Correct. Values of $\displaystyle{z}={2}\usepackage{cmbright}\Huge\$ and $\displaystyle{z}=-{3}\frac{{1}}{{2}}\usepackage{cmbright}\Huge\$ makes this equation true; $\displaystyle{z}=\frac{{1}}{{2}}\usepackage{cmbright}\Huge\$ and $\displaystyle{z}=-\frac{{1}}{{2}}\usepackage{cmbright}\Huge\$ are excluded from the domain because they would create denominators of 0.