Here's the Question: Find a set of points in the complex plane that satisfies:
$$|z-i|+|z+i| = 1$$ Now from triangle inequality I found: $$|z+z+i-i|=|2z|\geq1 $$ Which refers that there's no soluton and the set should be empty. But if I substitute z=x+iy and simplify then: $$ \Rightarrow \sqrt{ x^2 +(y-1)^2}+\sqrt {x^2 +(y+1)^2} = 1 $$ $$\Rightarrow \sqrt{ x^2 +(y-1)^2} = 1- \sqrt {x^2 +(y+1)^2}$$ Squaring both sides, $$\Rightarrow x^2 +(y-1)^2 = 1-2\sqrt{x^2 +(y+1)^2}+ x^2+(y+1)^2$$ Simplify and square again, $$\Rightarrow (4y+1)^2 =4(x^2+(y+1)^2)$$ $$\Rightarrow 16y^2+8y+1=4x^2+4y^2+8y+4$$ $$ \Rightarrow 12y^2-4x^2=3$$ Which implies a hyperbola. Now should I consider the points on the hyperbola as the expected points or there should be no points to satisfy the equation?
There is no such point because for all $z \in \mathbb{C}$ you have
$$|z-i|+|z+i| \geq |z-i - (z+i)| =|2i| = 2 > 1$$