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?

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    $\begingroup$ You got the modulus wrong, $|x+iy| = \sqrt{x^2 \color{red}{+} y^2}\,$. $\endgroup$ – dxiv May 11 '17 at 22:25
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    $\begingroup$ @dxiv Alright, thanks. $\endgroup$ – Mark Read May 11 '17 at 22:26

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 $cis \theta$:

$cis \theta = \cos \theta + i \sin \theta, \tag{5}$


$\vert cis \theta \vert^2 = cis \theta \overline{cis \theta}= (\cos \theta + i \sin \theta)(\cos \theta - i \sin \theta) = \cos^2 \theta + \sin^2 \theta = 1, \tag{6}$

from which

$\vert cis \theta \vert= 1 \tag{7}$

readily follows.


First of all, for a complex number; $$z = x + iy = r(cos\theta + i sin\theta) = r(cis\theta)$$ $$|z| = (\sqrt {x^2 + y^2})= r $$

Now, as your question refers 'geometric images', let me introduce you to the complex plane(2-d for now) also known as the Argand plane, wherein you can imagine the complex number to be plotted just like an implicit function, the only difference being that 'x' in the complex number refers to the x axis and the value with 'iota' i.e. 'y' refers to the y-axis.

Now, the question comes-- Why represent it as $|cis(\theta)| $? The answer is very simple. If you know a little of vectors and straight lines I suggest you understand this very carefully :-

For the complex number represented as z, |z| is it's distance from the origin.

Simply, from origin, the distance of point (x,y) would be -- $(\sqrt {(x-0)^2 + (y-0)^2}) = (\sqrt {x^2 + y^2})$

which actually is the modulus of the complex number and is also represented as '$r$'. Take a look at this image here.

Now, having this knowledge, coming back to your question; we are given $$|cis\theta| = 1$$ Hence by "things" we have -- modulus i.e. distance from origin of the complex number to be one. God, make it easy...

$$|cis\theta| = 1 \implies r=1 \implies \sqrt{x^2+y^2} = 1 \implies x^2+y^2 =1$$ and voila it's the equation of a circle having it's center at origin and radius as 1.

It can also be looked at geometrically-intuitively (sort of) , as what would you call a variable point that has distance from the center (origin in this case) as 1 units. I suggest you look at the image again . See when we say $cis\theta$ wwe mean the point represented by $cos\theta+sin\theta$. Here modulus is defined as one.

So, when $\theta$ is variable we have a point that traces the locus of a circle (as is represented in the image) whose modulus is one (i.e. the radius).

Also, make a point of visiting WIKI page (Stack exchange doesn't allow me to ppost more than two links :! ).


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