Exercise 9.24

Question

Exercise 24: For $(x,y)\ne(0,0)$, define $\v f=(f_1,f_2)$ by
    $$
        f_1(x,y)=\frac{x^2-y^2}{x^2+y^2},\qquad f_2(x,y)=\frac{xy}{x^2+y^2}.
    $$
    Compute the rank of $\v f'(x,y)$, and find the range of $\v f$.

analambanomenos · Accepted Answer

The Jacobian of $\v f$ is
    \begin{align*}
        \begin{vmatrix}
            D_1f_1(x,y) & D_2f_1(x,y) \
            D_1f_2(x,y) & D_2f_2(x,y) 
        \end{vmatrix}
        &= 
        \frac{1}{(x^2+y^2)^4}\begin{vmatrix}
            4xy^2 & -4x^2y \
            y(y^2-x^2) & x(x^2-y^2)
        \end{vmatrix} \
        &= \frac{4x^4y^2-4x^2y^4+4x^2y^4-4x^4y^2}{(x^2+y^2)^4} \
        &= 0
    \end{align*}
    so the rank of $\v f'$ is less than 2. Since $\v f$ is nonconstant, the rank of $\v f'$ must be more than 0, hence the rank of $\v f'$ is 1.

Converting to polar coordinates, we get
    $$
         \v f(r\cos\theta,r\sin\theta) = \bigg(\cos(2\theta),\frac{\sin(2\theta)}{2}\bigg)
    $$
    so we see that the range of $\v f$ is an ellipse centered at the origin and intersecting the coordinate axes at $(\pm 1,0)$ and $(0,\pm 1/2)$.

Exercise 9.24

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Exercise 9.24

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