Transformation of a Lagrangian $$L(\lambda,\mu,\dot{\lambda},\dot{\mu})=\frac{m}{2}(\lambda^2+\mu^2)(\dot{\lambda}^2+\dot{\mu}^2)-\alpha \lambda^2\mu^2,$$
I'm supposed to express this Lagrangian through 
$x=\lambda^2-\mu^2$
$y=2\lambda\mu$
My first thought was to use 
$x+\mu^2=\lambda^2$
by putting it into the second equation but then I get:
$y=2\mu\sqrt(x+\mu^2)$ 
and don´t know how to proceed.  
 A: 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
$$L = \frac{m}{2}\, (\lambda^2 + \zeta^2)\,(\dot{\lambda}^2 + \dot{\zeta}^2) - \alpha\, (\lambda \, \mu)^2 = \frac{m}{2}\,\frac{1}{4}\, (\dot{x}^2 + \dot{y}^{2}) - \alpha \frac{1}{4}\, y^2 =   \frac{m}{8}\, (\dot{x}^2 + \dot{y}^{2}) - \frac{\alpha}{4}\, y^2$$
