求解 θ=a^2-b/a^2+b

$$\theta = \frac { a ^ { 2 } - b } { a ^ { 2 } + b }$$

求解 a 的值 (复数求解)

$$\left\{\begin{matrix}a=-i\left(\theta -1\right)^{-\frac{1}{2}}\sqrt{b}\sqrt{\theta +1}\text{; }a=i\left(\theta -1\right)^{-\frac{1}{2}}\sqrt{b}\sqrt{\theta +1}\text{, }&b\neq 0\text{ and }\theta \neq 1\\a\neq 0\text{, }&b=0\text{ and }\theta =1\end{matrix}\right.$$

求解 a 的值

$$\left\{\begin{matrix}a=\sqrt{-\frac{b\left(\theta +1\right)}{\theta -1}}\text{; }a=-\sqrt{-\frac{b\left(\theta +1\right)}{\theta -1}}\text{, }&\left(\theta \leq -1\text{ and }b<0\right)\text{ or }\left(\theta >1\text{ and }b<0\right)\text{ or }\left(\theta \geq -1\text{ and }\theta <1\text{ and }b>0\right)\\a\neq 0\text{, }&b=0\text{ and }\theta =1\end{matrix}\right.$$

求解 b 的值

$$\left\{\begin{matrix}b=-\frac{\left(\theta -1\right)a^{2}}{\theta +1}\text{, }&a\neq 0\text{ and }\theta \neq -1\\b\neq 0\text{, }&a=0\text{ and }\theta =-1\end{matrix}\right.$$