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Nombres complexes : point de vue algébrique
Appliquer la formule du binôme de Newton - Exercice 1
12 min
25
Utiliser la formule du binôme pour développer :
Question 1
(
1
+
z
)
3
\left(1+z\right)^{3}
(
1
+
z
)
3
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
(
1
+
z
)
3
=
∑
k
=
0
n
(
3
k
)
1
k
z
3
−
k
\left(1+z\right)^{3} =\sum _{k=0}^{n}\left(\begin{array}{c} {3} \\ {k} \end{array}\right) 1^{k} z^{3-k}
(
1
+
z
)
3
=
k
=
0
∑
n
(
3
k
)
1
k
z
3
−
k
équivaut successivement à :
(
1
+
z
)
3
=
(
3
0
)
1
0
z
3
−
0
+
(
3
1
)
1
1
z
3
−
1
+
(
3
2
)
1
2
z
3
−
2
+
(
3
2
)
1
3
z
3
−
3
\left(1+z\right)^{3} =\left(\begin{array}{c} {3} \\ {0} \end{array}\right)1^{0} z^{3-0} +\left(\begin{array}{c} {3} \\ {1} \end{array}\right)1^{1} z^{3-1} +\left(\begin{array}{c} {3} \\ {2} \end{array}\right)1^{2} z^{3-2} +\left(\begin{array}{c} {3} \\ {2} \end{array}\right)1^{3} z^{3-3}
(
1
+
z
)
3
=
(
3
0
)
1
0
z
3
−
0
+
(
3
1
)
1
1
z
3
−
1
+
(
3
2
)
1
2
z
3
−
2
+
(
3
2
)
1
3
z
3
−
3
(
1
+
z
)
3
=
(
3
0
)
1
0
z
3
+
(
3
1
)
1
1
z
2
+
(
3
2
)
1
2
z
1
+
(
3
3
)
1
3
z
0
\left(1+z\right)^{3} =\red{\left(\begin{array}{c} {3} \\ {0} \end{array}\right)}1^{0} z^{3} +\blue{\left(\begin{array}{c} {3} \\ {1} \end{array}\right)}1^{1} z^{2} +\pink{\left(\begin{array}{c} {3} \\ {2} \end{array}\right)}1^{2} z^{1} +\purple{\left(\begin{array}{c} {3} \\ {3} \end{array}\right)}1^{3} z^{0}
(
1
+
z
)
3
=
(
3
0
)
1
0
z
3
+
(
3
1
)
1
1
z
2
+
(
3
2
)
1
2
z
1
+
(
3
3
)
1
3
z
0
(
1
+
z
)
3
=
1
×
1
0
×
z
3
+
3
×
1
1
×
z
2
+
3
×
1
2
×
z
1
+
1
×
1
3
×
z
0
\left(1+z\right)^{3} =\red{1}\times 1^{0} \times z^{3} +\blue{3}\times 1^{1} \times z^{2} +\pink{3}\times 1^{2} \times z^{1} +\purple{1}\times 1^{3} \times z^{0}
(
1
+
z
)
3
=
1
×
1
0
×
z
3
+
3
×
1
1
×
z
2
+
3
×
1
2
×
z
1
+
1
×
1
3
×
z
0
(
1
+
z
)
3
=
1
×
1
×
z
3
+
3
×
1
×
z
2
+
3
×
1
×
z
1
+
1
×
1
×
1
\left(1+z\right)^{3} =1\times 1\times z^{3} +3\times 1\times z^{2} +3\times 1\times z^{1} +1\times 1\times 1
(
1
+
z
)
3
=
1
×
1
×
z
3
+
3
×
1
×
z
2
+
3
×
1
×
z
1
+
1
×
1
×
1
Ainsi :
(
1
+
z
)
3
=
z
3
+
3
z
2
+
3
z
+
1
\left(1+z\right)^{3} =z^{3} +3z^{2} +3z+1
(
1
+
z
)
3
=
z
3
+
3
z
2
+
3
z
+
1
Question 2
(
2
+
z
)
4
\left(2+z\right)^{4}
(
2
+
z
)
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
(
2
+
z
)
4
=
∑
k
=
0
n
(
4
k
)
2
k
z
4
−
k
\left(2+z\right)^{4} =\sum _{k=0}^{n}\left(\begin{array}{c} {4} \\ {k} \end{array}\right) 2^{k} z^{4-k}
(
2
+
z
)
4
=
k
=
0
∑
n
(
4
k
)
2
k
z
4
−
k
équivaut successivement à :
(
2
+
z
)
4
=
(
4
0
)
2
0
z
4
+
(
4
1
)
2
1
z
4
−
1
+
(
4
2
)
2
2
z
4
−
2
+
(
4
3
)
2
3
z
4
