# Magnitude of roots of a quadratic function with complex coefficients

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?

• Maybe this helps : en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem Mar 18, 2018 at 19:24
• 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 $v\cdot (\cos(a+\pi)+i\cdot \sin(a+\pi))$ Mar 18, 2018 at 19:35
• @Peter: Thanks. Is it possible to give a sufficient condition (on magnitude of $c$) such that the roots have magnitude smaller than $1$? Mar 18, 2018 at 19:41
• Rouche's theorem should give such conditions. See my link above. Mar 18, 2018 at 19:42
• "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 $\,-(1 + \sqrt{3})/2 \lt -1\,$.
– dxiv
May 25, 2022 at 17:44

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

• Wish the downvoter had left a comment why.
– dxiv
Mar 19, 2018 at 15:54

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 $1$.