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Nombres complexes : point de vue algébrique
Appliquer la formule du binôme de Newton - Exercice 2
10 min
25
Utiliser la formule du binôme pour développer :
Question 1
(
4
+
i
)
3
\left(4+i\right)^{3}
(
4
+
i
)
3
Correction
Soient
a
a
a
et
b
b
b
deux nombres complexes. Pour tout entier naturel
n
n
n
, on a :
(
a
+
b
)
n
=
∑
k
=
0
n
(
n
k
)
a
k
b
n
−
k
\left(a+b\right)^{n} =\sum _{k=0}^{n}\left(\begin{array}{c} {n} \\ {k} \end{array}\right) a^{k} b^{n-k}
(
a
+
b
)
n
=
k
=
0
∑
n
(
n
k
)
a
k
b
n
−
k
i
3
=
i
2
×
i
=
(
−
1
)
×
i
=
−
i
i^{3} =i^{2} \times i=\left(-1\right)\times i=-i
i
3
=
i
2
×
i
=
(
−
1
)
×
i
=
−
i
(
4
+
i
)
3
=
∑
k
=
0
3
(
3
k
)
4
k
i
3
−
k
\left(4+i\right)^{3} =\sum _{k=0}^{3}\left(\begin{array}{c} {3} \\ {k} \end{array}\right) 4^{k} i^{3-k}
(
4
+
i
)
3
=
k
=
0
∑
3
(
3
k
)
4
k
i
3
−
k
équivaut successivement à :
(
4
+
i
)
3
=
(
3
0
)
4
0
i
3
−
0
+
(
3
1
)
4
1
i
3
−
1
+
(
3
2
)
4
2
i
3
−
2
+
(
3
3
)
4
3
i
3
−
3
\left(4+i\right)^{3} =\left(\begin{array}{c} {3} \\ {0} \end{array}\right)4^{0} i^{3-0} +\left(\begin{array}{c} {3} \\ {1} \end{array}\right)4^{1} i^{3-1} +\left(\begin{array}{c} {3} \\ {2} \end{array}\right)4^{2} i^{3-2} +\left(\begin{array}{c} {3} \\ {3} \end{array}\right)4^{3} i^{3-3}
(
4
+
i
)
3
=
(
3
0
)
4
0
i
3
−
0
+
(
3
1
)
4
1
i
3
−
1
+
(
3
2
)
4
2
i
3
−
2
+
(
3
3
)
4
3
i
3
−
3
(
4
+
i
)
3
=
(
3
0
)
4
0
i
3
+
(
3
1
)
4
1
i
2
+
(
3
2
)
4
2
i
1
+
(
3
3
)
4
3
i
0
\left(4+i\right)^{3} =\red{\left(\begin{array}{c} {3} \\ {0} \end{array}\right)}4^{0} i^{3} +\blue{\left(\begin{array}{c} {3} \\ {1} \end{array}\right)}4^{1} i^{2} +\pink{\left(\begin{array}{c} {3} \\ {2} \end{array}\right)}4^{2} i^{1} +\purple{\left(\begin{array}{c} {3} \\ {3} \end{array}\right)}4^{3} i^{0}
(
4
+
i
)
3
=
(
3
0
)
4
0
i
3
+
(
3
1
)
4
1
i
2
+
(
3
2
)
4
2
i
1
+
(
3
3
)
4
3
i
0
(
4
+
i
)
3
=
1
×
4
0
i
3
+
3
×
4
1
i
2
+
3
×
4
2
i
1
+
1
×
4
3
i
0
\left(4+i\right)^{3} =\red{1}\times 4^{0} i^{3} +\blue{3}\times 4^{1} i^{2} +\pink{3}\times 4^{2} i^{1} +\purple{1}\times 4^{3} i^{0}
(
4
+
i
)
3
=
1
×
4
0
i
3
+
3
×
4
1
i
2
+
3
×
4
2
i
1
+
1
×
4
3
i
0
(
4
+
i
)
3
=
1
×
1
×
(
−
i
)
+
3
×
4
×
(
−
1
)
+
3
×
16
