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Thread: F(x,y)=(x^{2}-y^{2}, 2xy)

  1. #1
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    F(x,y)=(x^{2}-y^{2}, 2xy)

    How do I show that the map $\displaystyle F(x,y)=(x^{2}-y^{2}, 2xy)$ is injective for y>0?
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  2. #2
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    Quote Originally Posted by Biscaim View Post
    How do I show that the map $\displaystyle F(x,y)=(x^{2}-y^{2}, 2xy)$ is injective for y>0?
    You need to show that for every $\displaystyle (x_1,y_1)$ and $\displaystyle (x_1,y_1)$ where y1 > 0 and y2 >0
    if $\displaystyle F(x_1,y_1)=F(x_2,y_2)$ then $\displaystyle (x_1,y_1)=(x_2,y_2)$

    $\displaystyle x_1^{2}-y_1^{2} = x_2^{2}-y_2^{2}$
    $\displaystyle 2x_1y_1 = 2x_2y_2$

    $\displaystyle y_1^{2}-y_2^{2} = x_1^{2}-x_2^{2} = x_1^{2}-\frac{x_1^2y_1^2}{y_2^2} = x_1^{2}\:\left(1-\frac{y_1^2}{y_2^2}\right) = x_1^{2}\:\frac{y_2^2-y_1^2}{y_2^2}$

    $\displaystyle (y_1^{2}-y_2^{2})\left(1+ \frac{x_1^2}{y_2^2}\right)=0$

    $\displaystyle y_1^{2}-y_2^{2}=0$

    $\displaystyle y_1-y_2=0$ since they are both > 0
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