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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 Mar 18 '18 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 Mar 18 '18 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 $1$? $\endgroup$ – user1101010 Mar 18 '18 at 19:41
  • $\begingroup$ Rouche's theorem should give such conditions. See my link above. $\endgroup$ – Peter Mar 18 '18 at 19:42
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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 $\,1\,$.

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  • $\begingroup$ Wish the downvoter had left a comment why. $\endgroup$ – dxiv Mar 19 '18 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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