This is the second part of the exercise $7$ on page $257$ of Analysis II of Amann and Escher.
Let $g:\Bbb R^4\to\Bbb R^3,\,(x,y,z,u)\mapsto (2(xz+yu),2(xu-yz),z^2+u^2-x^2-y^2)$. Verify that $g^{-1}(a)$ is an embedded curve in $\Bbb R^4$ for any $a\in\Bbb R^3\setminus\{0\}$.
I solved the exercise finding, explicitly, a function from $\varphi_a:(-\pi,\pi]\to\Bbb R^4$ such that it image is $g^{-1}(a)$, but it seems an overkill. Probably there is a more elegant solution or something that doesnt require to find explicitly this function.
This is my solution:
For any chosen $a\in\Bbb R^3\setminus\{0\}$ we have the system of equations $g_k=a_k$ for $g=(g_1,g_2,g_3)$ to see the form of $g^{-1}(a)$. If I set $v:=x+iy$ and $w:=z+iu$ then we need to solve $g(v,w)=(2\Re(v\bar w),-2\Im(v\bar w),|w|^2-|v|^2)=(a_1,a_2,a_3)$. Thus $v\bar w=\frac12(a_1-i a_2)$ and $|v|=\sqrt{|w|^2-a_3}$. And consequently $|w|\sqrt{|w|^2-a_3}=\frac12|a_1-i a_2|$ so $$ |w|^4-a_3|w|^2-\frac14|a_1-i a_2|^2=0\implies\\ |w|^2=\frac{a_3+\sqrt{a_3^2+a_1^2+ a_2^2}}{2},\quad |v|^2=\frac{-a_3+\sqrt{a_3^2+a_1^2+ a_2^2}}{2}\tag1 $$ Hence $|v||w|=\frac12\sqrt{a_1^2+a_2^2}$ and because $\Re(v\bar w)=|v||w|\cos\theta$ and $\Im(v\bar w)=|v||w|\sin\theta$ then we find the equations $$ \cos\theta=\frac{a_1}{\sqrt{a_1^2+a_2^2}},\quad\sin\theta=-\frac{a_2}{\sqrt{a_1^2+a_2^2}}\implies\theta={\rm atan2}(-a_2,a_1)\tag2 $$ where $\theta$ is the argument of $v\bar w$. Because $|v|$, $|w|$ and the angle $\theta$ (and the direction of the angle between them) are determined for any chosen $a:=(a_1,a_2,a_3)$ then the unique degree of freedom for the solutions of $g(v,w)=a$ is the direction of $v$, that is, the argument of $v$.
Setting $\alpha:=\arg v$ then it must exists a function $\varphi_a:(-\pi,\pi]\to\Bbb R^4$ such that ${\rm img}(\varphi_a)=g^{-1}(a)$. First observe that $v=|v|e^{i\alpha}=(x,y)$, thus $x=|v|\cos\alpha$ and $y=|v|\sin\alpha$. Also we have that $w=|w|e^{i(\alpha-\theta)}$ so $z=|w|\cos(\alpha-\theta)$ and $u=|w|\sin(\alpha-\theta)$. Thus the image of the function $\varphi_a:(-\pi,\pi]\to\Bbb R^4$ defined by $$ \varphi_a(\alpha):=\left(\sqrt{\frac{|a|-a_3}2}\cos\alpha,\sqrt{\frac{|a|-a_3}2}\sin\alpha,\sqrt{\frac{|a|+a_3}2}\cos(\alpha-\theta),\sqrt{\frac{|a|+a_3}2}\sin(\alpha-\theta)\right) $$ with $\theta:={\rm atan2}(-a_2,a_1)$ defines the embedded curve in $\Bbb R^4$ defined by $g^{-1}(a)$ for any chosen $a\in\Bbb R^3\setminus\{0\}$.
This exercise appeared in an introductory chapter to submanifolds so I guess that (probably) there is a better way to solve it using some of the stated theorems (such that the regular value theorem or the immersion theorem).
If possible provide a better way to solve this exercise (maybe just an sketch of the strategy to solve it). Thanks.
EDIT: reviewing the theory I guess that I should have used the regular value theorem.
Ok, I solved it using the regular value theorem. We have that
$$ [\partial g(\ldots)]=2\begin{bmatrix}z&u&x&y\\u&-z&-y&x\\-x&-y&z&u\end{bmatrix}\implies\begin{vmatrix}z&u&x\\u&-z&-y\\-x&-y&z\end{vmatrix}=-z(z^2+x^2+u^2+y^2) $$ and is easy to check that every determinant of any $3\times 3$ submatrix of $[\partial g(\ldots)]$ have the form $-c(x^2+y^2+z^2+u^2)$ for $c\in\{x,y,z,u\}$, and because it cannot be the case that $(x,y,z,u)=0$ for any chosen $a\in\Bbb R^3\setminus\{0\}$ then the rank of $[\partial g(\ldots)]$ is necessarily $3$ so $a$ is a regular value of $g$, and consequently $g^{-1}(a)$ is a $1$-dimensional $C^\infty$ submanifold of $\Bbb R^4$, that is, an embedded curve in $\Bbb R^4$.