# Find a set of points in the given complex plane

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?

• Note that the shortest path from $i$ to $-i$ is the segment connecting them. This segment has already length $2$. – trancelocation Mar 20 at 13:29
• What does that interpret? Is the last line(hyperbola) invalid then? – hasib ryan Mar 20 at 13:44
• @ hasib: I do not see how you obtained the equation of that hyperbola. Would be good to show your steps to see where something went wrong, as there cannot be any point satisfying the equation given at the beginning. – trancelocation Mar 20 at 13:50
There is no such point because for all $$z \in \mathbb{C}$$ you have
$$|z-i|+|z+i| \geq |z-i - (z+i)| =|2i| = 2 > 1$$