Suppose roots of a polynomial of degree 2 is given by $z = \frac{1}{2}(\phi_1 \pm \sqrt{\phi_1^2 + 4\phi_2})$ and has to lie within the unit circle (i.e. $|z| < 1$).
Case 1: when roots are distinct $\phi_1^2 + 4\phi_2 > 0$:
I was able to show that $\phi_2 - \phi_1 < 1$ and $\phi_2 + \phi_1 <1$. However, I still need to show that $\phi_2 > -1$.
Case 2: when roots are complex $\phi_1^2 + 4\phi_2 < 0$, I need to show $\phi_2 > -1$.
Any hints to get these inequalities?
$z_{1,2} = \frac{1}{2}\left(\phi_1 \pm \sqrt{\phi_1^2 + 4\phi_2}\right)\,$ are the roots of $\,z^2-\phi_1 z - \phi_2=0\,$, so $\,z_1z_2=-\phi_2\,$ by Vieta's relations, and therefore:
$$ |z_1|, |z_2| \lt 1 \quad\implies\quad |z_1z_2|=|\phi_2| \lt 1 \iff -1 \lt \phi_2\lt1 \tag{1} $$
It follows that $\,\phi_2 \in (-1,1)\,$ is a necessary condition for both roots to lie inside the unit circle.
For real roots, the condition $\,(1)\,$ alone is not sufficient, for example $\,z^2 - \frac{7}{3}z+\frac{2}{3}\,$ has $\,\phi_2 \gt -1 \,$ but one roots is $\,z=2\,$ outside the unit circle.
If the roots are complex, then they must be conjugate $\,z_2 = \overline{z_1}\,$, so $\,-\phi_2 = z_1 z_2 = |z_1|^2\,$ and therefore $\,|z_{1}| \lt 1 \iff \phi_2 \in (-1,0]\,$, so the condition $\,(1)\,$ is also sufficient.