Math help!

((z^2-4)/(z-3))/((z+2)/(z^2+z-12) = ((((z^2)-4)*((z^2)+z-12))/((z-3)*(z+2)))

Wait I don't get it... What are the restrictions?..

Oh wait nvrm got it

${z}^{2}-\frac{4}{z}-\frac{3}{z}+\frac{2}{{z}^{2}}+z-12$

${z}^{2}-\frac{7}{z}+\frac{2}{{z}^{2}}+z-12$

${z}^{2}+z-\frac{7}{z}+\frac{2}{{z}^{2}}-12$

The only thing that can be done is to simplify the expression.  The only restrictions would be if you subsitute $z$ for a number and then solve.  $z$ cannot be $0$ because if you divide $7$ by $0$ or $2$ by ${0}^{2}$ (or $0$), you would get $undefined$ as an answer.