−
3
+
(
4
4
)
2
4
z
4
−
4
\left(2+z\right)^{4} =\left(\begin{array}{c} {4} \\ {0} \end{array}\right)2^{0} z^{4} +\left(\begin{array}{c} {4} \\ {1} \end{array}\right)2^{1} z^{4-1} +\left(\begin{array}{c} {4} \\ {2} \end{array}\right)2^{2} z^{4-2} +\left(\begin{array}{c} {4} \\ {3} \end{array}\right)2^{3} z^{4-3} +\left(\begin{array}{c} {4} \\ {4} \end{array}\right)2^{4} z^{4-4}
(
2
+
z
)
4
=
(
4
0
)
2
0
z
4
+
(
4
1
)
2
1
z
4
−
1
+
(
4
2
)
2
2
z
4
−
2
+
(
4
3
)
2
3
z
4
−
3
+
(
4
4
)
2
4
z
4
−
4
(
2
+
z
)
4
=
(
4
0
)
2
0
z
4
+
(
4
1
)
2
1
z
3
+
(
4
2
)
2
2
z
2
+
(
4
3
)
2
3
z
1
+
(
4
4
)
2
4
z
0
\left(2+z\right)^{4} =\left(\begin{array}{c} {4} \\ {0} \end{array}\right)2^{0} z^{4} +\left(\begin{array}{c} {4} \\ {1} \end{array}\right)2^{1} z^{3} +\left(\begin{array}{c} {4} \\ {2} \end{array}\right)2^{2} z^{2} +\left(\begin{array}{c} {4} \\ {3} \end{array}\right)2^{3} z^{1} +\left(\begin{array}{c} {4} \\ {4} \end{array}\right)2^{4} z^{0}
(
2
+
z
)
4
=
(
4
0
)
2
0
z
4
+
(
4
1
)
2
1
z
3
+
(
4
2
)
2
2
z
2
+
(
4
3
)
2
3
z
1
+
(
4
4
)
2
4
z
0
(
2
+
z
)
4
=
(
4
0
)
2
0
z
4
+
(
4
1
)
2
1
z
3
+
(
4
2
)
2
2
z
2
+
(
4
3
)
2
3
z
1
+
(
4
4
)
2
4
z
0
\left(2+z\right)^{4} =\red{\left(\begin{array}{c} {4} \\ {0} \end{array}\right)}2^{0} z^{4} +\blue{\left(\begin{array}{c} {4} \\ {1} \end{array}\right)}2^{1} z^{3} +\pink{\left(\begin{array}{c} {4} \\ {2} \end{array}\right)}2^{2} z^{2} +\purple{\left(\begin{array}{c} {4} \\ {3} \end{array}\right)}2^{3} z^{1} +\green{\left(\begin{array}{c} {4} \\ {4} \end{array}\right)}2^{4} z^{0}
(
2
+
z
)
4
=
(
4
0
)
2
0
z
4
+
(
4
1
)
2
1
z
3
+
(
4
2
)
2
2
z
2
+
(
4
3
)
2
3
z
1
+
(
4
4
)
2
4
z
0
(
2
+
z
)
4
=
1
×
2
0
z
4
+
4
×
2
1
z
3
+
6
×
2
2
z
2
+
4
×
2
3
z
1
+
1
×
2
4
z
0
\left(2+z\right)^{4} =\red{1}\times 2^{0} z^{4} +\blue{4}\times 2^{1} z^{3} +\pink{6}\times 2^{2} z^{2} +\purple{4}\times 2^{3} z^{1} +\green{1}\times 2^{4} z^{0}
(
2
+
z
)
4
=
1
×
2
0
z
4
+
4
×
2
1
z
3
+
6
×
2
2
z
2
+
4
×
2
3
z
1
+
1
×
2
4
z
0
(
2
+
z
)
4
=
1
×
1
×
z
4
+
4
×
2
×
z
3
+
6
×
4
×
z
2
+
4
×
8
×
z
1
+
1
×
16
×
z
0
\left(2+z\right)^{4} =1\times 1\times z^{4} +4\times 2\times z^{3} +6\times 4\times z^{2} +4\times 8\times z^{1} +1\times 16\times z^{0}
(
2
+
z
)
4
=
1
×
1
×
z
4
+
4
×
2
×
z
3
+
6
×
4
×
z
2
+
4
×
8
×
z
1
+
1
×
16
×
z
0
Ainsi :
(
2
+
z
)
4
=
z
4
+
8
z
3
+
24
z
2
+
32
z
+
16
\left(2+z\right)^{4} =z^{4} +8z^{3} +24z^{2} +32z +16
(
2
+
z
)
4
=
z
4
+
8
z
3
+
24
z
2
+
32
z
+
16
Question 3
(
z
−
3
)
3
\left(z-3\right)^{3}
(
z
−
3
)
3
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
(
z
−
3
)
3
=
∑
k
=
0
3
(
3
k
)
z
k
(
−
3
)
3
−
k
