Need help with complex equation: |z−i|+|z+i|=2 I am trying to solve this equation: |z−i|+|z+i|=2 and don't know how to do it. This what I have:
$$\sqrt{(x+1)^2+y^2}+ \sqrt{(x-1)^2+y^2} = 2 /^2$$
$$(x+1)^2+y^2+(x-1)^2+y^2 + 2\sqrt{[(x+1)^2+y^2][(x-1)^2+y^2]} = 4$$
$$x^2 +2x + 1+y^2+x^2-2x + 1+y^2 + 2\sqrt{[(x^2 +2x + 1)+y^2][(x^2-2x + 1)+y^2]}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{[(x^2 +2x + 1)+y^2][(x^2-2x + 1)+y^2]}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{(x^2 +2x + 1)(x^2-2x + 1)+x^2y^2-2xy^2+y^2+x^2y^2-2xy^2+y^2+y^4}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{(x^2 +2x + 1)(x^2-2x + 1)+2x^2y^2-4xy^2+2y^2+y^4}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{x^4+2x^3+x^2-2x^3+4x^2-2x+x^2+2x+1+2x^2y^2-4xy^2+2y^2+y^4}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{x^4+6x^2+1+2x^2y^2-4xy^2+2y^2+y^4}=4$$
And I have no idea what to do next.
Any help would be much appreciated

From the beginning, move one of the absolute values to the other side. It will simplify the calculations:
$$|z+i|=2-|z-i|$$
Square this and you get $$|z+i|^2=(2-|z-i|)^2=4-4|z-i|+|z-i|^2$$
$$x^2+(y+1)^2=4-4|z-i|+x^2+(y-1)^2$$
$$x^2+y^2+2y+1=4-4|z-i|+y^2-2y+1$$
$$4y-4=-4|z-i|$$
$$y-1=-|z-1|$$
$$(y-1)^2=x^2+(y-1)^2$$
The only solution is $x=0$. This is independent of $y$. Next is to check if by squaring we add extra solutions. Not all $y$'s might be valid.
$$|y-1|+|y+1|=2$$
If $y\ge1$, the equation is $$2y=2$$So $y=1$. If $y\le -1$ then $$-(y-1)-(y+1)=2$$, meaning $y=-1$.In between, $$-(y-1)+y+1=2$$or $2=2$. The final answer is $z=\alpha i$, for $\alpha\in[-1,1]$
