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Les nombres complexes
Calculs algébriques - Exercice 3
7 min
10
Question 1
On considère les deux nombres complexes définis par
z
1
=
1
+
i
z_{1} =1+i
z
1
=
1
+
i
et
z
2
=
−
2
+
3
i
z_{2} =-2+3i
z
2
=
−
2
+
3
i
.
Calculez et donnez les résultats sous forme algébrique
z
1
−
2
z
2
z_{1} -2z_{2}
z
1
−
2
z
2
Correction
z
1
−
2
z
2
=
1
+
i
−
2
(
−
2
+
3
i
)
z_{1} -2z_{2} =1+i-2\left(-2+3i\right)
z
1
−
2
z
2
=
1
+
i
−
2
(
−
2
+
3
i
)
équivaut successivement à :
z
1
−
2
z
2
=
1
+
i
+
4
−
6
i
z_{1} -2z_{2} =1+i+4-6i
z
1
−
2
z
2
=
1
+
i
+
4
−
6
i
Ainsi :
z
1
−
2
z
2
=
5
−
5
i
z_{1} -2z_{2} =5-5i
z
1
−
2
z
2
=
5
−
5
i
Question 2
z
1
z
2
z_{1} z_{2}
z
1
z
2
Correction
z
1
z
2
=
(
1
+
i
)
(
−
2
+
3
i
)
z_{1} z_{2} =\left(1+i\right)\left(-2+3i\right)
z
1
z
2
=
(
1
+
i
)
(
−
2
+
3
i
)
équivaut successivement à :
z
1
z
2
=
−
2
+
3
i
−
2
i
+
3
i
2
z_{1} z_{2} =-2+3i-2i+3i^{2}
z
1
z
2
=
−
2
+
3
i
−
2
i
+
3
i
2
z
1
z
2
=
−
2
+
3
i
−
2
i
−
3
z_{1} z_{2} =-2+3i-2i-3
z
1
z
2
=
−
2
+
3
i
−
2
i
−
3
Ainsi :
z
1
z
2
=
−
5
+
i
z_{1} z_{2} =-5+i
z
1
z
2
=
−
5
+
i
Question 3
(
z
1
)
2
+
2
(
z
2
)
2
\left(z_{1} \right)^{2} +2\left(z_{2} \right)^{2}
(
z
1
)
2
+
2
(
z
2
)
2
Correction
(
z
1
)
2
+
2
(
z
2
)
2
=
(
1
+
i
)
2
+
2
(
−
2
+
3
i
)
2
\left(z_{1} \right)^{2} +2\left(z_{2} \right)^{2} =\left(1+i\right)^{2} +2\left(-2+3i\right)^{2}
(
z
1
)
2
+
2
(
z
2
)
2
=
(
1
+
i
)
2
+
2
(
−
2
+
3
i
)
2
équivaut successivement à :
(
z
1
)
2
+
2
(
z
2
)
2
=
1
+
2
i
+
i
2
+
2
(
4
−
12
i
+
9
i
2
)
\left(z_{1} \right)^{2} +2\left(z_{2} \right)^{2} =1+2i+i^{2} +2\left(4-12i+9i^{2} \right)
(
z
1
)
2
+
2
(
z
2
)
2
=
1
+
2
i
+
i
2
+
2
(
4
−
12
i
+
9
i
2
)
(
z
1
)
2
+
2
(
z
2
)
2
=
1
+
2
i
−
1
+
2
(
4
−
12
i
−
9
)
\left(z_{1} \right)^{2} +2\left(z_{2} \right)^{2} =1+2i-1+2\left(4-12i-9\right)
(
z
1
)
2
+
2
(
z
2
)
2
=
1
+
2
i
−
1
+
2
(
4
−
12
i
−
9
)
(
z
1
)
2
+
2
(
z
2
)
2
=
2
i
+
2
(
−
12
i
−
5
)
\left(z_{1} \right)^{2} +2\left(z_{2} \right)^{2} =2i+2\left(-12i-5\right)
(
z
1
)
2
+
2
(
z
2
)
2
=
2
i
+
2
(
−
12
i
−
5
)
(
z
1
)
2
+
2
(
z
2
)
2
=
2
i
−
24
i
−
10
\left(z_{1} \right)^{2} +2\left(z_{2} \right)^{2} =2i-24i-10
(
z
1
)
2
+
2
(
z
2
)
2
=
2
i
−
24
i
−
10
Ainsi :
(
z
1
)
2
+
2
(
z
2
)
2
=
−
10
−
22
i
\left(z_{1} \right)^{2} +2\left(z_{2} \right)^{2} =-10-22i
(
z
1
)
2
+
2
(
z
2
)
2
=
−
10
−
22
i
Question 4
z
1
+
z
2
z
2
\frac{z_{1} +z_{2} }{z_{2} }
z
2
z
1
+
