Range of $A$ in $z^3-Az^2+Az-1=0\;,$ If it has $3$ roots of equal magnitude. 
The ploynomial $z^3-Az^2+Az-1=0$ has three roots of equal modulus. Then Range of $A$ is
Where $z$ is a complex number.

$\bf{My\; Try::}$ Let $\alpha,\beta,\gamma$ be the roots of above equation.
So $$\alpha+\beta+\gamma= A\;\;,\alpha\beta+\beta\gamma+\gamma\alpha = A$$ and $\alpha\beta\gamma = 1$
Now Given $$|\alpha| = |\beta| = |\gamma| = k$$
So from $$\left|\alpha\beta\gamma\right| = 1\Rightarrow |\alpha||\beta||\gamma| = 1\Rightarrow k^3=1\Rightarrow k = 1$$
Now from $$|\alpha + \beta+\gamma|\leq |\alpha|+|\beta|+|\gamma| = 1+1+1 = 3$$
So we get $|A|\leq 3$
But answer given as $-1 \leq A\leq 3$
So plz help me how can I calculate lower bond of $A$
Thanks
 A: An engaging question worth answering, even though the solution to the problem is evident from A.G.'s hint.  So, lauds to A.G. and:
setting
$p_A(z) = z^3 - Az^2 + Az -1 \tag{1}$
for any $A \in \Bbb C$, we see that
$p_A(1) = 1^3 - A(1^2) + A(1) -1 = 0;  \tag{2}$
$1$ is always a zero of $p_A(z)$; let us denote the other two roots by $\rho_1, \rho_2 \in \Bbb C$; then as pointed out by juantheron in the text of the question,
$\rho_1 \rho_2 = \rho_1 \rho_2 (1) = 1.  \tag{3}$
Since
$\vert \rho_1 \vert = \vert \rho_2 \vert = 1, \tag{4}$
we see via (3) that we may take
$\rho_2^{-1} = \rho_1 = e^{i \theta} \tag{5}$
for some real $\theta \in [0, 2\pi)$; thus
$\rho_2 = \rho_1^{-1} = e^{-i \theta}; \tag{6}$
in fact, we may restrict $\theta \in [0, \pi] \subset [0, 2\pi)$, since the roles of $\rho_1$, $\rho_2$ interchange one with the other as $\theta$ passes through $\pi$.
We further see it follows that $A \in \Bbb R$; we need not assume this. Indeed,
$A = 1 + e^{i \theta} + e^{-i \theta}$
$= 1 + 2\cos \theta. \tag{7}$
Finally, as $\theta$ varies over $[0, \pi]$, $\cos \theta$ decreases monotonically 'twixt $1$ and $-1$; $A$ in turn, decreases from $3 = 1 + 2 \cos 0$ down to $-1 = 1 + 2 \cos \pi$; $A \in [-1, 3]$.
We note in closing that we also have the coefficient of $z$ in $p_A(z)$: 
$A = e^{i \theta} e^{-i \theta}+ (1)e^{i \theta}+ (1)e^{-i \theta}$
$= 1 + 2\cos \theta, \tag{8}$
consistent with the coefficient of $z^2$.
