Let be a function defined by $F(x,y)=(x^3-\alpha xy^2, \alpha x^2y-y^3)$. How can I define $\alpha$ such as $F$ is a conformal map? Let be a function defined by $F(x,y)=(x^3-\alpha xy^2, \alpha x^2y-y^3)$. How can I define $\alpha$ such as $F$ is a conformal map?
Also, let $F$ be a function $\mathbb{R}^2 \to \mathbb{R}^2$ and $\alpha$ be a constant over $\mathbb{R}$.
I understand the definition of a conformal map, but I can't figure how to tell if a function is conformal or not.
I'm leaving a big thank you in advance for any help.
 A: There is a slight error in the problem statement:  we only have a solution on $\Bbb R^2 \setminus \{ 0 \}$, so technically we should define $F:\Bbb R^2 \setminus \{ 0 \} \to \Bbb R^2$, as is made more clear below:
The mapping
$F(x, y) = (x^3 - \alpha x y^2, \alpha x^2 y - y^3) = (u(x, y), v(x, y)) \tag 1$
is conformal if and only if the function
$f(z) = f(x, y) = u(x, y) + i v(x, y) \tag 2$
is holomorphic and $f'(z) \ne 0$ (see this wikipedia entry), as will be case if $z = x + iy \ne 0$ and $u(x, y)$ and $v(x, y)$ satisfy the Cauchy-Riemmann equations
$u_x = v_y, \tag 3$
$u_y =  -v_x; \tag 4$
applying (3) to $u(x, y)$ and $v(x, y)$ as defined in (1) yields
$3x^2 - \alpha y^2 = \alpha x^2 - 3y^2; \tag 5$
from (5) we conclude that
$\alpha = 3; \tag 6$
indeed, (5) may be written
$3(x^2 + y^2) = \alpha (x^2 + y^2), \tag 7$
from which (6) immediately follows for any $(x, y) \ne (0, 0)$.  We then see that
$f(z) = (x^3 - 3 x y^2) + i(3x^2 y - y^3) = (x + iy)^3 = z^3, \tag 8$
a manifestly holomorphic function. 
The reader may easily check that (4) holds with this pair $u(x, y)$, $v(x, y)$ as well.  We see this is the only possible choice of $\alpha$; with $\alpha = 3$, the function $F(x, y)$ is therefore conformal except at $z = x + iy = 0$, and there is no alternative to the loss of conformality at $z = 0$.
Therefore, in any event,
$\alpha = 3. \tag 9$
