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Let $z_1$ and $z_2$ be two complex numbers of modulus $1$. Denote by $Re$, the real part of a complex number.

Using maple, I believe that the sign of $Re((1-z_1)(1-z_2))$ is equal to the sign of $Re((z_1-1)(z_2-1)(1-z_1z_2))$. Is there a nice way to prove this?

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  • $\begingroup$ I think you can use $\Re(z)=\frac{z+\overline{z}}{2}$ with $z$ being all those expressions and conclude something from there. $\endgroup$
    – UserX
    Nov 1, 2014 at 19:27

2 Answers 2

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Express the complex numbers as

$$z_1=a_1+ib_1$$

$$z_2=a_2+ib_2$$

Expanding the products should tell you if the result is true.

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  • $\begingroup$ Unfortunately, the signs of the resulting expressions seem quite hard to determine?! $\endgroup$
    – llort
    Nov 1, 2014 at 20:13
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For $z_{1,2} \approx 1$ and $\Im z_{1,2}>0$, we have that $1-z_1$, $1-z_2$, $1-z_1z_2$ are approximately multiples of $i$ with slightly negative real part, i.e. their arguments are slightly above $\frac\pi2$. We conclude that $(1-z_1)(1-z_2)$ has argument $\approx \pi$ and negative real part, whereas $(1-z_1)(1-z_2)(1-z_1z_2)$ has argument slightly above $\frac32\pi$, i.e. positive real part.

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    $\begingroup$ $1-z_1$ etc. are approximately multiples of $i$ with slightly positive real part, I think. Which unfortunately spoils your nice argument. $\endgroup$
    – TonyK
    Nov 1, 2014 at 19:53

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