Show the roots of the quadratic equation $z^2 +bz+ c = 0$ lie in or on the unit circle 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?
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,$$|z|^2=(b^2+(4c-b^2))/4=c\leq1$$ $$\implies |z|\leq1$$
We can consider that $c\geq 0$, because if $c&lt;0$ then the equation has real roots. Now you have that 

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

and so

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