<s>[INST] Show that the geometric images of any complex number defined by $cis \theta$ belong to a circumference of radius 1 
Show that the geometric images of any complex number defined by $cis
 \theta$ belong to the circumference centered at the origin with radius
  1.

My book states that this is proven because $|cis \theta|= 1$ . I tried:
$$|cis \theta| = \sqrt{\cos^2\theta+\sin^2\theta\cdot i^2} = \sqrt{\cos^2\theta-\sin^2\theta} = \sqrt{(\cos(2\theta)} = ??$$
What do I do next?
 [/INST] The modulus or norm of any complex number $z = a + bi$, $a, b \in \Bbb R$, is given by 
$\vert z \vert = \sqrt{a^2 + b^2}; \tag{1}$
using the identity
$x^2 - y^2 = (x + y)(x - y), \tag{2}$
which holds in any field (actually, in any commutative ring), and taking $x = a$, $y = bi$, and recalling that $i^2 = -1$, we find
$a^2 + b^2 = a^2 - (-b^2) = a^2 - (ib)^2 =$
$(a + ib)(a - ib) = z \bar z;  \tag{3}$
this may also be written
$\vert z \vert ^2 = z \bar z.  \tag{4}$
We apply the formula (4) to $cis \theta$:
$cis \theta = \cos \theta + i \sin \theta, \tag{5}$
whence
$\vert cis \theta \vert^2 = cis \theta \overline{cis \theta}= (\cos \theta + i \sin \theta)(\cos \theta - i \sin \theta) = \cos^2 \theta + \sin^2 \theta = 1, \tag{6}$
from which
$\vert cis \theta \vert= 1 \tag{7}$
readily follows.
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