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Nombres complexes
Passer de la forme exponentielle à la forme algébrique - Exercice 1
10 min
20
Ecrire sous forme algébrique les nombres complexes suivants :
Question 1
z
1
=
4
e
i
π
4
z_{1} =4e^{i\frac{\pi }{4} }
z
1
=
4
e
i
4
π
Correction
Soit
θ
{\color{blue}{\theta}}
θ
un réel .
e
i
θ
=
cos
(
θ
)
+
i
sin
(
θ
)
e^{i{\color{blue}{\theta}} } =\cos \left({\color{blue}{\theta}}\right)+i\sin \left({\color{blue}{\theta}}\right)
e
i
θ
=
cos
(
θ
)
+
i
sin
(
θ
)
z
1
=
4
e
i
π
4
z_{1} =4e^{i{\color{blue}{\frac{\pi }{4}}} }
z
1
=
4
e
i
4
π
équivaut successivement à :
z
1
=
4
×
(
cos
(
π
4
)
+
i
sin
(
π
4
)
)
z_{1} =4\times \left(\cos \left({\color{blue}{\frac{\pi }{4}}} \right)+i\sin \left({\color{blue}{\frac{\pi }{4}}} \right)\right)
z
1
=
4
×
(
cos
(
4
π
)
+
i
sin
(
4
π
)
)
z
1
=
4
×
cos
(
π
4
)
+
4
×
i
sin
(
π
4
)
z_{1} =4\times \cos \left(\frac{\pi }{4} \right)+4\times i\sin \left(\frac{\pi }{4} \right)
z
1
=
4
×
cos
(
4
π
)
+
4
×
i
sin
(
4
π
)
z
1
=
4
×
2
2
+
4
×
i
×
2
2
z_{1} =4\times \frac{\sqrt{2} }{2} +4\times i\times \frac{\sqrt{2} }{2}
z
1
=
4
×
2
2
+
4
×
i
×
2
2
z
1
=
4
×
2
2
+
4
×
2
2
i
z_{1} =\frac{4\times \sqrt{2} }{2} +\frac{4\times \sqrt{2} }{2} i
z
1
=
2
4
×
2
+
2
4
×
2
i
Ainsi :
z
1
=
2
2
+
2
2
i
z_{1} =2\sqrt{2} +2\sqrt{2} i
z
1
=
2
2
+
2
2
i
Question 2
z
2
=
2
e
i
π
3
z_{2} =2e^{i\frac{\pi }{3} }
z
2
=
2
e
i
3
π
Correction
Soit
θ
{\color{blue}{\theta}}
θ
un réel .
e
i
θ
=
cos
(
θ
)
+
i
sin
(
θ
)
e^{i{\color{blue}{\theta}} } =\cos \left({\color{blue}{\theta}}\right)+i\sin \left({\color{blue}{\theta}}\right)
e
i
θ
=
cos
(
θ
)
+
i
sin
(
θ
)
z
2
=
2
e
i
π
3
z_{2} =2e^{i{\color{blue}{\frac{\pi }{3}}} }
z
2
=
2
e
i
3
π
équivaut successivement à :
z
2
=
2
×
(
cos
(
π
3
)
+
i
sin
(
π
3
)
)
z_{2} =2\times \left(\cos \left({\color{blue}{\frac{\pi }{3}}} \right)+i\sin \left({\color{blue}{\frac{\pi }{3}}}\right)\right)
z
2
=
2
×
(
cos
(
3
π
)
+
i
sin
(
3
π
)
)
z
2
=
2
×
cos
(
π
3
)
+
2
×
i
sin
(
π
3
)
z_{2} =2\times \cos \left(\frac{\pi }{3} \right)+2\times i\sin \left(\frac{\pi }{3} \right)
z
2
=
2
×
cos
(
3
π
)
+
2
×
i
sin
(
3
π
)
z
2
=
2
×
1
2
+
2
×
i
×
3
2
z_{2} =2\times \frac{1}{2} +2\times i\times \frac{\sqrt{3} }{2}
z
2
=
2
×
2
1
+
2
×
i
×
2
3
Ainsi :
z
2
=
1
+
3
i
z_{2} =1+\sqrt{3} i
z
2
=
1
+
3
i
Question 3
z
3
=
5
e
−
i
π
2
z_{3} =5e^{-i\frac{\pi }{2} }
z
3
=
5
e
−
i
2
π
Correction
Soit
θ
{\color{blue}{\theta}}
θ
un réel .
