Regarding the roots of a quadratic polynomial Let $z^2-\alpha z+\beta$ be a complex quadratic polynomial where $|\alpha|<1$ and $|\beta|<1$. Then can we say that that if $z_1$ and $z_2$ are roots of this polynomial, then $|z_1|\leq 1$ and $|z_1|\leq 1$?
Is there any condition on $\alpha$, $\beta$ that I can put to get the desired claim?
I think $|\alpha|+|\beta|\leq 1$ would be such condition because of this post.
Could there be any other?
 A: According to Rouché's Theorem, if two functions $f$ and $g$ obey $|f|<|g|$ on the boundary of a domain then inside the domain $f$ and $f+g$ will have the same number of zeros. Here we can take $f=z^2$, g=$-\alpha z+\beta$ and note that $f$ has $2$ (multiple) roots  inside $|z|=1$, therefore $f+g=z^2 -\alpha z+\beta$ will have 2 roots there provided that $|f|<1$ that can be ensured if $|\alpha|+|\beta|<1$. A more accurate bound follows directly from the solution of the quadratic equation and can be written as:
$|\alpha+\sqrt{\alpha^2-4\beta}| \le2$. If we want to relax it a little we can also write $|\alpha|+|\alpha^2-4\beta|^{\frac{1}{2}}\le2$. By using Bernouly inequality $|\alpha^2-4\beta|^{\frac{1}{2}}\le|\alpha|+\frac{1}{2}|4\beta | $ we can also derive the condition $|\alpha|+|\beta|<1$.
A: We claim that if we look for a condition involving only the modulus of the coefficients, $\alpha, \beta$ it is indeed necessary and sufficient to have $|\alpha|+|\beta|\le 1$.
Fix $0<r<1$ and note that the equation $z^2+(c-r/c)iz+r=0$ has the roots $c/i, ri/c$ so in particular for $c=1$ where $|(c-r/c)i|+r=1$ we get a quadratic for which $|\alpha|+|\beta|=1$ but a root has modulus $1$ hence if we take $c=1+\epsilon$ then $|(c-r/c)i|+r=1+\epsilon + \epsilon r/(1-\epsilon)$ which can be made as close to $1$ as we wish by picking small enough $\epsilon$ and getting a quadratic of the requested type with a root outside the closed unit disc, so the condition $|\alpha|+|\beta|\le 1$ is necessary
For sufficiency, the easiest way is to notice that if $x=ce^{i\theta}, y=re^{i\alpha}/c, |x| \ge |y|$ are the two roots where $r=|\beta| <1$ and $c=|x| >0$ (assuming wlog the quadratic is not $z^2$ etc) one has
$1 \ge |\alpha|+|\beta|=|ce^{i\theta}+re^{i\alpha}/c|+r$, so by passing $r$ to the left and squaring we get
$c^2+r^2/c^2+2r\cos (\alpha-\theta) \le 1-2r+r^2$
and since $c^2+r^2/c^2+2r\cos (\alpha-\theta) \ge c^2+r^2/c^2-2r$, we obtain:
$c^2+r^2/c^2-2r \le 1-2r+r^2$ or $(c^2-r^2)(1-c^2) \ge 0$. Since $c^2>1$ implies $c^2 \le r^2<1$ which is contradictory, we clearly have $c \le 1$ hence $|y| \le |x|=c \le 1$ so we are done!
