If $az^2+bz+1=0$ where $a,b,z\in\mathbb{C}$ and $|a|=\frac{1}{2},$ have a root $\alpha$ such that $|\alpha|=1$ . Then $|a\bar{b}-b|$ equals
Try: let other root is $\beta$. Then $$\alpha+\beta=-\frac{b}{a}$$ and $$\alpha\beta=\frac{1}{a}$$
Now $\displaystyle |\alpha\beta|=\frac{1}{|a|}\Rightarrow |\beta|=2$.
Could some help me to solve it, Thanks
Hint:
$$a \,\alpha^2 + b \,\alpha + 1 = 0 \tag{1}$$
Multiply by $\,\bar a\,$ and use that $\,|a|=\dfrac{1}{2}\,$:
$$\dfrac{1}{4} \alpha^2 + \bar a b \,\alpha + \bar a=0$$
Take conjugates on both sides:
$$\dfrac{1}{4} \bar \alpha^2 + a \bar b \,\bar \alpha + a=0$$
Multiply by $\,\alpha^2\,$ and use that $\,|\alpha|=1\,$:
$$a \,\alpha^2 + a \bar b \,\alpha + \dfrac{1}{4} = 0 \tag{2}$$
Compare $(1)$ and $(2)$.