En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : Sur les calculs algébriquesCalcul avec les nombres complexes/Exercices/Sur les calculs algébriques », n'a pu être restituée correctement ci-dessus.
1° Calculez
(
1
2
+
i
3
)
4
{\displaystyle \left({\frac {1}{2}}+\mathrm {i} {\sqrt {3}}\right)^{4}}
2° Déterminez les nombres complexes
z
{\displaystyle z}
z
4
=
73
16
−
11
2
i
3
{\displaystyle z^{4}={\frac {73}{16}}-{\frac {11}{2}}\mathrm {i} {\sqrt {3}}}
Soit
f
(
z
)
=
z
2
+
z
+
1
z
3
+
z
2
+
1
{\displaystyle f(z)={\frac {z^{2}+z+1}{z^{3}+z^{2}+1}}}
Calculer
f
(
1
+
i
)
{\displaystyle f\left(1+\mathrm {i} \right)}
f
(
1
−
i
)
{\displaystyle f(1-\mathrm {i} )}
Solution
(
1
+
i
)
2
=
2
i
{\displaystyle \left(1+\mathrm {i} \right)^{2}=2\mathrm {i} }
(
1
+
i
)
3
=
2
i
−
2
{\displaystyle \left(1+\mathrm {i} \right)^{3}=2\mathrm {i} -2}
f
(
1
+
i
)
=
2
+
3
i
−
1
+
4
i
=
(
2
+
3
i
)
(
−
1
−
4
i
)
17
=
10
−
11
i
17
{\displaystyle f\left(1+\mathrm {i} \right)={\frac {2+3\mathrm {i} }{-1+4\mathrm {i} }}={\frac {\left(2+3\mathrm {i} \right)\left(-1-4\mathrm {i} \right)}{17}}={\frac {10-11\mathrm {i} }{17}}}
donc
f
(
1
−
i
)
=
10
+
11
i
17
{\displaystyle f\left(1-\mathrm {i} \right)={\frac {10+11\mathrm {i} }{17}}}
Soient
a
,
b
∈
Z
{\displaystyle a,b\in \mathbb {Z} }
n
∈
N
∗
{\displaystyle n\in \mathbb {N} ^{*}}
(
x
,
y
)
∈
Z
2
{\displaystyle \left(x,y\right)\in \mathbb {Z} ^{2}}
x
2
+
y
2
=
(
a
2
+
b
2
)
n
{\displaystyle x^{2}+y^{2}=\left(a^{2}+b^{2}\right)^{n}}
Solution
On peut choisir par exemple
x
+
i
y
=
(
a
+
i
b
)
n
{\displaystyle x+\mathrm {i} y=\left(a+\mathrm {i} b\right)^{n}}
x
=
∑
(
n
2
k
)
(
−
1
)
k
a
n
−
2
k
b
2
k
{\displaystyle x=\sum {\binom {n}{2k}}(-1)^{k}a^{n-2k}b^{2k}}
y
=
∑
(
n
2
k
+
1
)
(
−
1
)
k
a
n
−
2
k
−
1
b
2
k
+
1
{\displaystyle y=\sum {\binom {n}{2k+1}}(-1)^{k}a^{n-2k-1}b^{2k+1}}
Calculez
∑
k
=
0
n
−
1
i
k
{\displaystyle \sum _{k=0}^{n-1}\mathrm {i} ^{k}}
Solution
∑
k
=
0
n
−
1
i
k
=
1
−
i
n
1
−
i
=
{
0
si
n
≡
0
mod
4
1
si
n
≡
1
mod
4
1
+
i
si
n
≡
2
mod
4
i
si
n
≡
3
mod
4
.
