 Solution

Put $z=x+iy$. The expression whose complex argument is constrained becomesIts principal argument is $\pi /2$ exactly when its real part is zero and its imaginary part is strictly positive. HenceThe locus consists of the upper-right half of the upper hyperbola branch, $y=\sqrt{x^2+1}$ with $x>0$, and the lower-left half of the lower branch, $y=-\sqrt{x^2+1}$ with $x<0$. The vertices $z=\pm i$ are excluded because $z^2+1=0$ there and zero has no complex argument. The remaining halves have negative imaginary part and give argument $-\pi /2$, not $\pi /2$. The right panel of the original illustration marks the excluded vertices with open circles.