Exercice 2 (3,5 points)
Dans le plan complexe rapporté à un repère orthonormé direct ( O ; u ⃗ , v ⃗ ) (O; \vec{u}, \vec{v}) ( O ; u , v ) A , B , C , D A, B, C, D A , B , C , D Ω \Omega Ω
a = 1 + 2 i , b = a ˉ , c = 3 ( 3 + i ) 2 a = 1 + 2i, \quad b = \bar{a}, \quad c = \frac{3(3+i)}{2} a = 1 + 2 i , b = a ˉ , c = 2 3 ( 3 + i ) d = 3 ( 1 + i ) 2 , ω = 5 2 d = \frac{3(1+i)}{2}, \quad \omega = \frac{5}{2} d = 2 3 ( 1 + i ) , ω = 2 5
a) Vérifier que a + b = 2 a + b = 2 a + b = 2 P P P [ A B ] [AB] [ A B ] p = 1 p = 1 p = 1
Afficher la correction
On a :
a + b = a + a ˉ = 2 R e ( a ) = 2 ( 1 ) = 2 a + b = a + \bar{a} = 2\mathcal{R}e(a) = 2(1) = 2 a + b = a + a ˉ = 2 R e ( a ) = 2 ( 1 ) = 2 P P P [ A B ] [AB] [ A B ]
p = a + b 2 = 2 2 = 1 p = \frac{a+b}{2} = \frac{2}{2} = 1 p = 2 a + b = 2 2 = 1
1. b) Montrer que a a a b b b z 2 − 2 z + 5 = 0 dans C z^2 - 2z + 5 = 0 \quad \text{dans } \mathbb{C} z 2 − 2 z + 5 = 0 dans C
Afficher la correction
On a a + b = 2 a+b = 2 a + b = 2
a b = a ⋅ a ˉ = ∣ a ∣ 2 = ( 1 2 + 2 2 ) 2 = 1 + 4 = 5 ab = a \cdot \bar{a} = |a|^2 = (\sqrt{1^2 + 2^2})^2 = 1 + 4 = 5 ab = a ⋅ a ˉ = ∣ a ∣ 2 = ( 1 2 + 2 2 ) 2 = 1 + 4 = 5 Puisque la somme est a + b = 2 a+b = 2 a + b = 2 a b = 5 ab = 5 ab = 5 a a a b b b z 2 − ( a + b ) z + a b = 0 z^2 - (a+b)z + ab = 0 z 2 − ( a + b ) z + ab = 0
z 2 − 2 z + 5 = 0 z^2 - 2z + 5 = 0 z 2 − 2 z + 5 = 0
a) Vérifier que ∣ a − ω ∣ = ∣ b − ω ∣ = ∣ ω − c ∣ |a - \omega| = |b - \omega| = |\omega - c| ∣ a − ω ∣ = ∣ b − ω ∣ = ∣ ω − c ∣
Afficher la correction
Calculons chaque module :
ω − a = 5 2 − ( 1 + 2 i ) = 3 2 − 2 i ⟹ ∣ ω − a ∣ = 9 4 + 4 = 25 4 = 5 2 \omega - a = \frac{5}{2} - (1+2i) = \frac{3}{2} - 2i \implies |\omega - a| = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2} ω − a = 2 5 − ( 1 + 2 i ) = 2 3 − 2 i ⟹ ∣ ω − a ∣ = 4 9 + 4 = 4 25 = 2 5 ω − b = 5 2 − ( 1 − 2 i ) = 3 2 + 2 i ⟹ ∣ ω − b ∣ = 9 4 + 4 = 25 4 = 5 2 \omega - b = \frac{5}{2} - (1-2i) = \frac{3}{2} + 2i \implies |\omega - b| = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2} ω − b = 2 5 − ( 1 − 2 i ) = 2 3 + 2 i ⟹ ∣ ω − b ∣ = 4 9 + 4 = 4 25 = 2 5 ω − c = 5 2 − 9 + 3 i 2 = 5 − 9 − 3 i 2 = − 2 − 3 2 i ⟹ ∣ ω − c ∣ = 4 + 9 4 = 25 4 = 5 2 \omega - c = \frac{5}{2} - \frac{9+3i}{2} = \frac{5 - 9 - 3i}{2} = -2 - \frac{3}{2}i \implies |\omega - c| = \sqrt{4 + \frac{9}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2} ω − c = 2 5 − 2 9 + 3 i = 2 5 − 9 − 3 i = − 2 − 2 3 i ⟹ ∣ ω − c ∣ = 4 + 4 9 = 4 25 = 2 5
Alors on a bien :
