求解 n=frac{(Z_{alpha}+Z_{beta})^2+3}{1.5ln[(1+r)/(1-r)]}

$$n = \frac { ( Z _ { \alpha } + Z _ { \beta } ) ^ { 2 } + 3 } { 1.5 \ln [ ( 1 + r ) / ( 1 - r ) ] }$$

求解 Z_α 的值 (复数求解)

$$Z_{α}=\frac{-2Z_{β}-\sqrt{6\left(\ln(\frac{r+1}{1-r})n-2\right)}}{2}$$

$$Z_{α}=\frac{-2Z_{β}+\sqrt{6\left(\ln(\frac{r+1}{1-r})n-2\right)}}{2}\text{, }r\neq -1\text{ and }r\neq 1\text{ and }r\neq 0$$

求解 Z_β 的值 (复数求解)

$$Z_{β}=\frac{-2Z_{α}-\sqrt{6\left(\ln(\frac{r+1}{1-r})n-2\right)}}{2}$$

$$Z_{β}=\frac{-2Z_{α}+\sqrt{6\left(\ln(\frac{r+1}{1-r})n-2\right)}}{2}\text{, }r\neq -1\text{ and }r\neq 1\text{ and }r\neq 0$$

求解 Z_α 的值

$$\left\{\begin{matrix}Z_{α}=\frac{-2Z_{β}-\sqrt{6\left(\ln(\frac{r+1}{1-r})n-2\right)}}{2}\text{; }Z_{α}=\frac{-2Z_{β}+\sqrt{6\left(\ln(\frac{r+1}{1-r})n-2\right)}}{2}\text{, }&\left(r>-1\text{ and }r<0\text{ and }n\leq \frac{2}{\ln(\frac{r+1}{1-r})}\right)\text{ or }\left(r>0\text{ and }r<1\text{ and }n\geq \frac{2}{\ln(\frac{r+1}{1-r})}\right)\\Z_{α}=-Z_{β}\text{, }&r\neq 0\text{ and }n=\frac{2}{\ln(\frac{r+1}{1-r})}\text{ and }r>-1\text{ and }|r|<1\end{matrix}\right.$$

求解 Z_β 的值

$$\left\{\begin{matrix}Z_{β}=\frac{-2Z_{α}-\sqrt{6\left(\ln(\frac{r+1}{1-r})n-2\right)}}{2}\text{; }Z_{β}=\frac{-2Z_{α}+\sqrt{6\left(\ln(\frac{r+1}{1-r})n-2\right)}}{2}\text{, }&\left(r>-1\text{ and }r<0\text{ and }n\leq \frac{2}{\ln(\frac{r+1}{1-r})}\right)\text{ or }\left(r>0\text{ and }r<1\text{ and }n\geq \frac{2}{\ln(\frac{r+1}{1-r})}\right)\\Z_{β}=-Z_{α}\text{, }&r\neq 0\text{ and }n=\frac{2}{\ln(\frac{r+1}{1-r})}\text{ and }r>-1\text{ and }|r|<1\end{matrix}\right.$$