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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.

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1 Answer 1

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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 $(3)$.

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  • $\begingroup$ +1 Thanks for the answer! So I get $\frac 34$. I will accept in 24 hrs. $\endgroup$ Commented Apr 30, 2018 at 17:36
  • $\begingroup$ @SmarthBansal That's correct. Glad it helped. $\endgroup$
    – dxiv
    Commented Apr 30, 2018 at 17:37
  • $\begingroup$ Could using $e^{i\theta}$ notation help in solving this question?; (Just wondering) $\endgroup$ Commented Apr 30, 2018 at 17:39
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    $\begingroup$ @SmarthBansal I don't see offhand how that would work out. There are several ways to use the given condition that $\,|\alpha|=1\,$, and writing it in polar form is one such way, indeed. But in this case the more useful way is to use that $\,\overline \alpha = 1 / \alpha\,$. $\endgroup$
    – dxiv
    Commented Apr 30, 2018 at 17:45

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