Let $$\Lambda=\prod_{n=1}^N(a_n+ib_n)\not=0$$ where $a_n,b_n$ are real. Can we determine $\mathrm{sign}(\Lambda)$ when $\mathrm{Im}(\Lambda)=0$ and $a_n<0\;\forall n$?
The result can depend on $N$, but not the individual $a_n$s and $b_n$s.
The context for this problem is that I have the matrix product $$A=BC$$ where $A$ and $C$ are square and (in general) nonsymmetric, so $$\mathrm{det}(A)=\mathrm{det}(BC)=\mathrm{det}(B)\mathrm{det}(C).$$ I know that $\mathrm{det}(B)$ is positive and real and that $\mathrm{det}(C)$ is real (since the matrix $C$ is real, its determinant can only be real). Now I write $\mathrm{det}(A)$ as the product of the (complex) eigenvalues of $A.$ Furthermore, I know that all eigenvalues of $A$ have negative real part.
We can't know the sign. Assuming $a+b+c=abc$ we have that
$$(-1+ai)(-1+bi)(-1+ci)=ab+ac+bc-1.$$ Now, if $a=1,b=-1,c=0$ we get $-2<0$ and if $a=b=-2,c=-4/3$ we get $25/3>0.$