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.

I get the sense that you want an algebraic approach to transform the equation from the complex plane to Cartesian coordinates.  To this end, the key is to first square both sides:  $$\begin{align*} 16 &= (|z-1| + |z+1|)^2 \\ &= |z-1|^2 + 2|z-1||z+1| + |z+1|^2 \\ &= (x-1)^2 + y^2 + 2|z^2-1| + (x+1)^2 + y^2 \\ &= 2x^2 + 2 + 2y^2 + 2|(x+iy)^2 - 1| \\ &= 2\left(x^2 + y^2 + 1 + \sqrt{(x^2-y^2-1)^2 + (2xy)^2} \right). \end{align*}$$  Now dividing both sides by 2 and rearranging, we square again:  $$(8 - (x^2 + y^2 + 1))^2 = (x^2 - y^2 - 1)^2 + 4x^2 y^2.$$  Put all terms on one side and simplify:  $$\begin{align*} 0 &= (x^2-y^2-1)^2 - (x^2+y^2-7)^2 + 4x^2 y^2 \\ &= (-2y^2+6)(2x^2-8) + 4x^2 y^2 \\ &= -4(x^2-4)(y^2-3) + 4x^2 y^2 \\ &= 4(3x^2 + 4y^2 - 12). \end{align*} $$  This immediately gives the desired expression.  Note that the locus of points satisfying the equality is itself necessarily an equation, not an inequality:  the locus is the boundary of the ellipse, not its interior.
