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I recently bought H.S. Kasana's Complex Variables. It seems quite interesting, and a little harder for me than I had expected, though I should be able to get through it if I take my time.

Nevertheless, I am having some trouble with the very first example... It starts like this:

Under what conditions on complex constants $\alpha$ and $\beta$, will the quadratic equation $z^2 + \alpha z + \beta = 0$ have (a) real roots, (b) one of the roots on the unit circle $|z| = 1$?

Suppose the quadratic equation has real roots, say $z = x$. Then, $$x^2 + \alpha x + \beta = 0 \;\;\;\;\;\; \text{and} \;\;\;\;\;\; x^2 + \bar{\alpha} x + \bar{\beta} = 0$$ Eliminating x, ...

The first equation is simply the original, but where did the second one (with the conjugates) come from? At first I though they might both be valid because the real part is the same for both, but the imaginary part still isn't, so that's not it.

I understand the rest of the example fine, but I would love to understand why the second equation can be added, so that I could solve such a problem by myself in the future.

Thanks :)

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  • $\begingroup$ He is just taking complex conjugates on both sides of the first equation. Also, $\overline{0} = 0$. $\endgroup$ – Prahlad Vaidyanathan Jul 20 '15 at 13:13
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Remember that the complex conjugation has this two properties: $$\bar z+\bar w=\overline{z+w}$$ $$\bar z\bar w=\overline{zw}$$

So, if $x$ is a real root, we have $\bar x=x$ and $$0=\bar 0=\overline{x^2+\alpha x+\beta}=\overline {x^2}+\overline{\alpha x}+\bar \beta=x^2+\bar\alpha x+\bar\beta$$

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  • $\begingroup$ Thanks! I just realized he simply took the conjugate... I think the equations still looked so different to me, that it seemed like their roots might not be the same... Being silly $\endgroup$ – dusty Jul 20 '15 at 13:34
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Just complex conjugate $$ x^2 + \alpha x + \beta = 0 $$ to get $$ \overline{x^2 + \alpha x + \beta} = \overline{x^2} + \overline{\alpha} \overline{x} + \overline{\beta} = \overline{0} = 0. $$ And since $x$ is real, $$ x^2 + \overline{\alpha} x + \overline{\beta} = 0. $$

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