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Nombres complexes
Opérations sous forme exponentielle : le quotient (division) - Exercice 1
10 min
25
Simplifier les calculs suivants en les mettant sous la forme
r
e
i
θ
re^{i\theta }
r
e
i
θ
où
r
r
r
et
θ
\theta
θ
sont deux réels.
Question 1
z
1
=
4
e
i
π
3
2
e
i
π
4
z_{1} =\frac{4e^{i\frac{\pi }{3} } }{2e^{i\frac{\pi }{4} } }
z
1
=
2
e
i
4
π
4
e
i
3
π
Correction
Soient
r
\red{r}
r
,
r
′
{\color{blue}{r'}}
r
′
,
θ
\green{\theta }
θ
et
θ
′
\pink{\theta'}
θ
′
quatre réels avec
r
′
≠
0
{\color{blue}{r'\ne0}}
r
′
=
0
, on a alors :
r
e
i
θ
r
′
e
i
θ
′
=
r
r
′
×
e
i
(
θ
−
θ
′
)
\frac{\red{r}e^{i\green{\theta }} } {{\color{blue}{r'}}e^{i\pink{\theta '}}} =\frac{\red{r}}{{\color{blue}{r'}}}\times e^{i\left(\green{\theta} -\pink{\theta '}\right)}
r
′
e
i
θ
′
r
e
i
θ
=
r
′
r
×
e
i
(
θ
−
θ
′
)
z
1
=
4
e
i
π
3
2
e
i
π
4
z_{1} =\frac{\red{4}e^{i\green{\frac{\pi }{3} }} }{{\color{blue}{2}}e^{i\pink{\frac{\pi }{4} }} }
z
1
=
2
e
i
4
π
4
e
i
3
π
équivaut successivement à :
z
1
=
4
2
×
e
i
(
π
3
−
π
4
)
z_{1} =\frac{\red{4}}{{\color{blue}{2}}} \times e^{i\left(\green{\frac{\pi }{3}} -\pink{\frac{\pi }{4}} \right)}
z
1
=
2
4
×
e
i
(
3
π
−
4
π
)
z
1
=
2
e
i
(
π
3
−
π
4
)
z_{1} =2e^{i\left(\frac{\pi }{3} -\frac{\pi }{4} \right)}
z
1
=
2
e
i
(
3
π
−
4
π
)
z
1
=
2
e
i
(
4
×
π
4
×
3
−
π
×
3
4
×
3
)
z_{1} =2e^{i\left(\frac{4\times \pi }{4\times 3} -\frac{\pi \times 3}{4\times 3} \right)}
z
1
=
2
e
i
(
4
×
3
4
×
π
−
4
×
3
π
×
3
)
z
1
=
2
e
i
(
4
π
12
−
3
π
12
)
z_{1} =2e^{i\left(\frac{4\pi }{12} -\frac{3\pi }{12} \right)}
z
1
=
2
e
i
(
12
4
π
−
12
3
π
)
Ainsi :
z
1
=
2
e
i
π
12
z_{1} =2e^{i\frac{\pi }{12} }
z
1
=
2
e
i
12
π
Question 2
z
2
=
5
e
i
π
2
3
e
i
π
6
z_{2} =\frac{5e^{i\frac{\pi }{2} } }{3e^{i\frac{\pi }{6} } }
z
2
=
3
e
i
6
π
5
e
i
2
π
Correction
Soient
r
\red{r}
r
,
r
′
{\color{blue}{r'}}
r
′
,
θ
\green{\theta }
θ
et
θ
′
\pink{\theta'}
θ
′
quatre réels avec
r
′
≠
0
{\color{blue}{r'\ne0}}
r
′
=
0
, on a alors :
r
e
i
θ
r
′
e
i
θ
′
=
r
r
′
×
e
i
(
θ
−
θ
′
)
\frac{\red{r}e^{i\green{\theta }} } {{\color{blue}{r'}}e^{i\pink{\theta '}}} =\frac{\red{r}}{{\color{blue}{r'}}}\times e^{i\left(\green{\theta} -\pink{\theta '}\right)}
r
′
e
i
θ
′
r
e
i
θ
=
r
′
r
×
e
i
(
θ
−
θ
′
)
z
2
=
5
e
i
π
2
3
e
i
π
6
