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?
Best Answer
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 $|\alpha|+|\beta|<1$.