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Let $z_1,z_2 \in \mathbb{C}$. Write $z_1=e^{it_1}$ and $z_2=e^{it_2}$. (Let their modulus be 1).

Then $$|z_1-z_2|^2=(z_1-z_2) \overline{(z_1-z_2)}=(z_1-z_2)(\overline{z_1} - \overline{z_2})=(e^{it_1}-e^{it_2})(e^{-it_1}-e^{-it_2})$$

But it seems like the last part of the equation does not give a real number. So what am I missing here, or what have I done wrong? Please help.

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3 Answers 3

up vote 5 down vote accepted

Sure it does: you get $2 - e^{i(t_1 - t_2)} - e^{-i(t_1 - t_2)} = 2 - (w + \bar w)$, where $w = e^{i(t_1 - t_2)}$, and the sum of a complex number and its conjugate is a real number.

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Shucks. I did not see that. –  ireallydonknow Dec 31 '13 at 11:22
    
Such things happen to the best of us from time to time. When in doubt, write it out! –  heropup Dec 31 '13 at 11:28

No error

$$(e^{it_1}-e^{it_2})(e^{-it_1}-e^{-it_2})=1-e^{i(t_1-t_2)}-e^{i(t_2-t_1)}+1=2-2\cos (t_2-t_1) \in \Bbb R$$

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$z_1=e^{it_1}$ and $z_2=e^{it_2}$.
Then $$|z_1-z_2|^2=(z_1-z_2) \overline{(z_1-z_2)}=(z_1-z_2)(\overline{z_1} - \overline{z_2})=(e^{it_1}-e^{it_2})(e^{-it_1}-e^{-it_2})$$.
$e^{i(t_1-t_1)}-e^{i(t_1-t_2)}-e^{i(t_2-t_1)}+e^{i(t_2-t_2)}$
$e^{0}-e^{i(t_1-t_2)}-e^{i(t_2-t_1)}+e^{0}$
$1+1-e^{i(t_1-t_2)}-e^{i(t_2-t_1)}$
$2-e^{i(t_1-t_2)}-e^{i(t_2-t_1)}$ (and since the sum of a complex number and its conjugate is a real number e.g $$xx^{*}=|x|^2$$, where $x^{*}$ is the conjugate of $x$). Hence, it gives a real number.

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