Solution
= Solution
For $\nu=2$, $|r(iy)|^2=1+y^4/4$, so only $y=0$ is stable. For $\nu=3$, $|r(iy)|^2=1-y^4/12+y^6/36$, so the stable imaginary segment is $|y|\le\sqrt3$.
= Solution
For $\nu=2$, $|r(iy)|^2=1+y^4/4$, so only $y=0$ is stable. For $\nu=3$, $|r(iy)|^2=1-y^4/12+y^6/36$, so the stable imaginary segment is $|y|\le\sqrt3$.