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.

Note that for $z=x+iy$, we have the following three equalities:
$$|z|^2 = x^2+y^2$$
$$|z+1|^2 = |z|^2+1 + 2x $$
$$|z-1|^2 = |z|^2+1 - 2x$$
Now, from the given equation we have:
$$|z+1| = 4-|z-1|$$
By squaring once, we get:
$$|z+1|^2 = 16+|z-1|^2 - 8|z-1|.$$
From the second and third equalities above, we get:
$$4x = 16-8|z-1|$$
Hence:
$$2|z-1| = 4-x$$
Squaring AGAIN:
$$4|z-1|^2 = (4-x)^2$$
Using the third equality:
$$4|z|^2+4-8x = (4-x)^2$$
Finally, using the first equality gives:
$$4(x^2+y^2)+4-8x = (4-x)^2,$$
from where we have:
$$3x^2+4y^2 = 12.$$