e
i
θ
=
cos
(
θ
)
+
i
sin
(
θ
)
e^{i{\color{blue}{\theta}} } =\cos \left({\color{blue}{\theta}}\right)+i\sin \left({\color{blue}{\theta}}\right)
e
i
θ
=
cos
(
θ
)
+
i
sin
(
θ
)
z
3
=
5
e
−
i
π
2
z_{3} =5e^{-i{\color{blue}{\frac{\pi }{2} }}}
z
3
=
5
e
−
i
2
π
équivaut successivement à :
z
3
=
5
×
(
cos
(
−
π
2
)
+
i
sin
(
−
π
2
)
)
z_{3} =5\times \left(\cos \left({\color{blue}{-\frac{\pi }{2}}} \right)+i\sin \left({\color{blue}{-\frac{\pi }{2}}} \right)\right)
z
3
=
5
×
(
cos
(
−
2
π
)
+
i
sin
(
−
2
π
)
)
z
3
=
5
×
cos
(
−
π
2
)
+
5
×
i
sin
(
−
π
2
)
z_{3} =5\times \cos \left(-\frac{\pi }{2} \right)+5\times i\sin \left(-\frac{\pi }{2} \right)
z
3
=
5
×
cos
(
−
2
π
)
+
5
×
i
sin
(
−
2
π
)
z
3
=
5
×
0
+
5
×
i
×
(
−
1
)
z_{3} =5\times 0+5\times i\times \left(-1\right)
z
3
=
5
×
0
+
5
×
i
×
(
−
1
)
Ainsi :
z
3
=
−
5
i
z_{3} =-5i
z
3
=
−
5
i
Question 4
z
4
=
3
e
−
i
5
π
6
z_{4} =3e^{-i\frac{5\pi }{6} }
z
4
=
3
e
−
i
6
5
π
Correction
Soit
θ
{\color{blue}{\theta}}
θ
un réel .
e
i
θ
=
cos
(
θ
)
+
i
sin
(
θ
)
e^{i{\color{blue}{\theta}} } =\cos \left({\color{blue}{\theta}}\right)+i\sin \left({\color{blue}{\theta}}\right)
e
i
θ
=
cos
(
θ
)
+
i
sin
(
θ
)
z
4
=
3
e
−
i
5
π
6
z_{4} =3e^{-i{\color{blue}{\frac{5\pi }{6} }}}
z
4
=
3
e
−
i
6
5
π
équivaut successivement à :
z
4
=
3
×
(
cos
(
−
5
π
6
)
+
i
sin
(
−
5
π
6
)
)
z_{4} =3\times \left(\cos \left({\color{blue}{-\frac{5\pi }{6}}} \right)+i\sin \left({\color{blue}{-\frac{5\pi }{6}}} \right)\right)
z
4
=
3
×
(
cos
(
−
6
5
π
)
+
i
sin
(
−
6
5
π
)
)
z
4
=
3
×
cos
(
−
5
π
6
)
+
3
×
i
sin
(
−
5
π
6
)
z_{4} =3\times \cos \left(-\frac{5\pi }{6} \right)+3\times i\sin \left(-\frac{5\pi }{6} \right)
z
4
=
3
×
cos
(
−
6
5
π
)
+
3
×
i
sin
(
−
6
5
π
)
z
4
=
3
×
(
−
3
2
)
+
3
×
i
(
−
1
2
)
z_{4} =3\times \left(-\frac{\sqrt{3} }{2} \right)+3\times i\left(-\frac{1}{2} \right)
z
4
=
3
×
(
−
2
3
)
+
3
×
i
(
−
2
1
)
Ainsi :
z
4
=
−
3
3
2
−
3
2
i
z_{4} =-\frac{3\sqrt{3} }{2} -\frac{3}{2} i
z
4
=
−
2
3
3
−
2
3
i