×
i
+
64
\left(4+i\right)^{3} =1\times 1\times \left(-i\right)+3\times 4\times \left(-1\right)+3\times 16\times i+64
(
4
+
i
)
3
=
1
×
1
×
(
−
i
)
+
3
×
4
×
(
−
1
)
+
3
×
16
×
i
+
64
(
4
+
i
)
3
=
−
i
−
12
+
48
i
+
64
\left(4+i\right)^{3} =-i-12+48i+64
(
4
+
i
)
3
=
−
i
−
12
+
48
i
+
64
Ainsi :
(
4
+
i
)
3
=
52
+
47
i
\left(4+i\right)^{3} =52+47i
(
4
+
i
)
3
=
52
+
47
i
Question 2
(
3
+
i
)
4
\left(3+i\right)^{4}
(
3
+
i
)
4
Correction
Formule du bin
o
ˆ
me de Newton
\red{\text{Formule du binôme de Newton}}
Formule du bin
o
ˆ
me de Newton
Soient
a
a
a
et
b
b
b
deux nombres complexes. Pour tout entier naturel
n
n
n
, on a :
(
a
+
b
)
n
=
∑
k
=
0
n
(
n
k
)
a
k
b
n
−
k
\left(a+b\right)^{n} =\sum _{k=0}^{n}\left(\begin{array}{c} {n} \\ {k} \end{array}\right) a^{k} b^{n-k}
(
a
+
b
)
n
=
k
=
0
∑
n
(
n
k
)
a
k
b
n
−
k
(
3
+
i
)
4
=
∑
k
=
0
n
(
4
k
)
3
k
i
4
−
k
\left(3+i\right)^{4} =\sum _{k=0}^{n}\left(\begin{array}{c} {4} \\ {k} \end{array}\right) 3^{k} i^{4-k}
(
3
+
i
)
4
=
k
=
0
∑
n
(
4
k
)
3
k
i
4
−
k
équivaut successivement à :
(
3
+
i
)
4
=
(
4
0
)
3
0
i
4
+
(
4
1
)
3
1
i
4
−
1
+
(
4
2
)
3
2
i
4
−
2
+
(
4
3
)
3
3
i
4
−
3
+
(
4
4
)
3
4
i
4
−
4
\left(3+i\right)^{4} =\left(\begin{array}{c} {4} \\ {0} \end{array}\right)3^{0} i^{4} +\left(\begin{array}{c} {4} \\ {1} \end{array}\right)3^{1} i^{4-1} +\left(\begin{array}{c} {4} \\ {2} \end{array}\right)3^{2} i^{4-2} +\left(\begin{array}{c} {4} \\ {3} \end{array}\right)3^{3} i^{4-3} +\left(\begin{array}{c} {4} \\ {4} \end{array}\right)3^{4} i^{4-4}
(
3
+
i
)
4
=
(
4
0
)
3
0
i
4
+
(
4
1
)
3
1
i
4
−
1
+
(
4
2
)
3
2
i
4
−
2
+
(
4
3
)
3
3
i
4
−
3
+
(
4
4
)
3
4
i
4
−
4
(
3
+
i
)
4
=
(
4
0
)
3
0
i
4
+
(
4
1
)
3
1
i
3
+
(
4
2
)
3
2
i
2
+
(
4
3
)
3
3
i
1
+
(
4
4
)
3
4
i
0
\left(3+i\right)^{4} =\left(\begin{array}{c} {4} \\ {0} \end{array}\right)3^{0} i^{4} +\left(\begin{array}{c} {4} \\ {1} \end{array}\right)3^{1} i^{3} +\left(\begin{array}{c} {4} \\ {2} \end{array}\right)3^{2} i^{2} +\left(\begin{array}{c} {4} \\ {3} \end{array}\right)3^{3} i^{1} +\left(\begin{array}{c} {4} \\ {4} \end{array}\right)3^{4} i^{0}
(
3
+
i
)
4
=
(
4
0
)
3
0
i
4
+
(
4
1
)
3
1
i
3
+
(
4
2
)
3
2
i
2
+
(
4
3
)
3
3
i
1
+
(
4
4
)
3
4
i
0
(
