<s>[INST] 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
 [/INST] 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 $\alpha\in[-1,1]$
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