<s>[INST] Find all $z\in\Bbb C$ such that $|z+1|+ |z-1|=4$ I'd like to find all points of the complex plane which satisfy
$$|z+1| + |z-1| = 4. $$
I know this is an ellipsis with foci $1$ and $-1$, and I know that the answer is
$$3 x^2+4 y^2 \leq 12,$$
but I can't find a correct way of getting there.
First, I write $z$ as $x + i y$ and square both sides of the equation, then divide by 2 and get
$$x^2+y^2+1+\sqrt{(x-1)^2+y^2} \sqrt{(x+1)^2+y^2} =8.$$
Pass $x^2+y^2+1$ to the RHS (right hand side), then
$$\sqrt{(x-1)^2+y^2} \sqrt{(x+1)^2+y^2} =7-x^2-y^2.\tag{1} $$
Now, I would have to square both sides of the equation like this,
$$((x-1)^2+y^2)((x+1)^2+y^2) = (7-x^2-y^2)^2, \tag{2}$$
but the problem is that I cannot assure that the RHS is not negative, so there could be a value for $z$ such that (2) is satisfied but not (1), i.e it could exist $z=x + i y$ which satisfies
$$\sqrt{(x-1)^2+y^2} \sqrt{(x+1)^2+y^2} =-(7-x^2-y^2)\tag{3} $$
in which case also satisfy (2) but not (1)!
So I would get an incorrect solution.
 [/INST] As one comment suggested above, it is more efficient to 
subtract $|z-1|$ or $|z + 1 | $ from both sides first.
$$|z+1| + |z-1| = 4\\ \, \\
|z+ 1| = 4 - |z - 1 | \\ \, \\
|z+ 1|^2 = \left(4 - |z - 1 |\right)^2 \\ \, \\
|z+ 1|^2 = 16 - 8|z - 1 | + |z - 1 | ^2 \\ \, \\
|z+ 1|^2  - |z - 1 | ^2 = 16 - 8|z - 1 |  $$ Then let $z = x + iy$.
$$
|(x+iy) + 1|^2  - |(x+iy) - 1 | ^2 = 16 - 8|(x+iy) - 1 |  \\ \, \\
|(x+ 1) + iy|^2  - |(x-1) + iy | ^2 = 16 - 8|(x-1) + iy|  \\ \, \\
(x+1)^2 + y^2 - ( (x-1)^2 + y^2)  = 16 -8|(x-1) + iy| \\ \, \\
x^2 + 2x + 1 + y^2 -  (x-1)^2 - y^2  = 16 -8|(x-1) + iy|\\ \, \\
x^2 + 2x + 1 + y^2 - (x^2 -2x + 1) - y^2 =16 -8|(x-1) + iy| \\ \, \\
x^2 + 2x + 1 + y^2 -x^2 +2x - 1- y^2 =16 -8|(x-1) + iy|\\ \, \\
4x = 16 -8|(x-1) + iy|\\ \, \\
16-4x= 8|(x-1) + iy|\\ \, \\
4( 4 - x) = 8|(x-1) + iy|\\ \, \\
16( 4 - x)^2 = 64|(x-1) + iy|^2\\ \, \\
( 4 - x)^2 = \frac{64}{4} ( (x-1)^2 + y^2)\\ \, \\
16 -8x + x^2 = 4 (x-1)^2 +4y^2\\ \, \\
16 -8x + x^2 = 4x^2 -8x + 4  +4y^2\\ \, \\
16 = 3x^2 + 4y^2 + 4 \\ \, \\ 
3x^2 + 4y^2 = 12
$$ This is an equation of ellipse.
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