\left(z-3\right)^{3} =\sum _{k=0}^{3}\left(\begin{array}{c} {3} \\ {k} \end{array}\right) z^{k} \left(-3\right)^{3-k}
(
z
−
3
)
3
=
k
=
0
∑
3
(
3
k
)
z
k
(
−
3
)
3
−
k
équivaut successivement à :
(
z
−
3
)
3
=
(
3
0
)
z
0
(
−
3
)
3
−
0
+
(
3
1
)
z
1
(
−
3
)
3
−
1
+
(
3
2
)
z
2
(
−
3
)
3
−
2
+
(
3
3
)
z
3
(
−
3
)
3
−
3
\left(z-3\right)^{3} =\left(\begin{array}{c} {3} \\ {0} \end{array}\right)z^{0} \left(-3\right)^{3-0} +\left(\begin{array}{c} {3} \\ {1} \end{array}\right)z^{1} \left(-3\right)^{3-1} +\left(\begin{array}{c} {3} \\ {2} \end{array}\right)z^{2} \left(-3\right)^{3-2} +\left(\begin{array}{c} {3} \\ {3} \end{array}\right)z^{3} \left(-3\right)^{3-3}
(
z
−
3
)
3
=
(
3
0
)
z
0
(
−
3
)
3
−
0
+
(
3
1
)
z
1
(
−
3
)
3
−
1
+
(
3
2
)
z
2
(
−
3
)
3
−
2
+
(
3
3
)
z
3
(
−
3
)
3
−
3
(
z
−
3
)
3
=
(
3
0
)
z
0
(
−
3
)
3
+
(
3
1
)
z
1
(
−
3
)
2
+
(
3
2
)
z
2
(
−
3
)
1
+
(
3
3
)
z
3
(
−
3
)
0
\left(z-3\right)^{3} =\red{\left(\begin{array}{c} {3} \\ {0} \end{array}\right)}z^{0} \left(-3\right)^{3} +\blue{\left(\begin{array}{c} {3} \\ {1} \end{array}\right)}z^{1} \left(-3\right)^{2} +\pink{\left(\begin{array}{c} {3} \\ {2} \end{array}\right)}z^{2} \left(-3\right)^{1} +\purple{\left(\begin{array}{c} {3} \\ {3} \end{array}\right)}z^{3} \left(-3\right)^{0}
(
z
−
3
)
3
=
(
3
0
)
z
0
(
−
3
)
3
+
(
3
1
)
z
1
(
−
3
)
2
+
(
3
2
)
z
2
(
−
3
)
1
+
(
3
3
)
z
3
(
−
3
)
0
(
z
−
3
)
3
=
1
×
z
0
(
−
3
)
3
+
3
×
z
1
(
−
3
)
2
+
3
×
z
2
(
−
3
)
1
+
1
×
z
3
(
−
3
)
0
\left(z-3\right)^{3} =\red{1}\times z^{0} \left(-3\right)^{3} +\blue{3}\times z^{1} \left(-3\right)^{2} +\pink{3}\times z^{2} \left(-3\right)^{1} +\purple{1}\times z^{3} \left(-3\right)^{0}
(
z
−
3
)
3
=
1
×
z
0
(
−
3
)
3
+
3
×
z
1
(
−
3
)
2
+
3
×
z
2
(
−
3
)
1
+
1
×
z
3
(
−
3
)
0
(
z
−
3
)
3
=
1
×
z
0
(
−
3
)
3
+
3
×
z
1
(
−
3
)
2
+
3
×
z
2
(
−
3
)
1
+
1
×
z
3
(
−
3
)
0
\left(z-3\right)^{3} =1\times z^{0} \left(-3\right)^{3} +3\times z^{1} \left(-3\right)^{2} +3\times z^{2} \left(-3\right)^{1} +1\times z^{3} \left(-3\right)^{0}
(
z
−
3
)
3
=
1
×
z
0
(
−
3
)
3
+
3
×
z
1
(
−
3
)
2
+
3
×
z
2
(
−
3
)
1
+
1
×
z
3
(
−
3
)
0
(
z
−
3
)
3
=
1
×
1
×
(
−
27
)
+
3
×
z
×
9
+
3
×
z
2
×
(
−
3
)
+
1
×
z
3
×
1
\left(z-3\right)^{3} =1\times 1\times \left(-27\right)+3\times z\times 9+3\times z^{2} \times \left(-3\right)+1\times z^{3} \times 1
(
z
−
3
)
3
=
1
×
1
×
(
−
27
)
+
3
×
z
×
9
+
3
×
z
2
×
(
−
3
)
+
1
×
z
3
×
1
(
z
−
3
)
3
=
−
27
+
27
z
−
9
z
2
+
z
3
\left(z-3\right)^{3} =-27+27z-9z^{2} +z^{3}
(
z
−
3
)
3
=
−
27
+
27
z
−
9
z
2
+
z
3
Ainsi :
(
z
−
3
)
3
=
z
3
−
9
z
2
+
27
z
−
27
\left(z-3\right)^{3} =z^{3} -9z^{2} +27z-27
(
z
−
3
)
3
=
z
3
−
9
z
2
+
27
z
−
27