Show the roots of the quadratic equation $z^2 +bz+ c = 0$ lie in or on the unit circle

Question

So I need a little help with the following: Considering separately the cases of real and complex roots show that the roots of the quadratic equation $z^2 +bz+ c = 0$ lie in or on the unit circle (i.e. $|z_{i}|\leq 1$) if and only if $|c|\leq1$ and $|b|\leq 1+c$, where $b$ and $c$ are real coefficients.

I showed both sides of the relation for real roots, but now I am stuck on the complex case. Any ideas?

Answer 1

Since, roots are complex $b^2 \leq4c$. Hence, $c\geq0$. The roots are $$z_1,_2=(-b\pm i \sqrt{(4c-b^2)})/2$$We first assume roots lie in unit circle. Hence,$|z|^2\leq1$ which implies $$(b^2+(4c-b^2))/4\leq1\implies c\leq1, i.e, |c|\leq1$$ Also, $(1-c)^2\geq0\implies(1-c)^2+4c\geq4c\implies(1+c)^2\geq4c\implies(1+c)^2\geq b^2\implies|b|\leq 1+c$

Now, let's prove the other direction. Assume the conditions hold. Then,$$|z|^2=(b^2+(4c-b^2))/4=c\leq1$$ $$\implies |z|\leq1$$

Answer 2

We can consider that $c\geq 0$, because if $c<0$ then the equation has real roots. Now you have that

$$z=\cfrac{-b\pm i\sqrt{4c-b^2}}{2}$$

and so

$$|z|=\sqrt{\cfrac{b^2}{4}+\cfrac{4c-b^2}{4}}=\sqrt{c}\leq 1\Leftrightarrow c\leq 1$$

Answer 3

For the direct part, the equation $z^2+bz+c=0$ always has roots in complex field $\mathbb C$. Let $p$ and $q=\bar p$ are its roots, then product of roots is $p\bar p=c→|p|^2=|c|→|p|\le 1$. Thus, both roots lie within $|z|\le 1$.

For the converse part, assume that $z^2+bz+c=0$ has both its roots inside $|z|\le 1$. Let $p$ and $q$ be the roots s.t. $|p|\le 1$ and $|q|\le 1$ which gives $|p||q|\le 1$ or $|pq|\le 1$ i.e. $|c|\le 1$.

Also, if $p=u+iv$ then $q=u-iv$ then $|b|\le 1+c$ implies $2u\le 1+u^2 + v^2$ or $(u-1)^2+v^2\ge 0$ which holds for all real $u$ and $v$.