z
2
Correction
z
1
+
z
2
z
2
=
1
+
i
−
2
+
3
i
−
2
+
3
i
\frac{z_{1} +z_{2} }{z_{2} } =\frac{1+i-2+3i}{-2+3i}
z
2
z
1
+
z
2
=
−
2
+
3
i
1
+
i
−
2
+
3
i
z
1
+
z
2
z
2
=
−
1
+
4
i
−
2
+
3
i
\frac{z_{1} +z_{2} }{z_{2} } =\frac{-1+4i}{-2+3i}
z
2
z
1
+
z
2
=
−
2
+
3
i
−
1
+
4
i
z
1
+
z
2
z
2
=
(
−
1
+
4
i
)
(
−
2
−
3
i
)
(
−
2
+
3
i
)
(
−
2
−
3
i
)
\frac{z_{1} +z_{2} }{z_{2} } =\frac{\left(-1+4i\right)\left(-2-3i\right)}{\left(-2+3i\right)\left(-2-3i\right)}
z
2
z
1
+
z
2
=
(
−
2
+
3
i
)
(
−
2
−
3
i
)
(
−
1
+
4
i
)
(
−
2
−
3
i
)
z
1
+
z
2
z
2
=
2
+
3
i
−
8
i
−
12
i
2
2
2
+
3
2
\frac{z_{1} +z_{2} }{z_{2} } =\frac{2+3i-8i-12i^{2} }{2^{2} +3^{2} }
z
2
z
1
+
z
2
=
2
2
+
3
2
2
+
3
i
−
8
i
−
12
i
2
z
1
+
z
2
z
2
=
2
+
3
i
−
8
i
+
12
2
2
+
3
2
\frac{z_{1} +z_{2} }{z_{2} } =\frac{2+3i-8i+12}{2^{2} +3^{2} }
z
2
z
1
+
z
2
=
2
2
+
3
2
2
+
3
i
−
8
i
+
12
Ainsi :
z
1
+
z
2
z
2
=
14
13
−
5
13
i
\frac{z_{1} +z_{2} }{z_{2} } =\frac{14}{13} -\frac{5}{13} i
z
2
z
1
+
z
2
=
13
14
−
13
5
i
Question 5
z
1
−
2
z
2
z
1
+
z
2
\frac{z_{1} -2z_{2} }{z_{1} +z_{2} }
z
1
+
z
2
z
1
−
2
z
2
Correction
z
1
−
2
z
2
z
1
+
z
2
=
1
+
i
−
2
(
−
2
+
3
i
)
1
+
i
−
2
+
3
i
\frac{z_{1} -2z_{2} }{z_{1} +z_{2} } =\frac{1+i-2\left(-2+3i\right)}{1+i-2+3i}
z
1
+
z
2
z
1
−
2
z
2
=
1
+
i
−
2
+
3
i
1
+
i
−
2
(
−
2
+
3
i
)
z
1
−
2
z
2
z
1
+
z
2
=
1
+
i
+
4
−
6
i
−
1
+
4
i
\frac{z_{1} -2z_{2} }{z_{1} +z_{2} } =\frac{1+i+4-6i}{-1+4i}
z
1
+
z
2
z
1
−
2
z
2
=
−
1
+
4
i
1
+
i
+
4
−
6
i
z
1
−
2
z
2
z
1
+
z
2
=
5
−
5
i
−
1
+
4
i
\frac{z_{1} -2z_{2} }{z_{1} +z_{2} } =\frac{5-5i}{-1+4i}
z
1
+
z
2
z
1
−
2
z
2
=
−
1
+
4
i
5
−
5
i
z
1
−
2
z
2
z
1
+
z
2
=
(
5
−
5
i
)
(
−
1
−
4
i
)
(
−
1
+
4
i
)
(
−
1
−
4
i
)
\frac{z_{1} -2z_{2} }{z_{1} +z_{2} } =\frac{\left(5-5i\right)\left(-1-4i\right)}{\left(-1+4i\right)\left(-1-4i\right)}
z
1
+
z
2
z
1
−
2
z
2
=
(
−
1
+
4
i
)
(
−
1
−
4
i
)
(
5
−
5
i
)
(
−
1
−
4
i
)
z
1
−
2
z
2
z
1
+
z
2
=
−
5
−
20
i
+
5
i
+
20
i
2
1
2
+
4
2
\frac{z_{1} -2z_{2} }{z_{1} +z_{2} } =\frac{-5-20i+5i+20i^{2} }{1^{2} +4^{2} }
z
1
+
z
2
z
1
−
2
z
2
=
1
2
+
4
2
−
5
−
20
i
+
5
i
+
20
i
2
z
1
−
2
z
2
z
1
+
z
2
=
−
5
−
20
i
+
5
i
−
20
1
2
+
4
2
\frac{z_{1} -2z_{2} }{z_{1} +z_{2} } =\frac{-5-20i+5i-20}{1^{2} +4^{2} }
z
1
+
z
2
z
1
−
2
z
2
=
1
2
+
4
2
−
5
−
20
i
+
5
i
−
20
Ainsi :
z
1
−
2
z
2
z
1
+
z
2
=
−
25
17
−
15
17
i
\frac{z_{1} -2z_{2} }{z_{1} +z_{2} } =-\frac{25}{17} -\frac{15}{17} i
z
1
+
z
2
z
1
−
2
z
2
=
−
17
25
−
17
15
i