Suppose $c \in \mathbb C$ with $|c| < 1$. I constructed a quadratic function $t^2 – 2 c t + c = 0$. I want to know whether the magnitude of the roots are smaller than $1$. The answer for real $c$ is simple. If $c$ is real, then the roots are $c \pm \frac{\sqrt{4c^2 – 4c}}{2}$. Since $4c^2 – 4c < 0$, the second part is imaginary. So the magnitude will be $\sqrt{c^2 + \frac{4c-4c^2}{4}} = \sqrt{c} < 1$.
I got lost when considering $c$ is complex. Specifically, is the discriminant $4c^2 – 4c$ or $4|c|^2 – 4c$? How do we take the root of complex number?
Best Answer
(Too long for a comment.)
The equation can be written as $\,(t-c)^2 = c^2-c\,$ then by the triangle inequality with $\lambda=|c| \lt 1\,$:
$$ |t-c|^2 = |c|\,|1-c| \le |c|(1+|c|) \quad\implies\quad |t| \le |t-c|+|c| \le \lambda + \sqrt{\lambda(1+\lambda)} $$
Therefore $\,f(\lambda)=\lambda + \sqrt{\lambda(1+\lambda)}\,$ is an upper bound for the magnitude of roots $\,|t|\,$, but it does not insure that $\,|t| \le 1\,$ since $\,f(\lambda)\,$ can take values larger than $\,1\,$ e.g. $\,f(\lambda) \gt 1\,$ for $\,\forall \lambda \gt \frac{1}{3}\,$.
It also follows that $\,|c| \lt \frac{1}{3}\,$ is a sufficient condition for the roots to have magnitude less than $\,1\,$.
[ EDIT ] $\;$ My answer here covers this as the case $\,\beta = \gamma = c\,$. The necessary and sufficient conditions for both roots to be inside the unit circle, according to $\,(8)\,$ in that post:
$$ 2 \left(|c|^2 + |c^2 - c|\right) - 1 \;\lt\; |c|^2 \;\lt\; 1 $$
Here $\,|c| \lt 1\,$, so the second inequality is always satisfied. This leaves the condition:
$$ \begin{align} 2 \left(|c|^2 + |c^2 - c|\right) - 1 &\;\lt\; |c|^2 \quad\iff\quad |c|^2 + 2\,|c^2 - c| \lt 1 \end{align} $$
After straightforward algebraic manipulations, the condition reduces to:
$$ 3\,|c|^4 + 2\, \big(3 - 4\,\text{Re}(c)\big)\,|c|^2 \lt 1 $$
For real $\,c\,$, the condition reduces to $\,3 c^4 - 8 c^3 + 6 c^2 - 1 \lt 0 \iff c \in \left(- \dfrac{1}{3}, \,1\right)\,$.