{\displaystyle \sum _{k=0}^{n-1}\mathrm {i} ^{k}={\frac {1-\mathrm {i} ^{n}}{1-\mathrm {i} }}={\begin{cases}0&{\text{si }}n\equiv 0{\bmod {4}}\\1&{\text{si }}n\equiv 1{\bmod {4}}\\1+\mathrm {i} &{\text{si }}n\equiv 2{\bmod {4}}\\\mathrm {i} &{\text{si }}n\equiv 3{\bmod {4}}.\end{cases}}}
Donnez la forme cartésienne
a
+
i
b
{\displaystyle a+\mathrm {i} b}
(
1
+
i
)
2
{\displaystyle (1+\mathrm {i} )^{2}}
(
1
−
i
)
2
{\displaystyle (1-\mathrm {i} )^{2}}
(
2
+
i
)
2
{\displaystyle (2+\mathrm {i} )^{2}}
(
2
−
i
)
2
{\displaystyle (2-\mathrm {i} )^{2}}
(
3
+
4
i
)
2
{\displaystyle (3+4\mathrm {i} )^{2}}
(
3
−
4
i
)
2
{\displaystyle (3-4\mathrm {i} )^{2}}
(
1
+
i
)
(
1
−
i
)
−
1
{\displaystyle (1+\mathrm {i} )(1-\mathrm {i} )^{-1}}
(
1
−
i
)
(
1
+
i
)
−
1
{\displaystyle (1-\mathrm {i} )(1+\mathrm {i} )^{-1}}
(
3
+
4
i
)
(
1
+
i
)
−
1
{\displaystyle (3+4\mathrm {i} )(1+\mathrm {i} )^{-1}}
3
e
i
π
/
3
{\displaystyle 3\operatorname {e} ^{\mathrm {i} \pi /3}}
(
3
+
4
i
)
2
(
4
+
3
i
)
−
2
{\displaystyle (3+4\mathrm {i} )^{2}(4+3\mathrm {i} )^{-2}}
(
3
−
4
i
)
2
(
4
−
3
i
)
−
2
{\displaystyle (3-4\mathrm {i} )^{2}(4-3\mathrm {i} )^{-2}}
(
3
−
4
i
)
(
1
−
i
)
−
1
{\displaystyle (3-4\mathrm {i} )(1-\mathrm {i} )^{-1}}
Solution
(
1
+
i
)
2
=
2
i
{\displaystyle (1+\mathrm {i} )^{2}=2\mathrm {i} }
1
2
i
=
−
i
2
{\displaystyle {\frac {1}{2\mathrm {i} }}=-{\frac {\mathrm {i} }{2}}}
(
1
−
i
)
2
=
−
2
i
{\displaystyle (1-\mathrm {i} )^{2}=-2\mathrm {i} }
1
−
2
i
=
i
2
{\displaystyle {\frac {1}{-2\mathrm {i} }}={\frac {\mathrm {i} }{2}}}
(
2
+
i
)
2
=
3
+
4
i
{\displaystyle (2+\mathrm {i} )^{2}=3+4\mathrm {i} }
1
3
+
4
i
=
3
−
4
i
25
{\displaystyle {\frac {1}{3+4\mathrm {i} }}={\frac {3-4\mathrm {i} }{25}}}
(
2
−
i
)
2
=
3
−
4
i
{\displaystyle (2-\mathrm {i} )^{2}=3-4\mathrm {i} }
1
3
−
4
i
=
3
+
4
i
25
{\displaystyle {\frac {1}{3-4\mathrm {i} }}={\frac {3+4\mathrm {i} }{25}}}
(
3
+
4
i
)
2
=
−
7
+
24
i
{\displaystyle (3+4\mathrm {i} )^{2}=-7+24\mathrm {i} }
1
−
7
+
24
i
=
−
7
−
24
i
625
{\displaystyle {\frac {1}{-7+24\mathrm {i} }}={\frac {-7-24\mathrm {i} }{625}}}
(
3
−
4
i
)
2
=
−
7
−
24
i
{\displaystyle (3-4\mathrm {i} )^{2}=-7-24\mathrm {i} }
1
−
7
−
24
i
=
−
7
+
24
i
625
{\displaystyle {\frac {1}{-7-24\mathrm {i} }}={\frac {-7+24\mathrm {i} }{625}}}
(
1
+
i
)
(
1
−
i
)
−
1
=
(
1
+
i
)
2
2
=
i
{\displaystyle (1+\mathrm {i} )(1-\mathrm {i} )^{-1}={\frac {(1+\mathrm {i} )^{2}}{2}}=\mathrm {i} }
