Consider the function $q(z)=\frac{1}{2}(z+z^{-1})$ for all $z\in\mathbb{C}\setminus{\{0}\}$, how many points do you have $q^{-1}(z)$?

Question

Consider the function $q(z)=\frac{1}{2}(z+z^{-1})$ for all $z\in\mathbb{C}\setminus{\{0}\}$ show that if $z\in\mathbb{C}\setminus{\{-1,1}\}$ then $q^{-1}(z)$ It has at least two elements.

Idea. I think that if $z\in\mathbb{C}\setminus[-1,1]$ then one of the points must be in the unit disk and the other in the complement of the disk. But I don't see how to show it, can anyone help me?

Answer 1

If

$z \in q^{-1}(y), \tag 0$

then

$q(z) = \dfrac{1}{2}(z + z^{-1}) = y, \tag 1$

whence

$z + z^{-1} = 2y, \tag 2$

or

$z^2 - 2yz + 1 = 0; \tag 3$

the values of $z$ satisfying this quadratic equation are given by the well-known formula

$z = \dfrac{2y \pm \sqrt{4y^2 - 4}}{2} = \dfrac{2y \pm 2 \sqrt{y^2 - 1}}{2} = y \pm \sqrt{y^2 - 1}; \tag 4$

we thus see that if

$y \notin \{-1, 1\} \tag 5$

we have

$\vert q^{-1}(y) \vert = 2, \tag 6$

that is, there are precisely $2$ elements in each $q^{-1}(y)$.