Hyperbola as a Complex Locus

Locus of z whose difference of distances from z₁ and z₂ is constant 2a. The absolute value gives both branches. When 2a = 0, the locus degenerates to the perpendicular bisector. The condition 2a < |z₁−z₂| ensures a valid hyperbola.

The hyperbola is defined by a difference of distances rather than a sum. The complex form makes this explicit — and the absolute value sign automatically captures both branches.

The Locus Condition

$$\bigl||z - z_1| - |z - z_2|\bigr| = 2a, \qquad 0 < 2a < |z_1 - z_2|$$

The absolute value gives two cases:

• $|z - z_1| - |z - z_2| = 2a$ (nearer to $z_2$, i.e., the branch closer to $z_1$)
• $|z - z_1| - |z - z_2| = -2a$ (nearer to $z_1$, the other branch)

Both branches are captured together.

Reducing to Cartesian Form

Place foci at $z_1 = c$ and $z_2 = -c$ with $2c = |z_1 - z_2|$. Take one branch:

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

Move the second radical and square:

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

Simplifying:

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

For this to be valid (right side must be non-negative), we need $cx + a^2 \leq 0$. Squaring:

$$c^2x^2 + 2a^2cx + a^4 = a^2(x^2 + 2cx + c^2 + y^2)$$ $$(c^2 - a^2)x^2 - a^2y^2 = a^2(c^2 - a^2)$$

Setting $b^2 = c^2 - a^2$ (positive since $c > a$):

$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$

The standard Cartesian hyperbola. The other branch follows identically. $\blacksquare$

Comparison with the Ellipse

When $2a = 0$, the hyperbola degenerates to the perpendicular bisector of $z_1 z_2$ — which is the $k = 1$ case of the Apollonius circle, now arrived at from a different direction.