Why the imaginary half circles is this plot?

Question

I plotted the square roots of the Oppermann boundaries and got this:

Why the imaginary half circles?

Answer 1

It's enough consider the case of $\sqrt{k^2-k}$, since the second is just the $k\mapsto -k$ reflection of the first.

Writing $\sqrt{k^2-k}=x+I y$ and squaring both sides gives $x^2-y^2+(2xy)i=k^2-k.$ Since the plot is for real $k$, we can match the real and imaginary parts of the two sides and conclude that $x^2-y^2=k^2-k$ and $xy=0$.

The latter requires that at least one of $x,y$ must vanish for any given $k$ (which is apparent in your graphs above.) If $x=0$, then $y^2=k-k^2$; but $y^2$ can't be negative, so $k$ must satisfy $k-k^2\geq 0$ aka $0\leq k \leq 1$. This is the domain of interest, so this expression for $y$ is what we want.

But we can complete the square to rewrite this equality as $y^2+(k-\frac{1}{2})^2=\frac{1}{4}$, which we may recognize as the equation of a circle of radius $1/2$ centered at $(k,y)=(1/2,0)$. Since only the positive root is taken, the plot is indeed a semicircle.