सोडोवचें n=(t^2)^-2*(-1)^2/t^4

$$n = \frac { ( t ^ { 2 } ) ^ { - 2 } \cdot ( - 1 ) ^ { 2 } } { t ^ { 4 } }$$

t खातीर सोडोवचें (जटील सोल्यूशन)

$$t\in \sqrt{2}\left(\frac{1}{2}+\frac{1}{2}i\right)n^{-\frac{1}{8}},n^{-\frac{1}{8}},in^{-\frac{1}{8}},\sqrt{2}\left(-\frac{1}{2}+\frac{1}{2}i\right)n^{-\frac{1}{8}},-n^{-\frac{1}{8}},\sqrt{2}\left(-\frac{1}{2}-\frac{1}{2}i\right)n^{-\frac{1}{8}},-in^{-\frac{1}{8}},\sqrt{2}\left(\frac{1}{2}-\frac{1}{2}i\right)n^{-\frac{1}{8}}$$

$$n\neq 0$$

n खातीर सोडोवचें

$$n=\frac{1}{t^{8}}$$

$$t\neq 0$$

t खातीर सोडोवचें

$$t=\frac{1}{\sqrt[8]{n}}$$

$$t=-\frac{1}{\sqrt[8]{n}}\text{, }n>0$$