What is the plane graph of $|z-1|+|z-5| < 4$?

Question

What is the plane graph of $|z-1|+|z-5| < 4$ ?

What i know is that there is nothing for $y\geq4$ or $y \leq -4$ or $x \geq 5$ or $x \leq 1$.

Trying to let $z=x+y i$ such that $x,y \in \mathbb{R}$ did not help either.

Answer 1

$\left| x+i y-5\right| +\left| x+i y-1\right| <4$

$\sqrt{(x-5)^2+y^2}+\sqrt{(x-1)^2+y^2}<4$

$2 \sqrt{(x-5)^2+y^2} \sqrt{(x-1)^2+y^2}+2x^2-12 x+2 y^2+26<16$

$\sqrt{(x-5)^2+y^2} \sqrt{(x-1)^2+y^2}<-x^2+6 x-y^2-5$

$\left((x-5)^2+y^2\right) \left((x-1)^2+y^2\right)<\left(-x^2+6 x-y^2-5\right)^2$

$x^4-12 x^3+2 x^2 y^2+46 x^2-12 x y^2-60 x+y^4+26 y^2+25<x^4-12 x^3+2 x^2 y^2+46 x^2-12 x y^2-60 x+y^4+10 y^2+25$

$16 y^2<0$

There is no $z$ which satisfies the relation

Hope this helps

Answer 2

Consider two points in cartesian plane, $A(1,0)$ and $B(5,0)$. You have to find a locus of point $P$ such that sum of distance from these two points is less than $4$.

$$PA+PB < 4$$

Since the points themselves are at a distance $4$, ie $AB = 4$ , you can never have $PA+PB < 4$, because triangle inequality tells :

$$PA+PB \ge AB$$