Solucionar R+sqrt{R^2-L^2/4}=L

$$R + \sqrt { R ^ { 2 } - \frac { L ^ { 2 } } { 4 } } = L$$

Resolver para L (solución compleja)

$$\left\{\begin{matrix}L=0\text{, }&arg(R)\geq \pi \text{ or }R=0\\L=\frac{8R}{5}\text{, }&arg(2L-2R)<\pi \end{matrix}\right,$$

Resolver para R (solución compleja)

$$\left\{\begin{matrix}R=\frac{5L}{8}\text{, }&L=0\text{ or }arg(L)<\pi \\R\in \mathrm{C}\text{, }&L=0\text{ and }arg(R)\geq \pi \end{matrix}\right,$$

Resolver para L

$$\left\{\begin{matrix}L=0\text{, }&R\leq 0\\L=\frac{8R}{5}\text{, }&R\geq 0\end{matrix}\right,$$

Resolver para R

$$\left\{\begin{matrix}R=\frac{5L}{8}\text{, }&L>0\\R\leq 0\text{, }&L=0\end{matrix}\right,$$