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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?

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  • $\begingroup$ Maybe this helps : en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem $\endgroup$
    – Peter
    Commented Mar 18, 2018 at 19:24
  • $\begingroup$ For the root of a complex number. If $z=x+iy$, then the square roots have absolute value $v=(x^2+y^2)^{1/4}$ and the argument is $a=1/2\cdot \arctan(\frac{y}{x})$ (if $x\ne 0$ , if x=0, the argument is $a=\frac{\pi}{4}$). Then, the roots are $v\cdot(\cos(a)+i\cdot \sin(a))$ and $v\cdot (\cos(a+\pi)+i\cdot \sin(a+\pi))$ $\endgroup$
    – Peter
    Commented Mar 18, 2018 at 19:35
  • $\begingroup$ @Peter: Thanks. Is it possible to give a sufficient condition (on magnitude of $c$) such that the roots have magnitude smaller than $1$? $\endgroup$ Commented Mar 18, 2018 at 19:41
  • $\begingroup$ Rouche's theorem should give such conditions. See my link above. $\endgroup$
    – Peter
    Commented Mar 18, 2018 at 19:42
  • $\begingroup$ "If $c$ is real ... $4c^2 - 4c < 0$" $\;$ That's only true if $\,c \in (0,1)\,$. For example, if $\,c = -1/2\,$ then $\,|c| \lt 1\,$ but the equation $\,t^2 + t - 1/2 = 0\,$ has a real root $\,-(1 + \sqrt{3})/2 \lt -1\,$. $\endgroup$
    – dxiv
    Commented May 25, 2022 at 17:44

2 Answers 2

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(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 $\,3 c^4 - 8 c^3 + 6 c^2 - 1 \lt 0 \iff c \in \left(- \dfrac{1}{3}, \,1\right)\,$.

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  • $\begingroup$ Wish the downvoter had left a comment why. $\endgroup$
    – dxiv
    Commented Mar 19, 2018 at 15:54
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I found a counterexample : Let $c=0.7+0.6i$ , then $|c|<1$ , but the polynomial $t^2-2ct+c=0$ has a root with absolute value greater than $1$.

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