ដោះស្រាយ z=sqrt{x^2+y^2},x=e^2t,y=e^-2t2z/2t

$$z = \sqrt { x ^ { 2 } + y ^ { 2 } } , x = e ^ { 2 t } , y = e ^ { - 2 t } \frac { 2 z } { 2 t }$$

ដោះស្រាយសម្រាប់ x, y, z (complex solution)

$$\left\{\begin{matrix}x=e^{2t}\text{, }y=e^{2t}\left(t^{2}e^{4t}-1\right)^{-\frac{1}{2}}\text{, }z=\sqrt{\frac{t^{2}e^{8t}}{t^{2}e^{4t}-1}}\text{, }&t^{2}e^{4t}-1\neq 0\text{ and }|-\frac{arg(\frac{t^{2}e^{8t}}{t^{2}e^{4t}-1})}{2}+arg(te^{4t}\left(t^{2}e^{4t}-1\right)^{-\frac{1}{2}})|<\pi \text{ and }t\neq 0\text{ and }t^{2}e^{4t}\neq 1\\x=e^{2t}\text{, }y=-e^{2t}\left(t^{2}e^{4t}-1\right)^{-\frac{1}{2}}\text{, }z=\sqrt{\frac{t^{2}e^{8t}}{t^{2}e^{4t}-1}}\text{, }&t^{2}e^{4t}-1\neq 0\text{ and }|-\frac{arg(\frac{t^{2}e^{8t}}{t^{2}e^{4t}-1})}{2}+arg(-te^{4t}\left(t^{2}e^{4t}-1\right)^{-\frac{1}{2}})|<\pi \text{ and }t\neq 0\text{ and }t^{2}e^{4t}\neq 1\end{matrix}\right.$$

ដោះស្រាយសម្រាប់ x, y, z

$$\left\{\begin{matrix}x=e^{2t}\text{, }y=-\frac{e^{2t}}{\sqrt{t^{2}e^{4t}-1}}\text{, }z=e^{4t}\sqrt{\frac{t^{2}}{t^{2}e^{4t}-1}}\text{, }&t^{2}e^{4t}\neq 1\text{ and }\frac{e^{4t}}{t^{2}e^{4t}-1}\geq 0\text{ and }t^{2}e^{4t}-1>0\text{ and }t<0\\x=e^{2t}\text{, }y=\frac{e^{2t}}{\sqrt{t^{2}e^{4t}-1}}\text{, }z=e^{4t}\sqrt{\frac{t^{2}}{t^{2}e^{4t}-1}}\text{, }&t^{2}e^{4t}\neq 1\text{ and }\frac{e^{4t}}{t^{2}e^{4t}-1}\geq 0\text{ and }t^{2}e^{4t}-1>0\text{ and }t>0\end{matrix}\right.$$