This function will be undefined wherever the denominator = 0
So, setting (r - 1)^2 + (r + 1)^2 = 0 (and expanding) we have
r^2 - 2r + 1 + r^2 + 2r + 1 = 0
2r^2 + 2 = 0 divide through by 2
r^2 + 1 = 0
And this never = 0
So this is defined everywhere, i.e., ( -∞, ∞ )
