Proposition 1.4.1 If $z_1=\cos \alpha + i \sin \alpha$ and $z_2=\cos \beta + i \sin \beta$ , then $z_1z_2=\cos (\alpha + \beta) + i \sin (\alpha + \beta )$ .

Another formulation of this proposition follows:

$\displaystyle \forall z_1 \in \mathbb{C}, \forall z_2 \in \mathbb{C}, \; \opera... ...\nolimits (z_1) + \operatorname{arg}\nolimits (z_2) + 2k \pi, k \in \mathbb{Z}.$

Proof. Let $z_1 =\cos \theta_1 + i \sin \theta_1$ and $z_2= \cos \theta_2 + i \sin \theta_2$ . Then:

$\displaystyle z_1z_2$	$\displaystyle = (\cos \theta_1 + i \sin \theta_1)(\cos \theta_2 + i \sin \theta_2)$
$\displaystyle \quad$	$\displaystyle = \left( \cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2 + i \; (\cos \theta_1 \sin \theta_2 + \sin \theta_1 \cos \theta_2 )\right)$
$\displaystyle \quad$	$\displaystyle = \cos (\theta_1+\theta_2) + i \; \sin (\theta_1+\theta_2).$

$\qedsymbol$

Corollary 1.4.2

$\displaystyle \forall z_1 \in \mathbb{C}, \forall z_2 \in \mathbb{C} - \{0 \}, ... ...\nolimits (z_1) - \operatorname{arg}\nolimits (z_2) + 2k \pi, k \in \mathbb{Z}.$

Proof. For any $z_1 \in \mathbb{C}$ and $z_2 \in \mathbb{C} - \{ 0 \}$ , we have:

$\displaystyle \frac {z_1}{z_2} \cdot z_2 = z_1 \Longrightarrow \operatorname{ar... ...orname{arg}\nolimits (z_2) \right)= \operatorname{arg}\nolimits (z_1)+ 2 k \pi.$

The result follows. $\qedsymbol$

**Figure 8:** Abraham DeMoivre 1667-1754.
$\begin{figure}\mbox{ \epsfig{file= De_Moivre.eps,height=5cm}}\end{figure}$

Corollary 1.4.3 (De Moivre's formula) If $z=\cos \alpha + i \sin \alpha$ and is any natural number, then $z^n=\cos n \alpha + i \sin n \alpha$ .

Another formulation of this proposition follows:

$\displaystyle \forall z \in \mathbb{C} - \{ 0 \}, \forall n \in \mathbb{N}, \; ... ...\nolimits (z^n) = n \operatorname{arg}\nolimits (z) + 2k \pi, k \in \mathbb{Z}.$

Proof. We prove it by induction on

(i) For

, we have

and $\operatorname{arg}\nolimits (z^0)=0+2k \pi = 0 \cdot \operatorname{arg}\nolimits (z)+2k \pi.$
(ii) Suppose that for some natural number

, the equation $\operatorname{arg}\nolimits {z^n} = n \; \operatorname{arg}\nolimits (z) + 2k \pi, \; k \in \mathbb{Z}$ holds. Then, by Proposition 4.1 we have:

$\displaystyle \operatorname{arg}\nolimits (z^{n+1})=\operatorname{arg}\nolimits... ...orname{arg}\nolimits (z)+ 2k \pi= (n+1)\operatorname{arg}\nolimits (z)+ 2k \pi.$

$\qedsymbol$

Example 1.4.4 Let $z=\frac 12 + i \frac {\sqrt{3}}{2} = \cos \frac {\pi}{3} + i \sin \frac {\pi}{3}$ .

Then: $z^{26}= \cos \frac {26 \pi}{3} + i \sin \frac {26 \pi}{3} = \cos \frac {2\pi}{3} + i \sin \frac {2\pi}{3} =- \frac 12 + i \frac {\sqrt{3}}{2}$ .

This formula can be generalized to any integer .

Corollary 1.4.5

$\displaystyle \forall z \in \mathbb{C} - \{ 0 \}, \forall n \in \mathbb{Z}, \; ... ...\nolimits (z^n) = n \operatorname{arg}\nolimits (z) + 2k \pi, k \in \mathbb{Z}.$

Proof. For non negative

, this is exactly De Moivre's formula. We consider only the case where

is a negative integer (whence

$\displaystyle \operatorname{arg}\nolimits (z^n) = \operatorname{arg}\nolimits \... ...eratorname{arg}\nolimits (z)+2k \pi = n \operatorname{arg}\nolimits (z)+2k \pi.$

$\qedsymbol$

Example 1.4.6 Let $z=\sqrt{3} + i=2 \; (\cos \pi/6 +i \; \sin \pi /6)$ . Then

$\displaystyle z^{-4}$	$\displaystyle =2^{-4} \; \left( \cos \frac {4 \pi}{6} + i \; \sin \frac {4 \pi}... ...\frac {1}{16} \left( \cos \frac {2 \pi}{3} + i \; \sin \frac {3 \pi}{3} \right)$
$\displaystyle \quad$	$\displaystyle = \frac {1}{16} \left( -\frac 12 + i \frac {\sqrt{3}}{2} \right) =\frac {1}{32} \left( -1+i \; \sqrt{3} \right).$

Example 1.4.7 Let $z_0=-1+i \sqrt{3}$ . Find all the integers

such that

is real.

We have:

$\displaystyle z_0=-1+i \sqrt{3} = 2 \left( - \frac 12 + i \frac {\sqrt{3}}{2} \right) = 2 \left( \cos \frac {2 \pi}{3} + i \sin \frac {2 \pi}{3} \right).$

By De Moivre's formula (v.s. 4.3), we have:

$\displaystyle z_0^n = 2^n \left( \cos \frac {2n \pi}{3} + i \sin \frac {2n \pi}{3} \right).$

By 3.8, we have:

$\displaystyle z_0^n \in \mathbb{R} \Longleftrightarrow \exists k \in \mathbb{Z} , \; \frac {2n \pi}{3} = k \pi$

i.e.

$\displaystyle 2n=3k$

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