Polar form.

We denote by $ \mathbb{U}$ the set of all complex numbers whose absolute value is equal to 1.

$\displaystyle \mathbb{U} = \{ z \in \mathbb{C} \; \vert \; \vert z\vert=1 \}$    

For example, $ i \in \mathbb{U}$ and $ \frac 45 + \frac 34 i \in \mathbb{U}$ , but $ 1+i \not\in \mathbb{U}$ .

The image of $ \mathbb{U}$ in the Cauchy-Argand plane is the unit circle: if $ z=x+iy$ , with real $ x,y$ , then $ z
\in \mathbb{U}$ if, and only if $ x^2+y^2=1$ .

Proposition 1.3.1   Let $ z
\in \mathbb{U}$ ; then $ z^{-1}$ exists and is an element of $ \mathbb{U}$ .

We leave the proof to the reader; use Corollary 1.18.

Proposition 1.3.2   For any $ z_1 \in \mathbb{U}$ and $ z_2 \in \mathbb{U}$ , we have $ z_1z_2 \in \mathbb{U}$ .

Proof. For any $ z_1 \in \mathbb{U}$ and $ z_2 \in \mathbb{U}$ , we have:

$\displaystyle \vert z_1z_2\vert=\vert z_1\vert \; \vert z_2\vert=1 \cdot 1=1,$    

whence $ z_1z_2 \in \mathbb{U}$ . $ \qedsymbol$

And now we will see how complex numbers are tied with trigonometry.

Theorem 1.3.3   Let $ z
\in \mathbb{U}$ . There exists a real number $ \theta$ such that $ z= \cos \theta + i \sin \theta$ .

Proof. Denote $ z=x+iy$ , where $ x$ and $ y$ are real numbers. Then $ z
\in \mathbb{U}$ if, and only if, $ x^2+y^2=1$ , i.e. the image of $ z$ in the complex plane is a point on the unit-circle. For each point on the unit-circle, there exist a real number $ \theta$ such that the coordinates of this point are $ (\cos \theta, \; \sin \theta )$ , whence the result. $ \qedsymbol$

Definition 1.3.4   The number $ \theta$ is called an argument of $ z$ and is denoted $ \operatorname{arg}\nolimits (z)$ .

Note that this argument is defined up to an additional $ 2k \pi, k \in \mathbb{Z}$ .

Example 1.3.5  

$\displaystyle \frac 12 + i \frac {\sqrt{3}}{2}$ $\displaystyle = \cos \frac {\pi }{3} + i \sin \frac {\pi }{3} \Longrightarrow \...
...2 + i \frac {\sqrt{3}}{2} \right) = \frac {\pi }{3} + 2k \pi, k \in \mathbb{Z}.$    
$\displaystyle i$ $\displaystyle = \cos \frac {\pi }{2} + i \sin \frac {\pi }{2} \Longrightarrow \operatorname{arg}\nolimits (i) = \frac {\pi }{2} + 2k \pi, k \in \mathbb{Z}.$    

In Fig 5, a value of the argument of the complex number corresponding to a point is diplayed in green.

Figure 5: The unit circle in Cauchy-Argand plane.
\begin{figure}\mbox{\epsfig{file=UnitCircle.eps,height=5cm}}\end{figure}

We can now generalize this to any non zero complex number. First note that for any $ z \in \mathbb{C}^*$ , we have $ \frac {z}{\vert z\vert} \in \mathbb{U}$ , as the following holds:

$\displaystyle \lvert \frac {z}{\vert z\vert} \rvert = \frac {\lvert z\rvert }{\lvert z\rvert }=1.$    

Definition 1.3.6   For any $ z \in \mathbb{C}^*$ ,

$\displaystyle \operatorname{arg}\nolimits (z)=\operatorname{arg}\nolimits \left(\frac {z}{\vert z\vert} \right).$    

Figure 6: The trigonometric form of a non zero complex number.
\begin{figure}\mbox{\epsfig{file=UnitCircle2.eps,height=4.5cm}}\end{figure}

Example 1.3.7  
  1. Take $ z=1+i$ . The absolute value of $ z$ is given by $ \lvert z\rvert =\sqrt{1^1+1^2}=\sqrt{2}$ . Then we have:

    $\displaystyle z= \sqrt{2}\; \left( \frac {1}{\sqrt{2}}+i \; \frac {1}{\sqrt{2}} \right) = \sqrt{2}\; ( \cos \frac {\pi}{4} + i \; \sin \frac {\pi}{4}).$    

    It follows that $ \operatorname{arg}\nolimits (z)=\frac {\pi}{4} + 2k \pi$ .
  2. Now take $ z=-1+i \sqrt{3}$ , then $ \lvert z\rvert =\sqrt{1+3}=2$ and the following hold:

    $\displaystyle z=2 \; \left( - \frac 12 + i \; \frac{\sqrt{3}}{2} \right)= 2 \left( \cos \frac {2 \pi}{3} + i \; \sin \frac {2 \pi}{3} \right).$    

    It follows that $ \operatorname{arg}\nolimits (z)= \frac {2 \pi}{3} + 2k \pi$ .

The following proposition is easy to understand. It explains the specific role of the coordinate axes when representing complex numbers in the Cauchy-Argand plane.

Proposition 1.3.8  
(i)
$ z \in \mathbb{R} \Leftrightarrow \operatorname{arg}\nolimits (z)=k \pi, \; k \in \mathbb{Z}$ .
(ii)
$ z$ is pure imaginary $ \Leftrightarrow \operatorname{arg}\nolimits (z)= \frac {\pi}{2} +k \pi, \; k \in \mathbb{Z}$ .

Figure 7: Particular cases.
\begin{figure}\mbox{\epsfig{file=images.eps,height=4cm}}\end{figure}

Proof. Let $ z \in \mathbb{C}^*$ ; denote $ \operatorname{arg}\nolimits (z)= \theta + 2k \pi, \; k \in \mathbb{Z}$ . Then:
(i)
$ z \in \mathbb{R} \Leftrightarrow \operatorname{Im}\nolimits (z) =0 \Leftrighta...
...t \sin \theta =0 \Leftrightarrow \sin \theta =0
\Leftrightarrow \theta = 2k \pi$ .
(ii)
Please do it yourself.
$ \qedsymbol$

Noah Dana-Picard 2007-12-24