given two complex number $z,w$ such number that $|z|\le1,|w|\le1$ and $|z+iw|=|z-i\overline{w}|=2$, then find the possible values of $z$
i tryed to use triangular inequality and got that
$$|z+iw|\le|z|+|iw|=|z|+|i||w|=|z|+|w|\\ |z-i\overline{w}|\le|z|+|-i\overline{w}|=|z|+|-i||\overline{w}|=|z|+|w|$$ since $|z|\le1,|w|\le1$ then $$|z+iw|\le|z|+|w|\le2\\ |z-i\overline{w}|\le|z|+|w|\le2$$
then since $|z+iw|=|z-i\overline{w}|=2$ then $|z|+|w|=2$ and $|z|=|w|=1$ because if $|z|<1$ then $|w|>1$ and if $|w|<1$ then $|z|>1$.
with the hint in the coment i find that the equality holds for $$|z+iw|\rightarrow z=iw\\ |z-i\overline{w}|\rightarrow z=-i\overline{w}$$ then $$iw=-i\overline{w}\\ w^2=-|w|^2=-1\\ w=\pm i\\ z=iw=\pm i^2=\mp 1$$