Sketching on the complex plane $|z + 1| - |z - 1| = 4$

Question

Sketch the set of points that satisfy $|z + 1| - |z - 1| = 4$ on the complex plane.

Wolfram alpha gives me an empty graph, I end up with the equation of an ellipse but with the condition $x>4$. The ellipse is the same as $|z + 1| + |z - 1| = 4$ but I had to be careful with the square roots.

So is it true that no complex number satisfies this?

Answer 1

Triangle inequality:

$$\lvert z+1\rvert = \lvert (z-1)+2\rvert \leqslant \lvert z-1\rvert + 2 \implies \lvert z+1\rvert - \lvert z-1\rvert \leqslant 2.$$

So is it true that no complex number satisfies this?

Yes, that is true.

Answer 2

That is not true. Take $z=2$, then $$|z+1|+|z-1|=3+1=4$$

I believe this is the equation of a hyperbola, not an ellipse. The subtraction means it will be a hypomolobro

I can't comment or upvote, but whoever just said it's a hyperbola is my hero. #upvotes

Answer 3

It must be a hyperbola, because it states that the difference of distances from the foci 1 and -1 is constant (equal to 4). Be careful, squaring tends to introduce additional solutions if it removes the signs.

EDIT:

Sorry, I wrote too quickly without looking at the position of the foci. The definition with difference of distances between foci implies a hyperbola, but the right side is incompatible! A hyperbola always passes between the foci, and in this case, there's no space for that (because the distance between foci is 2, which is less than 4). This is why there's no solutions at all!

All the additional solutions (the "ellipse") come as extra false solutions that are the artifact of squaring. Each squaring folds negative and positive parts and may create false matching of LHS and RHS.