For the function $w=\frac{1}{z}$ find the image of the parabola $y=x^2.$
Attempt:
If $w=u+vi$ and $z=x+yi,$ then $w=\frac{1}{z}=\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}$.
Define $u=\frac{x}{x^2+y^2}, v=-\frac{y}{x^2+y^2}.$
Making the substitution $y=x^2$ in $u,v$ then $u^2+v^2=\frac{1}{x^2+x^4}$.
I do not know what I can do from here.
Could someone help please?
The answer of the book is $u^2=\frac{-v^3}{(v+1)}$
Hint: eliminate $\,x\,$ between $\,u = \dfrac{1}{x(x^2+1)}\,$ and $\,v=-\dfrac{1}{x^2+1}\,$.
[ EDIT ] A couple of ways to do the elimination:
divide the equations to get $\,x = -v/u\,$, then substitute in either equation;
get $\,x^2 = -1 - 1/v\,$ from the second equation, then substitute in $\,u^2 = v^2 / x^2\,$.