Sketch the set of points in the complex plane satisfying

Question

$|z+1|-|z-1|=4$

Solution: Using reverse triangle inequality: $||z+1|-|z-1||\le|(z+1)-(z-1)|=|2|=2$

Thus, there are no points, because $|z+1|-|z-1|$ can't be 4.

Is this correct?

Answer 1

Yes, that's correct. You can also do it with the triangle inequality, rather than the reverse triangle inequality. Note that $|z+1| = |(z-1)+2| \leq |z-1| + |2| = |z - 1| + 2$ so

$$|z+1| - |z-1| \leq |z-1| + 2 - |z - 1| = 2.$$

Therefore $|z + 1| - |z - 1| \neq 4$.