Answer: E

Small Hint:

Let $t=\frac{b}{a}$ after noting that the denominators must be nonzero

Big Hint:

The resulting quadratic $t^2+t+1=0$ has negative discriminant

Solution:

The denominators require $a\ne0$ and $a+b\ne0.$ Setting $t=\frac{b}{a}$ and cross-multiplying gives $$(1+t)^2=t,$$ or $t^2+t+1=0.$ Its discriminant is $-3,$ so $t$ is not real. Therefore $a$ and $b$ cannot both be real. Depending on the nonzero complex scale chosen for $a,$ one can be real and the other nonreal, or both can be nonreal.

Thus, the correct answer is E.