If $aZ^2 + bZ + 1 = 0$ and $|a|=0.5$. What is the value of $|ab' - b|$?

Question

If $aZ^2 + bZ + 1 = 0$ where a,b,Z are complex numbers ; $|a|=\frac 12$ and have a root $\alpha$ such that $ |\alpha|=1$, then what is the value of $|ab' - b|$?
(I use $Q'$ to represent the conjugate of $Q$)

Since $\alpha$ is a root. I might as well put it in place of $Z$.
$$a\alpha^2 + b\alpha + 1 = 0$$ It does seem that conjugating the original equation might help. We get
$$a'\alpha'^2 + b'\alpha' + 1 = 0$$
This doesn't seem it could go further.
We could try putting $ \alpha = e^{i\theta}$ since it's magnitude is one. But again I don't see a way to proceed.

All help will be appreciated.

Answer 1

It does seem that conjugating the original equation might help.

That's the right idea, indeed:

$$\overline a \,\overline \alpha^2 + \overline b \overline \alpha + 1 = 0 \tag{1}$$

Multiplying $(1)$ by $\alpha^2$ and using that $\alpha \overline \alpha = |\alpha|^2= 1\,$:

$$\alpha^2 + \overline b \alpha + \overline a = 0 \tag{2}$$

Multiplying $(2)$ by $a$ and subtracting from the original equation $a\alpha^2 + b\alpha + 1 = 0$:

$$ (b-a \overline b)\alpha+ 1 - |a|^2 = 0 \quad\iff\quad b- a \overline b = \frac{1}{\alpha}\left( -1 + |a|^2\right) \tag{3} $$

Then all that's left is to take the modulus on both sides of $(3)$.