show that :

1) $\displaystyle \overline {e^ { i\theta}} = {e^ {-i\theta}} $

proof:

$\displaystyle \overline {e^ {i\theta}} = cos\theta - isin\theta = cos(-\theta) + isin (-\theta ) = {e^ {-i\theta}} $

2)$\displaystyle \mid {e^ { i\theta}}\mid =1 $

proof:

$\displaystyle \mid {e^ { i\theta}}\mid =\sqrt{{cos^2}\theta +{sin^2}\theta} = \sqrt {1}=1 $