3
+
i
)
4
=
(
4
0
)
3
0
i
4
+
(
4
1
)
3
1
i
3
+
(
4
2
)
3
2
i
2
+
(
4
3
)
3
3
i
1
+
(
4
4
)
3
4
i
0
\left(3+i\right)^{4} =\red{\left(\begin{array}{c} {4} \\ {0} \end{array}\right)}3^{0} i^{4} +\blue{\left(\begin{array}{c} {4} \\ {1} \end{array}\right)}3^{1} i^{3} +\pink{\left(\begin{array}{c} {4} \\ {2} \end{array}\right)}3^{2} i^{2} +\purple{\left(\begin{array}{c} {4} \\ {3} \end{array}\right)}3^{3} i^{1} +\green{\left(\begin{array}{c} {4} \\ {4} \end{array}\right)}3^{4} i^{0}
(
3
+
i
)
4
=
(
4
0
)
3
0
i
4
+
(
4
1
)
3
1
i
3
+
(
4
2
)
3
2
i
2
+
(
4
3
)
3
3
i
1
+
(
4
4
)
3
4
i
0
(
3
+
i
)
4
=
1
×
3
0
i
4
+
4
×
3
1
i
3
+
6
×
3
2
i
2
+
4
×
3
3
i
1
+
1
×
3
4
i
0
\left(3+i\right)^{4} =\red{1}\times 3^{0} i^{4} +\blue{4}\times 3^{1} i^{3} +\pink{6}\times 3^{2} i^{2} +\purple{4}\times 3^{3} i^{1} +\green{1}\times 3^{4} i^{0}
(
3
+
i
)
4
=
1
×
3
0
i
4
+
4
×
3
1
i
3
+
6
×
3
2
i
2
+
4
×
3
3
i
1
+
1
×
3
4
i
0
(
3
+
i
)
4
=
1
×
1
×
i
4
+
4
×
3
×
i
3
+
6
×
9
×
i
2
+
4
×
27
×
i
1
+
1
×
81
×
i
0
\left(3+i\right)^{4} =1\times 1\times i^{4} +4\times 3\times i^{3} +6\times 9\times i^{2} +4\times 27\times i^{1} +1\times 81\times i^{0}
(
3
+
i
)
4
=
1
×
1
×
i
4
+
4
×
3
×
i
3
+
6
×
9
×
i
2
+
4
×
27
×
i
1
+
1
×
81
×
i
0
(
3
+
i
)
4
=
1
×
1
×
i
2
×
i
2
+
4
×
3
×
i
2
×
i
+
6
×
9
×
i
2
+
4
×
27
×
i
1
+
1
×
81
×
i
0
\left(3+i\right)^{4} =1\times 1\times i^{2}\times i^{2} +4\times 3\times i^{2}\times i +6\times 9\times i^{2} +4\times 27\times i^{1} +1\times 81\times i^{0}
(
3
+
i
)
4
=
1
×
1
×
i
2
×
i
2
+
4
×
3
×
i
2
×
i
+
6
×
9
×
i
2
+
4
×
27
×
i
1
+
1
×
81
×
i
0
(
3
+
i
)
4
=
1
×
1
×
(
−
1
)
×
(
−
1
)
+
4
×
3
×
(
−
1
)
×
i
+
6
×
9
×
(
−
1
)
+
4
×
27
×
i
1
+
1
×
81
×
i
0
\left(3+i\right)^{4} =1\times 1\times \left(-1\right)\times \left(-1\right) +4\times 3\times \left(-1\right)\times i +6\times 9\times\left(-1\right) +4\times 27\times i^{1} +1\times 81\times i^{0}
(
3
+
i
)
4
=
1
×
1
×
(
−
1
)
×
(
−
1
)
+
4
×
3
×
(
−
1
)
×
i
+
6
×
9
×
(
−
1
)
+
4
×
27
×
i
1
+
1
×
81
×
i
0
(
3
+
i
)
4
=
1
−
12
i
−
54
+
108
i
+
81
\left(3+i\right)^{4} =1-12i-54+108i+81
(
3
+
i
)
4
=
1
−
12
i
−
54
+
108
i
+
81
Ainsi :
(
3
+
i
)
4
=
28
+
96
i
\left(3+i\right)^{4} =28+96i
(
3
+
i
)
4
=
28
+
96
i