This problem is taken from Rudin analysis , Chapter 9 page No : $242$ . Problem No.$24$
For $(x,y) \neq (0,0)$ , define $f = (f_1 , f_2)$ by $f_1 (x,y) = \frac{x^2-y^2}{x^2+y^2}$ ,$f_2 = \frac{xy}{x^2+y^2}$
Compute the rank of $ f'(x,y)$ and find the range of $f$
My attempt : i found the answer and my doubt is in red line and red circle
im not getting the red line and red circle ?
I need help !
You seem to be confused with the following terminology:
This really just means that $x$ is taken to $y$ under the function $f$ under discussion; in other words $f(x)=y$. In your case, "$(1,0)$ is its own image" means nothing but $f(1,0)=(1,0)$ which you can check easily: $$f(1,0)=(f_1(1,0),f_2(1,0))=\left(\frac{1^2-0^2}{1^2+0^2},\frac{1\cdot0}{1^2+0^2}\right)=(1,0).$$ The statement for $(-1,0)$ is the same thing. The part in the red circle just says that $$\begin{split}f\left(1,\pm\sqrt{\frac{1-u}{1+u}}\right)&=\left(\frac{1^2-(1-u)/(1+u)}{1^2+(1-u)/(1+u)},\frac{\pm\sqrt{(1-u)/(1+u)}}{1^2+(1-u)/(1+u)}\right)\\ &=\left(u,\pm\frac12\sqrt{(1-u)(1+u)}\right)\end{split}$$ where the second coordinate of the RHS is the same as $v$ in the line immediately above the red circle.