∣ a − ω ∣ = ∣ b − ω ∣ = ∣ ω − c ∣ = 5 2 |a - \omega| = |b - \omega| = |\omega - c| = \frac{5}{2} ∣ a − ω ∣ = ∣ b − ω ∣ = ∣ ω − c ∣ = 2 5 (Note : ∣ a − ω ∣ |a - \omega| ∣ a − ω ∣ ∣ ω − a ∣ |\omega - a| ∣ ω − a ∣
2. b) Déduire que ω \omega ω A B C ABC A B C
Afficher la correction
On a :
∣ a − ω ∣ = ∣ b − ω ∣ = ∣ c − ω ∣ |a - \omega| = |b - \omega| = |c - \omega| ∣ a − ω ∣ = ∣ b − ω ∣ = ∣ c − ω ∣ Donc Ω A = Ω B = Ω C \Omega A = \Omega B = \Omega C Ω A = Ω B = Ω C A A A B B B C C C Ω \Omega Ω R = 5 2 R = \frac{5}{2} R = 2 5 Ω \Omega Ω A B C ABC A B C
a) Vérifier que d − c a − b = 3 4 i \frac{d - c}{a - b} = \frac{3}{4} i a − b d − c = 4 3 i
Afficher la correction
Calculons séparément :
d − c = 3 ( 1 + i ) 2 − 3 ( 3 + i ) 2 = 3 + 3 i − 9 − 3 i 2 = − 6 2 = − 3 d - c = \frac{3(1+i)}{2} - \frac{3(3+i)}{2} = \frac{3 + 3i - 9 - 3i}{2} = \frac{-6}{2} = -3 d − c = 2 3 ( 1 + i ) − 2 3 ( 3 + i ) = 2 3 + 3 i − 9 − 3 i = 2 − 6 = − 3 a − b = ( 1 + 2 i ) − ( 1 − 2 i ) = 4 i a - b = (1+2i) - (1-2i) = 4i a − b = ( 1 + 2 i ) − ( 1 − 2 i ) = 4 i
Donc :
d − c a − b = − 3 4 i = − 3 i 4 i 2 = − 3 i − 4 = 3 4 i \frac{d-c}{a-b} = \frac{-3}{4i} = \frac{-3i}{4i^2} = \frac{-3i}{-4} = \frac{3}{4}i a − b d − c = 4 i − 3 = 4 i 2 − 3 i = − 4 − 3 i = 4 3 i
3. b) Montrer que d − b = ( c − a ) e i π 2 d - b = (c - a)e^{\frac{i\pi}{2}} d − b = ( c − a ) e 2 iπ ( D B ) (DB) ( D B ) ( A C ) (AC) ( A C )
Afficher la correction
Calculons d − b d-b d − b c − a c-a c − a
d − b = 3 + 3 i 2 − ( 1 − 2 i ) = 3 + 3 i − 2 + 4 i 2 = 1 + 7 i 2 d - b = \frac{3+3i}{2} - (1-2i) = \frac{3+3i - 2 + 4i}{2} = \frac{1+7i}{2} d − b = 2 3 + 3 i − ( 1 − 2 i ) = 2 3 + 3 i − 2 + 4 i = 2 1 + 7 i c − a = 9 + 3 i 2 − ( 1 + 2 i ) = 9 + 3 i − 2 − 4 i 2 = 7 − i 2 c - a = \frac{9+3i}{2} - (1+2i) = \frac{9+3i - 2 - 4i}{2} = \frac{7-i}{2} c − a = 2 9 + 3 i − ( 1 + 2 i ) = 2 9 + 3 i − 2 − 4 i = 2 7 − i
Faisons le rapport :
d − b c − a = 1 + 7 i 7 − i = ( 1 + 7 i ) ( 7 + i ) 7 2 + ( − 1 ) 2 = 7 + i + 49 i − 7 50 = 50 i 50 = i = e i π 2 \begin{aligned}
\frac{d-b}{c-a} &= \frac{1+7i}{7-i} \\
&= \frac{(1+7i)(7+i)}{7^2 + (-1)^2} \\
&= \frac{7 + i + 49i - 7}{50} \\
&= \frac{50i}{50} \\
&= i \\
&= e^{i\frac{\pi}{2}}
\end{aligned} c − a d − b = 7 − i 1 + 7 i = 7 2 + ( − 1 ) 2 ( 1 + 7 i ) ( 7 + i ) = 50 7 + i + 49 i − 7 = 50 50 i = i = e i 2 π Alors : d − b = ( c − a ) e i π 2 d - b = (c - a)e^{i\frac{\pi}{2}} d − b = ( c − a ) e i 2 π
Pour la déduction :
( A C → , B D → ) ≡ arg ( d − b c − a ) ( m o d 2 π ) ≡ arg ( i ) ( m o d 2 π ) ≡ π 2 ( m o d 2 π ) \begin{aligned}
(\overrightarrow{AC}, \overrightarrow{BD}) &\equiv \arg\left(\frac{d-b}{c-a}\right) \pmod{2\pi} \\
