Mathematics

Problem

z

satisfies the inequality

|z - 1| < |z + 1|

, then one has a.

|z - 2 - i| < |z + 2 - i|, i = \sqrt{-1}

|\arg(z + i)| < \frac{\pi}{2}, i = \sqrt{-1}

\operatorname{Re}(z) < 0

\operatorname{Im}(i\bar{z}) > 0, i = \sqrt{-1}