Complex numbers and function

Question

What domain of the z-plane is represented by $$|z+2|+|z-2|\lt 4$$ Please give me a hint... Can I use the formula ? $$|z_1|+|z_2|\geqslant |z_1+z_2|$$

Answer 1

Suppose $z_1,z_2\in\mathbb C$ are complex numbers and let $|z|$ denote the modulus of $z$. Then, the triangle inequality states that

$$|z_1+z_2|\le |z_1| + |z_2|$$

similarly, we have

$$|z_1-z_2|=|z_1+(-z_2)|\le|z_1| + |z_2|$$

so let $z_1=z+2, z_2=z-2$ and observe that

$$4=|z_1-z_2|=|(z+2)-(z-2)|\le |z+2| + |z-2|=|z_1| + |z_2|$$

therefore the domain is the empty set.

Answer 2

Note that the LHS of the given equation $$|z+2|+|z-2|\lt 4$$ represents the sum of the distances of the point $z$ to the two fixed points at $(-2,0)$ and $(2,0)$, which has to be less than 4 according to the RHS.

Given that the distance between the two fixed points $(-2,0)$ and $(2,0)$ is already 4, which is the minimum of the LHS, i.e

$$|z+2|+|z-2|\ge |(z+2)-(z-2)| = 4$$

Thus, the given equation represents an empty domain.