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Question:- If equation $z^2+\alpha z + \beta=0$ has a real root, prove that $$(\alpha\bar{\beta}-\beta\bar{\alpha})(\bar{\alpha}-\alpha)=(\beta-\bar{\beta})^2$$


I tried goofing around with the discriminant but was unable to come with anything good. Just a hint towards a solution, might work.

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    $\begingroup$ If $z$ is a real root, then $z-\bar{z}=0$. Try taking the conjugate of the quadratic and doing some algebraic manipulations to remove reference to $z$. $\endgroup$ Jul 9 '16 at 17:06
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Eliminate $r$ between

$$r^2+r\alpha+\beta=0$$ and $$r^2+r\bar\alpha+\bar\beta=0.$$

By Cramer,

$$r^2=-\frac{\left|\begin{matrix}\beta&\alpha\\\bar\beta&\bar\alpha\end{matrix}\right|}{\left|\begin{matrix}1&\alpha\\1&\bar\alpha\end{matrix}\right|},$$ $$r=-\frac{\left|\begin{matrix}1&\beta\\1&\bar\beta\end{matrix}\right|}{\left|\begin{matrix}1&\alpha\\1&\bar\alpha\end{matrix}\right|}.$$

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Let the roots be $-x,-y$ with $x$ real. Then $\alpha=x+y$ and $\beta=xy$, hence

$$ \alpha\bar{\beta}-\beta\bar{\alpha}=(x+y)x\bar{y}-xy(x+\bar{y})=x^2(\bar{y}-y) \\ (\bar{\alpha}-\alpha)=\bar{y}-y \\ (\beta-\bar{\beta})^2 =x^2 (y-\bar{y})^2 \\ $$

It is easy to see that the product of the first two is the last.

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  • $\begingroup$ So $\alpha$ and $\beta$ are both real (since they are sums and products of real numbers) and hence both sides are trivially $0$? $\endgroup$
    – Zain Patel
    Jul 9 '16 at 17:20
  • $\begingroup$ @ZainPatel Only one root is real, the other one could be complex. I only assumed that $x$ is real, $y$ could be complex. In that case, $\alpha$ and $\beta$ cannot be real. $\endgroup$
    – N. S.
    Jul 9 '16 at 17:22
  • $\begingroup$ Ah, of course. Silly of me, thank you! $\endgroup$
    – Zain Patel
    Jul 9 '16 at 17:22
  • $\begingroup$ Is it not true that the roots of a quadratic occur in pairs (l.e. if one is complex, the other is just its conjugate, which is also complex)? $\endgroup$
    – Mick
    Jul 9 '16 at 19:05
  • $\begingroup$ @Mick No, it is not true. It is only true for quadratics with real coefficients. $\endgroup$
    – N. S.
    Jul 9 '16 at 23:27
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Let $x$ be a real root of the given equation. Then we have \begin{align*} x^2+\alpha x+\beta&=0\\ x^2+\overline\alpha x+\overline\beta&=0 \end{align*} and after subtracting we get \begin{gather*} (\alpha-\overline\alpha)x+(\beta-\overline\beta)=0\\ x=-\frac{\beta-\overline\beta}{\alpha-\overline\alpha} \end{gather*} Now let us plug this into the original equation \begin{align*} x^2&=-\alpha x -\beta\\ \frac{(\beta-\overline\beta)^2}{(\alpha-\overline\alpha)^2} &= \frac{\alpha(\beta-\overline\beta)}{\alpha-\overline\alpha} - \beta\\ \frac{(\beta-\overline\beta)^2}{(\alpha-\overline\alpha)^2} &= \frac{\alpha(\beta-\overline\beta)}{\alpha-\overline\alpha} - \frac{\beta(\alpha-\overline\alpha)}{\alpha-\overline\alpha}\\ \frac{(\beta-\overline\beta)^2}{(\alpha-\overline\alpha)^2} &= \frac{\beta\overline\alpha-\alpha\overline\beta}{\alpha-\overline\alpha} \\ (\beta-\overline\beta)^2 &= (\beta\overline\alpha-\alpha\overline\beta)(\alpha-\overline\alpha) \end{align*} This is basically what we wanted to prove. (In the given problem, the sign in both brackets is changed to opposite, which does not change the expression.)

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