Transformation of a Lagrangian

This is the answer that physshyp had in mind but felt like not writing down. Define the complex variables $zeta = lambda + i, mu$ and $z = x + i, y$. Then begin{align}zeta^2 =& (lambda + i, mu)^2= (lambda + i, mu)(lambda + i, mu) =& lambda^2 + i, lambda, mu + i , mu , lambda + (i, mu)^2 = lambda^2 + 2, i, lambda, mu - , mu^2 =& (lambda^2 - mu^2) + i (2 , lambda , mu) end{align} Consequently, since begin{align} &x = lambda^2 - mu^2 &y = 2, lambda mu end{align} we have $$z = x + i, y = (lambda^2 - mu^2) + i (2 , lambda , mu) = (lambda + i, mu)^2 = zeta^2$$ So in complex numbers, $$z = zeta^2$$ Now, it is easy to differentiate the change of variables and get $$dot{z} = 2, zeta, dot{zeta}$$ Then, by taking absolute value squared of complex numbers $$|dot{z}|^2 = 4, |zeta|^2, |dot{zeta}|^2$$ If you expand in real coordinates, recalling the definition of absolute value squared of complex numbers $$dot{x}^2 + dot{y}^{2} = |dot{z}|^2 = 4, |zeta|^2, |dot{zeta}|^2 = 4 , (lambda^2 + zeta^2),(dot{lambda}^2 + dot{zeta}^2)$$ The latter expression is the first term of the Lagrangian and combined with the fact that $y = 2, lambda, mu$ we get the desired change of variables in the Lagrangian function $$L = frac{m}{2}, (lambda^2 + zeta^2),(dot{lambda}^2 + dot{zeta}^2) - alpha, (lambda , mu)^2 = frac{m}{2},frac{1}{4}, (dot{x}^2 + dot{y}^{2}) - alpha frac{1}{4}, y^2 = frac{m}{8}, (dot{x}^2 + dot{y}^{2}) - frac{alpha}{4}, y^2$$