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Prove that for any complex numbers: $a,b,c$ , $|a|=|b|=|c|=1$ and $abc \neq -1$ number $$x = \frac{a+b+c+ab+ac+bc}{1+abc}$$ is a real number.

I wanted to calculate $2 \cdot Im(x) = x- \overline x$ and show that it's equal to zero:

$$\frac{a+b+c+ab+ac+bc}{1+abc} - \frac{\overline{a}+\overline{b}+\overline{c}+\overline{ab}+\overline{ac}+\overline{bc}}{1+\overline{abc}} =$$ $$=a+b+c+ab+ac+bc-\overline{a}-\overline{b}-\overline{c}-\overline{ab}-\overline{ac}-\overline{bc} = $$ $$=2(Im(a) + Im(b)+Im(c)+Im(ab)+Im(ac)+Im(bc))=$$$$=2(Im(a+b+c+ab+bc+ac))$$But I have no idea what to do next or what am I missing.

Restless
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Vojtie
  • 143

3 Answers3

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$$x=\frac{a+b+c+ab+bc+ca}{1+abc}$$ $$=\frac{abc}{abc}\cdot\frac{a+b+c+ab+bc+ca}{1+abc}$$ $$=\frac{\frac{a+b+c+ab+bc+ca}{abc}}{\frac{1+abc}{abc}}$$ $$=\frac{\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}+\frac{1}{c}+\frac{1}{a}+\frac{1}{b}}{\frac{1}{abc}+1}$$ $$=\frac{\overline{a}+\overline{b}+\overline{c}+\overline{ab}+\overline{bc}+\overline{ca}}{1+\overline{abc}}$$ $$\overline{\left(\frac{a+b+c+ab+bc+ca}{1+abc}\right)}=\overline{x}$$

using if $|a|=1$ then $a\overline{a}=1$, and if $x=\overline{x}$, then $x\in\mathbb{R}$.

JMP
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Consider $(a+b+c+ab+ac+bc)(1+\overline {abc})$.

When you multiply out remember that $a\overline {a}=1$.

2

$$x = \frac{(a+1)(b+1)(c+1)-(1+abc)}{1+abc} = \frac{(a+1)(b+1)(c+1)}{1+abc}-1$$

and using the property $a \bar{a} = |a|^2 = 1$ repeatedly:

$$\frac{(a+1)(b+1)(c+1)}{1+abc} = \frac{(1+1/a)(1+1/b)(1+1/c)}{1/abc+1} = \frac{(1+\bar{a})(1+\bar{b})(1+\bar{c})}{\overline{abc}+1}$$

so $x = \bar{x}$ (why?) and hence $x$ is real.

Toby Mak
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