<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] Let $C$ be the solutions. Since $C = -C$ so we can look for solutions with non negative real part.
Note that $2 = |2i| = |z+i - (z-i)| \le |z+i|+|z-i|$.
Let $f(x) = |x+iy+i| + |x+iy-i|$ and note that $f$ is continuous everywhere, differentiable for $ x > 0$ and $f'(x) >0$ for such $x$. It follows from the mean value theorem that $f(x) > f(0) \ge 2$ for $x>0$. Hence it follows that $x=0$.
It is not hard to see that $|iy+i| + |iy-i| = |y+1|+|y-1| = \max(2,2|y|)$, hence
we must have $|y| \le 1$, that is $C \subset [-i,i]$.
It is straightforward to verify that $[-i,i] \subset C$.
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