z_{2} =\frac{\red{5}e^{i\green{\frac{\pi }{2} }} }{{\color{blue}{3}}e^{i\pink{\frac{\pi }{6} }} }
z
2
=
3
e
i
6
π
5
e
i
2
π
équivaut successivement à :
z
2
=
5
3
×
e
i
(
π
2
−
π
6
)
z_{2} =\frac{\red{5}}{{\color{blue}{3}}} \times e^{i\left(\green{\frac{\pi }{2}} -\pink{\frac{\pi }{6}} \right)}
z
2
=
3
5
×
e
i
(
2
π
−
6
π
)
z
2
=
5
3
e
i
(
3
×
π
3
×
2
−
π
6
)
z_{2} =\frac{5}{3} e^{i\left(\frac{3\times \pi }{3\times 2} -\frac{\pi }{6} \right)}
z
2
=
3
5
e
i
(
3
×
2
3
×
π
−
6
π
)
z
2
=
5
3
e
i
(
3
π
6
−
π
6
)
z_{2} =\frac{5}{3} e^{i\left(\frac{3\pi }{6} -\frac{\pi }{6} \right)}
z
2
=
3
5
e
i
(
6
3
π
−
6
π
)
z
2
=
5
3
e
i
2
π
6
z_{2} =\frac{5}{3} e^{i\frac{2\pi }{6} }
z
2
=
3
5
e
i
6
2
π
z
2
=
5
3
e
i
2
π
2
×
3
z_{2} =\frac{5}{3} e^{i\frac{2\pi }{2\times 3} }
z
2
=
3
5
e
i
2
×
3
2
π
Ainsi :
z
2
=
5
3
e
i
π
3
z_{2} =\frac{5}{3} e^{i\frac{\pi }{3} }
z
2
=
3
5
e
i
3
π
Question 3
z
3
=
25
e
i
π
4
5
e
−
i
5
π
6
z_{3} =\frac{25e^{i\frac{\pi }{4} } }{5e^{-i\frac{5\pi }{6} } }
z
3
=
5
e
−
i
6
5
π
25
e
i
4
π
Correction
Soient
r
\red{r}
r
,
r
′
{\color{blue}{r'}}
r
′
,
θ
\green{\theta }
θ
et
θ
′
\pink{\theta'}
θ
′
quatre réels avec
r
′
≠
0
{\color{blue}{r'\ne0}}
r
′
=
0
, on a alors :
r
e
i
θ
r
′
e
i
θ
′
=
r
r
′
×
e
i
(
θ
−
θ
′
)
\frac{\red{r}e^{i\green{\theta }} } {{\color{blue}{r'}}e^{i\pink{\theta '}}} =\frac{\red{r}}{{\color{blue}{r'}}}\times e^{i\left(\green{\theta} -\pink{\theta '}\right)}
r
′
e
i
θ
′
r
e
i
θ
=
r
′
r
×
e
i
(
θ
−
θ
′
)
z
3
=
25
e
i
π
4
5
e
−
i
5
π
6
z_{3} =\frac{\red{25}e^{i\green{\frac{\pi }{4} }} }{{\color{blue}{5}}e^{\pink{-}i\pink{\frac{5\pi }{6} }} }
z
3
=
5
e
−
i
6
5
π
25
e
i
4
π
équivaut successivement à :
z
3
=
25
5
×
e
i
(
π
4
−
(
−
5
π
6
)
)
z_{3} =\frac{\red{25}}{{\color{blue}{5}}} \times e^{i\left(\green{\frac{\pi }{4}} -\left(\pink{-\frac{5\pi }{6}} \right)\right)}
z
3
=
5
25
×
e
i
(
4
π
−
(
−
6
5
π
)
)
z
3
=
5
e
i
(
π
4
+
5
π
6
)
z_{3} =5 e^{i\left(\frac{ \pi }{4} +\frac{5\pi }{6} \right)}
z
3
=
5
e
i
(
4
π
+
6
5
π
)
z
3
=
5
e
i
(
3
×
π
3
×
4
+
2
×
5
π
2
×
6
)
z_{3} =5e^{i\left(\frac{3\times \pi }{3\times 4} +\frac{2\times 5\pi }{2\times 6} \right)}
z
3
=
5
e
i
(
3
×
4
3
×
π
+
2
×
6
2
×
5
π
)
z
3
=
5
e
i
(
3
π
12
+
10
π
12
)
z_{3} =5e^{i\left(\frac{3\pi }{12} +\frac{10\pi }{12} \right)}
z
3
=
5
e
i
(
12
3
π
+
12
10
π
)
Ainsi :
z
3
=
5
e
i
13
π
12
z_{3} =5e^{i\frac{13\pi }{12} }
z
3
=
5
e
i
12
13
π