1
i
=
−
i
{\displaystyle {\frac {1}{\mathrm {i} }}=-\mathrm {i} }
(
1
−
i
)
(
1
+
i
)
−
1
=
−
i
{\displaystyle (1-\mathrm {i} )(1+\mathrm {i} )^{-1}=-\mathrm {i} }
1
−
i
=
i
{\displaystyle {\frac {1}{-\mathrm {i} }}=\mathrm {i} }
(
3
+
4
i
)
(
1
+
i
)
−
1
=
(
3
+
4
i
)
1
−
i
2
=
7
+
i
2
{\displaystyle (3+4\mathrm {i} )(1+\mathrm {i} )^{-1}=(3+4\mathrm {i} ){\frac {1-\mathrm {i} }{2}}={\frac {7+\mathrm {i} }{2}}}
2
7
+
i
=
7
−
i
25
{\displaystyle {\frac {2}{7+\mathrm {i} }}={\frac {7-\mathrm {i} }{25}}}
3
e
i
π
/
3
=
3
2
(
1
+
i
3
)
{\displaystyle 3\operatorname {e} ^{\mathrm {i} \pi /3}={\frac {3}{2}}(1+\mathrm {i} {\sqrt {3}})}
1
3
e
−
i
π
/
3
=
1
6
(
1
−
i
3
)
{\displaystyle {\frac {1}{3}}\operatorname {e} ^{-\mathrm {i} \pi /3}={\frac {1}{6}}(1-\mathrm {i} {\sqrt {3}})}
(
3
+
4
i
)
2
(
4
+
3
i
)
−
2
=
(
−
7
+
24
i
)
i
−
2
(
3
−
4
i
)
−
2
=
(
−
7
+
24
i
)
7
−
24
i
625
=
527
+
336
i
625
{\displaystyle (3+4\mathrm {i} )^{2}(4+3\mathrm {i} )^{-2}=(-7+24\mathrm {i} )\mathrm {i} ^{-2}(3-4\mathrm {i} )^{-2}=(-7+24\mathrm {i} ){\frac {7-24\mathrm {i} }{625}}={\frac {527+336\mathrm {i} }{625}}}
625
527
+
336
i
=
527
−
336
i
625
{\displaystyle {\frac {625}{527+336\mathrm {i} }}={\frac {527-336\mathrm {i} }{625}}}
(
3
−
4
i
)
2
(
4
−
3
i
)
−
2
=
527
−
336
i
625
{\displaystyle (3-4\mathrm {i} )^{2}(4-3\mathrm {i} )^{-2}={\frac {527-336\mathrm {i} }{625}}}
625
527
−
336
i
=
527
+
336
i
625
{\displaystyle {\frac {625}{527-336\mathrm {i} }}={\frac {527+336\mathrm {i} }{625}}}
(
3
−
4
i
)
(
1
−
i
)
−
1
=
7
−
i
2
{\displaystyle (3-4\mathrm {i} )(1-\mathrm {i} )^{-1}={\frac {7-\mathrm {i} }{2}}}
2
7
−
i
=
7
+
i
25
{\displaystyle {\frac {2}{7-\mathrm {i} }}={\frac {7+\mathrm {i} }{25}}}
Donnez la forme polaire
r
e
i
θ
{\displaystyle r\operatorname {e} ^{\mathrm {i} \theta }}
1
+
i
{\displaystyle 1+\mathrm {i} }
(
1
+
i
)
2
{\displaystyle (1+\mathrm {i} )^{2}}
(
1
+
i
)
(
1
−
i
)
−
1
{\displaystyle (1+\mathrm {i} )(1-\mathrm {i} )^{-1}}
1
+
i
3
{\displaystyle 1+\mathrm {i} {\sqrt {3}}}
3
+
i
{\displaystyle {\sqrt {3}}+\mathrm {i} }
(
1
+
i
3
)
(
1
+
i
)
−
1
{\displaystyle (1+\mathrm {i} {\sqrt {3}})(1+\mathrm {i} )^{-1}}
(
1
+
i
3
)
(
3
+
i
)
−
1
{\displaystyle (1+\mathrm {i} {\sqrt {3}})({\sqrt {3}}+\mathrm {i} )^{-1}}
−
3
+
i
{\displaystyle -{\sqrt {3}}+\mathrm {i} }
−
1
−
i
{\displaystyle -1-\mathrm {i} }
Solution
1
+
i
=
2
e
i
π
/
4
