1) Définition
Tout nombre complexe z z z , de module r r r et d’argument θ \theta θ , s’écrit sous forme : z = r e i θ z = re^{i\theta} z = r e i θ .
Cette écriture est appelée notation exponentielle du nombre complexe z z z .
Exemple
Soit z = 2 + i 2 z = \sqrt{2} + i\sqrt{2} z = 2 + i 2 , on a ∣ z ∣ = 2 |z| = 2 ∣ z ∣ = 2 donc
z = 2 ( 2 2 + i 2 2 ) = 2 ( cos π 4 + i . sin π 4 ) \begin{align*}
z &= 2\left(\dfrac{\sqrt{2}}{2} + i\dfrac{\sqrt{2}}{2}\right)\\
&=2(\cos\dfrac\pi4+i.\sin\dfrac\pi4)
\end{align*} z = 2 ( 2 2 + i 2 2 ) = 2 ( cos 4 π + i . sin 4 π )
Donc ∣ z ∣ = 2 |z| = 2 ∣ z ∣ = 2 et arg ( z ) ≡ π 4 [ 2 π ] \arg(z) \equiv \dfrac{\pi}{4} ~[2\pi] arg ( z ) ≡ 4 π [ 2 π ] ,
alors : z = 2 e i π 4 z = 2e^{i\dfrac{\pi}{4}} z = 2 e i 4 π est la notation exponentielle de z z z .
Propriété
Soient r r r et θ \theta θ deux nombres réels.
Si r ≥ 0 r \ge 0 r ≥ 0 et z = r e i θ z = re^{i\theta} z = r e i θ , alors ∣ z ∣ = r |z| = r ∣ z ∣ = r et arg ( z ) ≡ θ ( m o d 2 π ) \arg(z) \equiv \theta \pmod{2\pi} arg ( z ) ≡ θ ( mod 2 π ) .
2) Opérations sur la notation exponentielle
Propriété
Soient r , r ′ ∈ R + ∗ r, r' \in \R_+^* r , r ′ ∈ R + ∗ et θ , θ ′ ∈ R \theta, \theta' \in \R θ , θ ′ ∈ R , on a :
r e i θ ‾ = r e − i θ \overline{re^{i\theta}} = re^{-i\theta} r e i θ = r e − i θ
− r e i θ = r e i ( θ + π ) -re^{i\theta} = re^{i(\theta + \pi)} − r e i θ = r e i ( θ + π )
1 r e i θ = 1 r ⋅ e − i θ \dfrac{1}{re^{i\theta}} = \dfrac{1}{r} \cdot e^{-i\theta} r e i θ 1 = r 1 ⋅ e − i θ
( r e i θ ) n = r n ⋅ e i n θ \left(re^{i\theta}\right)^n = r^n \cdot e^{in\theta} ( r e i θ ) n = r n ⋅ e in θ
r e i θ × r ′ e i θ ′ = r r ′ ⋅ e i ( θ + θ ′ ) re^{i\theta} \times r'e^{i\theta'} = rr' \cdot e^{i(\theta + \theta')} r e i θ × r ′ e i θ ′ = r r ′ ⋅ e i ( θ + θ ′ )
r e i θ r ′ e i θ ′ = r r ′ ⋅ e i ( θ − θ ′ ) \dfrac{re^{i\theta}}{r'e^{i\theta'}} = \dfrac{r}{r'} \cdot e^{i(\theta - \theta')} r ′ e i θ ′ r e i θ = r ′ r ⋅ e i ( θ − θ ′ )
Exercice
Donner la forme exponentielle des nombres complexes suivants :
