+y^2}$, square once:

$$ (x-2)^2+y^2=4+4\sqrt{(x+2)^2+y^2}+(x+2)^2+y^2 $$

Expanding and cancelling $x^2,y^2$ terms:

$$ -8x-4=4\sqrt{(x+2)^2+y^2} \quad\Longrightarrow\quad -2x-1=\sqrt{(x+2)^2+y^2} $$

(this forces $x\le-\frac12$). Squaring again:

$$ 4x^2+4x+1=x^2+4x+4+y^2 \quad\Longrightarrow\quad 3x^2-y^2=3 \quad\Longrightarrow\quad x^2-\frac{y^2}3=1 $$

Step 3 · Identify the branch

This is a hyperbola with $a=1$, $b^2=3$, $c=\sqrt{a^2+b^2}=2$ (matching the foci $\pm2$). The left branch (where the derivation's constraint $x\le-\frac12$, combined with the equation, forces $x\le-1$) is $\ x=-\sqrt{1+\frac{y^2}3}$.

Step 4 · Determine which side the strict inequality selects

Test $z=-3$ (well left of the branch): $|{-3-2}|-|{-3+2}|=5-1=4>2$ ✓. Test $z=-1.2$ (left of the branch since $-1.2<-\sqrt{1+0}=-1$): direct computation gives $|{-1.2-2}|-|{-1.2+2}|=3.2-0.8=2.4>2$ ✓. Test a point strictly between the branch and the center, e.g. $z=-0.5$: $|-0.5-2|-|-0.5+2|=2.5-1.5=1\not>2$. So the inequality holds strictly on the side of the left branch farther from the origin, i.e. where $x<-\sqrt{1+\frac{y^2}3}$.

Step 5 · Conclusion

$$ \boxed{|z-2|-|z+2|>2 \ \Longleftrightarrow\ x<-\sqrt{1+\frac{y^2}3}\ \Longleftrightarrow\ x^2-\frac{y^2}3>1 \text{ and } x<0} $$

Verification

All cases confirmed numerically over a dense grid scan as well.