{\displaystyle 1+\mathrm {i} ={\sqrt {2}}\operatorname {e} ^{\mathrm {i} \pi /4}}
2
e
i
π
/
4
¯
=
2
e
−
i
π
/
4
{\displaystyle {\overline {{\sqrt {2}}\operatorname {e} ^{\mathrm {i} \pi /4}}}={\sqrt {2}}\operatorname {e} ^{-\mathrm {i} \pi /4}}
1
2
e
i
π
/
4
=
1
2
e
−
i
π
/
4
{\displaystyle {\frac {1}{{\sqrt {2}}\operatorname {e} ^{\mathrm {i} \pi /4}}}={\frac {1}{\sqrt {2}}}\operatorname {e} ^{-\mathrm {i} \pi /4}}
(
1
+
i
)
2
=
2
e
i
π
/
2
{\displaystyle (1+\mathrm {i} )^{2}=2\operatorname {e} ^{\mathrm {i} \pi /2}}
2
e
i
π
/
2
¯
=
2
e
−
i
π
/
2
{\displaystyle {\overline {2\operatorname {e} ^{\mathrm {i} \pi /2}}}=2\operatorname {e} ^{-\mathrm {i} \pi /2}}
1
2
e
i
π
/
2
=
1
2
e
−
i
π
/
2
{\displaystyle {\frac {1}{2\operatorname {e} ^{\mathrm {i} \pi /2}}}={\frac {1}{2}}\operatorname {e} ^{-\mathrm {i} \pi /2}}
(
1
+
i
)
(
1
−
i
)
−
1
=
e
i
π
/
2
{\displaystyle (1+\mathrm {i} )(1-\mathrm {i} )^{-1}=\operatorname {e} ^{\mathrm {i} \pi /2}}
e
i
π
/
2
¯
=
1
e
i
π
/
2
=
e
−
i
π
/
2
{\displaystyle {\overline {\operatorname {e} ^{\mathrm {i} \pi /2}}}={\frac {1}{\operatorname {e} ^{\mathrm {i} \pi /2}}}=\operatorname {e} ^{-\mathrm {i} \pi /2}}
1
+
i
3
=
2
e
i
π
/
3
{\displaystyle 1+\mathrm {i} {\sqrt {3}}=2\operatorname {e} ^{\mathrm {i} \pi /3}}
2
e
i
π
/
3
¯
=
2
e
−
i
π
/
3
{\displaystyle {\overline {2\operatorname {e} ^{\mathrm {i} \pi /3}}}=2\operatorname {e} ^{-\mathrm {i} \pi /3}}
1
2
e
i
π
/
3
=
1
2
e
−
i
π
/
3
{\displaystyle {\frac {1}{2\operatorname {e} ^{\mathrm {i} \pi /3}}}={\frac {1}{2}}\operatorname {e} ^{-\mathrm {i} \pi /3}}
3
+
i
=
2
e
i
π
/
6
{\displaystyle {\sqrt {3}}+\mathrm {i} =2\operatorname {e} ^{\mathrm {i} \pi /6}}
2
e
i
π
/
6
¯
=
2
e
−
i
π
/
6
{\displaystyle {\overline {2\operatorname {e} ^{\mathrm {i} \pi /6}}}=2\operatorname {e} ^{-\mathrm {i} \pi /6}}
1
2
e
i
π
/
6
=
1
2
e
−
i
π
/
6
{\displaystyle {\frac {1}{2\operatorname {e} ^{\mathrm {i} \pi /6}}}={\frac {1}{2}}\operatorname {e} ^{-\mathrm {i} \pi /6}}
(
1
+
i
3
)
(
1
+
i
)
−
1
=
2
e
i
π
/
12
{\displaystyle (1+\mathrm {i} {\sqrt {3}})(1+\mathrm {i} )^{-1}={\sqrt {2}}\operatorname {e} ^{\mathrm {i} \pi /12}}
2
e
i
π
/
12
¯
=
2
e
−
i
π
/
12
{\displaystyle {\overline {{\sqrt {2}}\operatorname {e} ^{\mathrm {i} \pi /12}}}={\sqrt {2}}\operatorname {e} ^{-\mathrm {i} \pi /12}}
1
2
e
i
π
/
12
=
1
2
e
−
i
π
/
12
{\displaystyle {\frac {1}{{\sqrt {2}}\operatorname {e} ^{\mathrm {i} \pi /12}}}={\frac {1}{\sqrt {2}}}\operatorname {e} ^{-\mathrm {i} \pi /12}}