z 1 = 2 + 2 i z_1 = 2 + 2i\quad z 1 = 2 + 2 i ; z 2 = 1 − i 3 \quad z_2 = 1 - i\sqrt{3} z 2 = 1 − i 3
z 1 × z 2 z_1 \times z_2 \quad z 1 × z 2 ; z 1 z 2 \quad\dfrac{z_1}{z_2} z 2 z 1
Correction
Pour z 1 = 2 + 2 i z_1 = 2 + 2i z 1 = 2 + 2 i :
∣ z 1 ∣ = 2 2 + 2 2 = 2 2 |z_1| = \sqrt{2^2 + 2^2} = 2\sqrt{2} ∣ z 1 ∣ = 2 2 + 2 2 = 2 2
donc
z 1 = 2 2 ( 1 2 + i 1 2 ) = 2 2 ( cos π 4 + i sin π 4 ) = 2 2 e i π 4 \begin{align*}
z_1 &= 2\sqrt{2} \left( \dfrac{1}{\sqrt{2}} + i \dfrac{1}{\sqrt{2}} \right) \\
&= 2\sqrt{2} \left( \cos \dfrac{\pi}{4} + i \sin \dfrac{\pi}{4} \right) \\
&= 2\sqrt{2} e^{i \dfrac{\pi}{4}}
\end{align*} z 1 = 2 2 ( 2 1 + i 2 1 ) = 2 2 ( cos 4 π + i sin 4 π ) = 2 2 e i 4 π
Pour z 2 = 1 − i 3 z_2 = 1 - i\sqrt{3} z 2 = 1 − i 3 :
∣ z 2 ∣ = 1 2 + ( 3 ) 2 = 2 |z_2| = \sqrt{1^2 + (\sqrt{3})^2} = 2 ∣ z 2 ∣ = 1 2 + ( 3 ) 2 = 2
donc
z 2 = 2 ( 1 2 − i 3 2 ) = 2 ( cos ( − π 3 ) + i sin ( − π 3 ) ) = 2 e − i π 3 \begin{align*}
z_2 &= 2 \left( \dfrac{1}{2} - i \dfrac{\sqrt{3}}{2} \right)\\
& = 2 \left( \cos \left( -\dfrac{\pi}{3} \right) + i \sin \left( -\dfrac{\pi}{3} \right) \right) \\
&= 2e^{-i\dfrac{\pi}{3}}
\end{align*} z 2 = 2 ( 2 1 − i 2 3 ) = 2 ( cos ( − 3 π ) + i sin ( − 3 π ) ) = 2 e − i 3 π
Produit z 1 × z 2 z_1 \times z_2 z 1 × z 2 :
z 1 × z 2 = ( 2 2 e i π 4 ) ⋅ ( 2 e − i π 3 ) = 4 2 e i ( π 4 − π 3 ) = 4 2 e i π 12 \begin{align*}
z_1 \times z_2 &= (2\sqrt{2} e^{i \dfrac{\pi}{4}}) \cdot (2 e^{-i \dfrac{\pi}{3}}) \\
&= 4\sqrt{2} e^{i\left(\dfrac{\pi}{4} - \dfrac{\pi}{3}\right)} \\
&= 4\sqrt{2} e^{i\dfrac{\pi}{12}}
\end{align*} z 1 × z 2 = ( 2 2 e i 4 π ) ⋅ ( 2 e − i 3 π ) = 4 2 e i ( 4 π − 3 π ) = 4 2 e i 12 π
Quotient z 1 z 2 \dfrac{z_1}{z_2} z 2 z 1 :
z 1 z 2 = 2 2 e i π 4 2 e − i π 3 = 2 e i ( π 4 − − π 3 ) = 2 e i 7 π 12 \begin{align*}
\dfrac{z_1}{z_2} &= \dfrac{2\sqrt{2} e^{i \dfrac{\pi}{4}}}{2 e^{-i \dfrac{\pi}{3}}} \\
&= \sqrt{2} e^{i \left( \dfrac{\pi}{4} - \dfrac{-\pi}{3} \right)} \\
&= \sqrt{2} e^{i \dfrac{7\pi}{12}}
\end{align*} z 2 z 1 = 2 e − i 3 π 2 2 e i 4 π = 2 e i ( 4 π − 3 − π ) = 2 e i 12 7 π