Sketching on the complex plane $|z + 1| - |z - 1| = 4$ 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 $x&gt;4$. 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?
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.
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