Answer
There are two equations in cylindrical coordinates:
${r_1} = {f_1}\left( {\theta ,z} \right) = \cos \theta + \sin \theta + \sqrt {1 + \sin 2\theta + {z^2}} $
${r_2} = {f_2}\left( {\theta ,z} \right) = \cos \theta + \sin \theta - \sqrt {1 + \sin 2\theta + {z^2}} $
Work Step by Step
In cylindrical coordinates we have
$x = r\cos \theta $, ${\ }$ $y = r\sin \theta $, and ${r^2} = {x^2} + {y^2}$.
So, the equation becomes
${r^2} - {z^2} = 2r\left( {\cos \theta + \sin \theta } \right)$
${r^2} - 2\left( {\cos \theta + \sin \theta } \right)r - {z^2} = 0$
Notice that this is a quadratic equation. The solutions are
${r_{1,2}} = \frac{{2\left( {\cos \theta + \sin \theta } \right) \pm \sqrt {4{{\left( {\cos \theta + \sin \theta } \right)}^2} - 4\cdot1\cdot\left( { - {z^2}} \right)} }}{{2\cdot1}}$
${r_{1,2}} = \cos \theta + \sin \theta \pm \sqrt {{{\left( {\cos \theta + \sin \theta } \right)}^2} + {z^2}} $
${r_{1,2}} = \cos \theta + \sin \theta \pm \sqrt {{{\cos }^2}\theta + 2\cos \theta \sin \theta + {{\sin }^2}\theta + {z^2}} $
Since ${\cos ^2}\theta + {\sin ^2}\theta = 1$ and $\sin 2\theta = 2\cos \theta \sin \theta $, so
${r_{1,2}} = \cos \theta + \sin \theta \pm \sqrt {1 + \sin 2\theta + {z^2}} $
It follows that
${r_1} = {f_1}\left( {\theta ,z} \right) = \cos \theta + \sin \theta + \sqrt {1 + \sin 2\theta + {z^2}} $
${r_2} = {f_2}\left( {\theta ,z} \right) = \cos \theta + \sin \theta - \sqrt {1 + \sin 2\theta + {z^2}} $
