Nombres complexes

Opérations sous forme exponentielle : le produit (multiplication) - Exercice 1

5 min
15
Simplifier les calculs suivants en les mettant sous la forme reiθre^{i\theta }rr et θ\theta sont deux réels.
Question 1

z1=3eiπ2×2eiπ4z_{1} =3e^{i\frac{\pi }{2} } \times 2e^{i\frac{\pi }{4} }

Correction
Soient r\red{r}, r{\color{blue}{r'}}, θ\green{\theta } et θ\pink{\theta'} quatre réels, on a alors :
  • reiθ×reiθ=r×r×ei(θ+θ)\red{r}e^{i\green{\theta }} \times {\color{blue}{r'}}e^{i\pink{\theta '}} =\red{r}\times {\color{blue}{r'}}\times e^{i\left(\green{\theta} +\pink{\theta '}\right)}
  • z1=3eiπ2×2eiπ4z_{1} =\red{3}e^{i\green{\frac{\pi }{2}}} \times {\color{blue}{2}}e^{i\pink{\frac{\pi }{4}}} équivaut successivement à :
    z1=3×2eiπ2×eiπ4z_{1} =\red{3}\times {\color{blue}{2}}e^{i\green{\frac{\pi }{2}}} \times e^{i\pink{\frac{\pi }{4}}}
    z1=6ei(π2+π4)z_{1} =6e^{i\left(\green{\frac{\pi }{2}} +\pink{\frac{\pi }{4}} \right)}
    z1=6ei(2×π2×2+π4)z_{1} =6e^{i\left(\frac{2\times \pi }{2\times 2} +\frac{\pi }{4} \right)}
    z1=6ei(2π4+π4)z_{1} =6e^{i\left(\frac{2\pi }{4} +\frac{\pi }{4} \right)}
    Ainsi :
    z1=6ei3π4z_{1} =6e^{i\frac{3\pi }{4} }

    Question 2

    z2=4eiπ3×5ei5π6z_{2} =4e^{i\frac{\pi }{3} } \times 5e^{i\frac{5\pi }{6} }

    Correction
    Soient r\red{r}, r{\color{blue}{r'}}, θ\green{\theta } et θ\pink{\theta'} quatre réels, on a alors :
  • reiθ×reiθ=r×r×ei(θ+θ)\red{r}e^{i\green{\theta }} \times {\color{blue}{r'}}e^{i\pink{\theta '}} =\red{r}\times {\color{blue}{r'}}\times e^{i\left(\green{\theta} +\pink{\theta '}\right)}
  • z2=4eiπ3×5ei5π6z_{2} =\red{4}e^{i\green{\frac{\pi }{3}}} \times {\color{blue}{5}}e^{i\pink{\frac{5\pi }{6}}} équivaut successivement à :
    z2=4×5eiπ3×ei5π6z_{2} =\red{4}\times {\color{blue}{5}}e^{i\green{\frac{\pi }{3}}} \times e^{i\pink{\frac{5\pi }{6}}}
    z2=20ei(π3+5π6)z_{2} =20e^{i\left(\green{\frac{\pi }{3}} +\pink{\frac{5\pi }{6}} \right)}
    z2=20ei(2×π2×3+5π6)z_{2} =20e^{i\left(\frac{2\times \pi }{2\times 3} +\frac{5\pi }{6} \right)}
    z2=20ei(2π6+5π6)z_{2} =20e^{i\left(\frac{2\pi }{6} +\frac{5\pi }{6} \right)}
    Ainsi :
    z2=20ei7π6z_{2} =20e^{i\frac{7\pi }{6} }
    Question 3

    z3=6ei3π4×4eiπ6z_{3} =6e^{i\frac{3\pi }{4} } \times 4e^{-i\frac{\pi }{6} }

    Correction
    Soient r\red{r}, r{\color{blue}{r'}}, θ\green{\theta } et θ\pink{\theta'} quatre réels, on a alors :
  • reiθ×reiθ=r×r×ei(θ+θ)\red{r}e^{i\green{\theta }} \times {\color{blue}{r'}}e^{i\pink{\theta '}} =\red{r}\times {\color{blue}{r'}}\times e^{i\left(\green{\theta} +\pink{\theta '}\right)}
  • z3=6ei3π4×4eiπ6z_{3} =\red{6}e^{i\green{\frac{3\pi }{4}}} \times {\color{blue}{4}}e^{\pink{-}i\pink{\frac{\pi }{6}}} équivaut successivement à :
    z3=6×4ei3π4×eiπ6z_{3} =\red{6}\times {\color{blue}{4}}e^{i\green{\frac{3\pi }{4}}} \times e^{\pink{-}i\pink{\frac{\pi }{6}}}
    z3=24ei(3π4+(π6))z_{3} =24e^{i\left(\green{\frac{3\pi }{4}} +\left(\pink{-\frac{\pi }{6}} \right)\right)}
    z3=24ei(3π4π6)z_{3} =24e^{i\left(\frac{3\pi }{4} -\frac{\pi }{6} \right)}
    z3=24ei(3×3π3×42×π2×6)z_{3} =24e^{i\left(\frac{3\times 3\pi }{3\times 4} -\frac{2\times \pi }{2\times 6} \right)}
    z3=24ei(9π122π12)z_{3} =24e^{i\left(\frac{9\pi }{12} -\frac{2\pi }{12} \right)}
    Ainsi :
    z3=24ei7π12z_{3} =24e^{i\frac{7\pi }{12} }