When $c$ is real and in the interval $[-1,1]$, the roots $z$ of $z^2-2cz+1=0$ have $|z|=1$. 
When $c$ is real and in the interval $[-1,1]$, the roots $z$ of $z^2-2cz+1=0$ have $|z|=1$; when $c$ is any other complex number, this equation has one root $z_1$ with the property that $|z_1|>1$ and a second root $z_2$ having $|z_2|<1$. Corroborate these statements and, in the latter case, identify $z_1$ and $z_2$ explicitly.
(Hint. Recall Exercise $4.12$ The determinations of $z_1$ and $z_2$ will vary, depending on whether $\operatorname{Re}c>0$, $\operatorname{Re}c=0$, or $\operatorname{Re}c<0$).

Exercise $4.12$ is: Complex Analysis - Bruce P . Palka - Ex: 4.12
When I solve $z^2-2cz+1=0$,  I get $z=c\pm\sqrt{c^2-1} $, but I do not know how to use the help and find the remaining roots, could someone help me please? Thank you very much.
 A: If $|c|\leq1$ then $z=c\pm\sqrt{1-c^2}i$.
Thus, we see that $$|z|=\sqrt{c^2+\left(\sqrt{1-c^2}\right)^2}=1.$$
If $|c|>1$ then for $c>1$ we have $|c+\sqrt{c^2+1}|>1$ and since $z_1z_2=1$,
for the second root $z_2$ we have $|z_2|<1$.
For $c<-1$ we have $|c-\sqrt{c^2-1}|>1$ and $|c+\sqrt{c^2-1}|<1$.
A: Hint:  $z=0$ is not a root, so dividing by $z \ne 0$ the equation becomes $z+1/z = 2c\,$. A root $z_1$ has modulus $1$ iff $\bar z_1 = 1 / z_1\,$, which then gives $z_1 + \bar z_1 = 2c\,$. Draw your conclusions on $c$ from here.
A: By Vieta's formulas, $z_1z_2=1$ $\implies$ $|z_1||z_2|=1$ $\implies$ either both $|z_1|=|z_2|=1$, or one of them is $>1$ and the other $<1$.
In the case $|z_1|=|z_2|=1$, since $z_1\bar{z}_1=z_1z_2=1$, we can easily see that $z_2=\bar{z}_1$, and so by Vieta's formulas again $2c=z_1+z_2=z_1+\bar{z}_1$ is a real number. Therefore for all non-real values of $c$, the absolute values of $z_1$ and $z_2$ fall into the second alternative above.
For real $c$, let's start with the case $c\in(-1,1)$. In this case $c^2-1<0$, so $\sqrt{c^2-1}$ is the square root of a negative number, and thus it's a pure imaginary number. Therefore the solutions of the quadratic equation are $z_{1,2}=c\pm\sqrt{c^2-1}=c\pm i\sqrt{1-c^2}$, with $\operatorname{Re}(z_{1,2})=c$ and $\operatorname{Im}(z_{1,2})=\pm\sqrt{1-c^2}$, and the absolute value of both of them is
$$|z_{1,2}|=\sqrt{c^2+(1-c^2)}=1,$$
as desired.
The cases of $c=-1$ and $c=1$ can be handled by directly plugging them in. Note that in these cases you end up with complete squares.
And finally, if $c$ is real and $|c|>1$, then both roots are real, and one of them is clearly greater than $1$ (so the other one is less than $1$).