(
1
+
i
3
)
(
3
+
i
)
−
1
=
e
i
π
/
6
{\displaystyle (1+\mathrm {i} {\sqrt {3}})({\sqrt {3}}+\mathrm {i} )^{-1}=\operatorname {e} ^{\mathrm {i} \pi /6}}
e
i
π
/
6
¯
=
1
e
i
π
/
6
=
e
−
i
π
/
6
{\displaystyle {\overline {\operatorname {e} ^{\mathrm {i} \pi /6}}}={\frac {1}{\operatorname {e} ^{\mathrm {i} \pi /6}}}=\operatorname {e} ^{-\mathrm {i} \pi /6}}
−
3
+
i
=
2
e
5
i
π
/
6
{\displaystyle -{\sqrt {3}}+\mathrm {i} =2\operatorname {e} ^{5\mathrm {i} \pi /6}}
2
e
5
i
π
/
6
¯
=
2
e
−
5
i
π
/
6
{\displaystyle {\overline {2\operatorname {e} ^{5\mathrm {i} \pi /6}}}=2\operatorname {e} ^{-5\mathrm {i} \pi /6}}
1
2
e
5
i
π
/
6
=
1
2
e
−
5
i
π
/
6
{\displaystyle {\frac {1}{2\operatorname {e} ^{5\mathrm {i} \pi /6}}}={\frac {1}{2}}\operatorname {e} ^{-5\mathrm {i} \pi /6}}
−
1
−
i
=
2
e
−
3
i
π
/
4
{\displaystyle -1-\mathrm {i} ={\sqrt {2}}\operatorname {e} ^{-3\mathrm {i} \pi /4}}
2
e
−
3
i
π
/
4
¯
=
2
e
3
i
π
/
4
{\displaystyle {\overline {{\sqrt {2}}\operatorname {e} ^{-3\mathrm {i} \pi /4}}}={\sqrt {2}}\operatorname {e} ^{3\mathrm {i} \pi /4}}
1
2
e
−
3
i
π
/
4
=
1
2
e
3
i
π
/
4
{\displaystyle {\frac {1}{{\sqrt {2}}\operatorname {e} ^{-3\mathrm {i} \pi /4}}}={\frac {1}{\sqrt {2}}}\operatorname {e} ^{3\mathrm {i} \pi /4}}
Donnez les racines carrées des nombres précédents et la forme cartésienne des racines carrées des nombres suivants, de celles de leurs conjugués et de celles de leurs inverses :
8
+
6
i
{\displaystyle 8+6\mathrm {i} }
5
+
12
i
{\displaystyle 5+12\mathrm {i} }
9
+
40
i
{\displaystyle 9+40\mathrm {i} }
16
+
30
i
{\displaystyle 16+30\mathrm {i} }
Soit
z
∈
C
∖
R
−
{\displaystyle z\in \mathbb {C} \setminus \mathbb {R} _{-}}
v
=
z
+
|
z
|
{\displaystyle v=z+|z|}
Montrer que les racines carrées de
z
{\displaystyle z}
±
v
2
Re
(
v
)
{\displaystyle \pm {\frac {v}{\sqrt {2\operatorname {Re} (v)}}}}
![{\displaystyle \pm {\sqrt[{4}]{2}}\operatorname {e} ^{\mathrm {i} \pi /8}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d458c4bfd6df4237b3adf480ff725dd534aa954)
![{\displaystyle \pm {\sqrt[{4}]{2}}\operatorname {e} ^{\mathrm {i} \pi /24}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d670fb8cd299556d9aa8a2b38d63269c91fa61b4)
![{\displaystyle \pm {\sqrt[{4}]{2}}\operatorname {e} ^{-3\mathrm {i} \pi /8}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a404555a47a5fbd266f53b05975a2ea71e7d835)
