Equivalent Impedance
You are very close. You'd done well up to the point where you just stopped. And I've waited long enough. I'm now going to assume that you either (1) aren't very familiar with complex fractions and don't understand how to use the value "1" composed of the complex conjugate of the denominator to normalize the expression into real and imaginary terms; or else, (2) you do know how, but you don't know why you'd bother with normalizing the expression into real and imaginary terms.

To put a fraction with complex expressions in the numerator and denominator, multiply it by another fraction formed by an identical numerator and denominator made up of the complex conjugate of the denominator. I'll use my own expression, which is the inversion of yours:

$$\begin{align*}
10\:\Omega &amp;=\frac{-A\,B+j\,B}{1+j\left(A+B\right)}\\\\
&amp;=\frac{-A\,B+j\,B}{1+j\left(A+B\right)}\cdot\frac{1-j\left(A+B\right)}{1-j\left(A+B\right)}\\\\
&amp;=\frac{B^2+j\,B+j\,A^2\,B+j\,A\,B^2}{1+\left(A+B\right)^2}\\\\
&amp;=\frac{B^2}{1+\left(A+B\right)^2}+j\,\frac{B\left(A^2+A\,B+1\right)}{1+\left(A+B\right)^2}
\end{align*}$$

But, this has to mean that \$10\:\Omega=\frac{B^2}{1+\left(A+B\right)^2}\:\Omega\$ and also that \$j\,0\:\Omega=j\,\frac{B\left(A^2+A\,B+1\right)}{1+\left(A+B\right)^2}\:\Omega\$. This provides you with two equations and two unknowns. You should be able to solve those for the values of \$A\$ and \$B\$.