For |z+1|>√2 AND Re(z)<0 the argument is less than π/4
For |z+1|=√2 AND Re(z)<0 the argument is exactly π/4
For |z+1|<√2 AND Re(z)<0 the argument is greater than π/4
The line Re(z)=0 is divided into two general cases and a degenerate case.
If |Im(z)| > 1, then the argument is ambiguous (0 or 2π) and the inequality is not defined.
If |Im(z)| < 1, then the argument is π.
If |Im(z)| = 1, then you have Arg(0) which is undefined.
The argument for Re(z)>0 is well defined and satisfies the inequality.