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I don't understand how you would take the conjugate of a quadratic equation and how it would be useful to solve this question.

I would normally show it by saying if $b$ is real, then it is equal to $\alpha$ times $\beta$, say $a$ is equal to $-(\alpha + \beta)$ and then just show it, but this doesn't involve conjugates as far as I can tell.

Siong Thye Goh
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Xemor
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2 Answers2

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Note that the conjugate of $z^2$ is $\bar z^2$

For real numbers $a$ and $b$ the conjugate of $z^2 + az + b$ is $\bar z^2 + a\bar z + b$

Therefore if z is a solution of $z^2 + az + b=0$ so is $\bar z$, which means the complex solutions appear in conjugate pairs.

Mohammad Riazi-Kermani
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  • +1: nicely phrased. Minor quibble: your answer overloads the $z$ variable. I would have phrased things differently : let $f(z) = z^2 + az + b$ and suppose that $f(z_0) = 0$. This implies that $\overline{f(z_0)} = 0.$ However, [as indicated in your analysis] $\overline{f(z_0)} = f\left( ~\overline{z_0} ~\right).$ Therefore, the assumption that $f(z_0) = 0$ implies that $f\left( ~\overline{z_0} ~\right) = 0.$ – user2661923 Oct 30 '21 at 17:34
  • Good comment, thanks. – Mohammad Riazi-Kermani Oct 30 '21 at 18:05
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You have that $z^2 + az + b = 0$. The given proposition only holds for real $a$ and real $b$. This needs to be stated.

You need to either show or assume the following properties of complex conjugation:

$\overline {z + w} = \overline z + \overline w$ (conjugate of sum is sum of conjugates),

$\overline {zw} = \overline z\cdot \overline w$ (conjugate of product is product of conjugates),

and

$z = w \implies \overline z = \overline w$ (equality implies equality of conjugates)

All of these are trivial to show (if you need to) by letting $z = x + yi$ and $w = a + bi$. The second one needs some algebraic expansion.

A trivial corollary of the second (when $z = w$) is that $\overline z^2 = (\overline z)^2$

Also, the first and second one can be extended to the sum and product (respectively) of an arbitrary number of terms. You'll be doing that with the first. Justifying this is easy with the associative property of addition.

Anyway, once all that preliminary work done and established as lemmas (or assumed), you can simply invoke the third lemma to take the conjugate of both sides of the quadratic. Then invoke the first two lemmas to immediately give $(\overline z)^2 + a\overline z + b = 0$

From that, you can immediately conclude that $\overline z$ is a solution to the quadratic as well. This leaves two possibilities: $z = \overline z$ which implies real root(s) or $z \neq \overline z$ which implies conjugate non-real complex roots. Thus the proof is done.

Deepak
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