<s>[INST] Complex roots of the polynomial $bz^{2}+2az+b$ are on the unit circle I want to show that the roots of the polynomial $bz^{2}+2az+b$ ($a,b$ are real) when $\left|\frac{a}{b}\right|\leq1$ (which is equivalent to the discriminant not being positive) are on the unit circle. The roots are given by $$z_{\pm}=\frac{-a\pm\sqrt{a^{2}-b^{2}}}{b}=-\frac{a}{b}\pm\sqrt{\left(\frac{a}{b}\right)^{2}-1}$$ I tried messing around with the modulus of this to reach the required result but without success. 
Following the hint setting: $$z_{+}=-\frac{a}{b}+\sqrt{\left(\frac{a}{b}\right)^{2}-1}=-\frac{a}{b}+i\sqrt{1-\left(\frac{a}{b}\right)^{2}}$$
$$z_{-}=-\frac{a}{b}-i\sqrt{1-\left(\frac{a}{b}\right)^{2}}=-\frac{a}{b}-i\sqrt{1-\left(\frac{a}{b}\right)^{2}}$$ 
We get that:
$$\left|z_{+}\right|^{2}=z_{+}\cdot\overline{z_{+}}=z_{+}\cdot z_{-}=\left(-\frac{a}{b}+i\sqrt{1-\left(\frac{a}{b}\right)^{2}}\right)\left(-\frac{a}{b}-i\sqrt{1-\left(\frac{a}{b}\right)^{2}}\right)$$
$$=\left(\left(\frac{a}{b}\right)^{2}-i^{2}\left(1-\left(\frac{a}{b}\right)^{2}\right)\right)=\left(\left(\frac{a}{b}\right)^{2}+\left(1-\left(\frac{a}{b}\right)^{2}\right)\right)=1$$
 [/INST] Let $z = \rho (\cos \theta + i \sin \theta)$ be a root of polynomial $z^2+cz+1$, where $c\triangleq 2a/b$. Then:
$$\rho^2\sin(2\theta)+c\rho\sin\theta = 0$$
and
$$\rho^2\cos(2\theta)+c\rho\cos\theta + 1 = 0.$$
Multiplying the first relation by $\cos\theta$ and the second by $\sin \theta$, and then subtracting, we get:
$$(\rho^2-1)\sin\theta = 0.$$
Hence $z$ either has modulus $1$ or it is real (excluded by the assumption that $|c|<2$ or making $z$ trivially $1$ or $-1$ if $|c|=2$ is assumed).
</s>