&\equiv \arg(i) \pmod{2\pi} \\
&\equiv \frac{\pi}{2} \pmod{2\pi}
\end{aligned} ( A C , B D ) ≡ arg ( c − a d − b ) ( mod 2 π ) ≡ arg ( i ) ( mod 2 π ) ≡ 2 π ( mod 2 π ) Donc les droites ( D B ) (DB) ( D B ) ( A C ) (AC) ( A C ) ( A C ) ⊥ ( B D ) (AC) \perp (BD) ( A C ) ⊥ ( B D )
Soit h h h C C C 2 3 \frac{2}{3} 3 2 M M M z z z M ′ M' M ′ z ′ z' z ′ h ( P ) = G h(P) = G h ( P ) = G
4. a) Vérifier que z ′ = 2 3 z + 1 3 c z' = \frac{2}{3}z + \frac{1}{3}c z ′ = 3 2 z + 3 1 c
Afficher la correction
L'homothétie h h h C ( c ) C(c) C ( c ) k = 2 3 k = \frac{2}{3} k = 3 2
h ( M ) = M ′ ⟺ z ′ − c = k ( z − c ) ⟺ z ′ = 2 3 ( z − c ) + c ⟺ z ′ = 2 3 z − 2 3 c + c ⟺ z ′ = 2 3 z + 1 3 c \begin{aligned}
h(M) = M' &\iff z' - c = k(z - c) \\
&\iff z' = \frac{2}{3}(z - c) + c \\
&\iff z' = \frac{2}{3}z - \frac{2}{3}c + c \\
&\iff z' = \frac{2}{3}z + \frac{1}{3}c
\end{aligned} h ( M ) = M ′ ⟺ z ′ − c = k ( z − c ) ⟺ z ′ = 3 2 ( z − c ) + c ⟺ z ′ = 3 2 z − 3 2 c + c ⟺ z ′ = 3 2 z + 3 1 c (En utilisant la valeur de c c c z ′ = 2 3 z + 3 + i 2 = 2 3 z + 3 2 + 1 2 i z' = \frac{2}{3}z + \frac{3+i}{2} = \frac{2}{3}z + \frac{3}{2} + \frac{1}{2}i z ′ = 3 2 z + 2 3 + i = 3 2 z + 2 3 + 2 1 i
4. b) Montrer que l'affixe du point G G G g = 13 6 + 1 2 i g = \frac{13}{6} + \frac{1}{2}i g = 6 13 + 2 1 i
Afficher la correction
On a h ( P ) = G h(P) = G h ( P ) = G p = 1 p=1 p = 1
g = 2 3 p + 1 3 c = 2 3 ( 1 ) + ( 3 2 + 1 2 i ) = 2 3 + 3 2 + 1 2 i = 4 6 + 9 6 + 1 2 i = 13 6 + 1 2 i \begin{aligned}
g &= \frac{2}{3}p + \frac{1}{3}c \\
&= \frac{2}{3}(1) + \left(\frac{3}{2} + \frac{1}{2}i\right) \\
&= \frac{2}{3} + \frac{3}{2} + \frac{1}{2}i \\
&= \frac{4}{6} + \frac{9}{6} + \frac{1}{2}i \\
&= \frac{13}{6} + \frac{1}{2}i
\end{aligned} g = 3 2 p + 3 1 c = 3 2 ( 1 ) + ( 2 3 + 2 1 i ) = 3 2 + 2 3 + 2 1 i = 6 4 + 6 9 + 2 1 i = 6 13 + 2 1 i
Montrer que les points Ω \Omega Ω G G G D D D
Afficher la correction
Calculons les vecteurs Ω G → \overrightarrow{\Omega G} Ω G Ω D → \overrightarrow{\Omega D} Ω D
g − ω = ( 13 6 + 1 2 i ) − 5 2 = 13 6 − 15 6 + 1 2 i = − 2 6 + 1 2 i = − 1 3 + 1 2 i \begin{aligned}
g - \omega &= \left(\frac{13}{6} + \frac{1}{2}i\right) - \frac{5}{2} \\
&= \frac{13}{6} - \frac{15}{6} + \frac{1}{2}i \\
&= -\frac{2}{6} + \frac{1}{2}i \\
&= -\frac{1}{3} + \frac{1}{2}i
\end{aligned} g − ω = ( 6 13 + 2 1 i ) − 2 5 = 6 13 − 6 15 + 2 1 i = − 6 2 + 2 1 i = − 3 1 + 2 1 i Et pour d − ω d - \omega d − ω
d − ω = 3 ( 1 + i ) 2 − 5 2 = 3 + 3 i − 5 2 = − 1 + 3 2 i \begin{aligned}
d - \omega &= \frac{3(1+i)}{2} - \frac{5}{2} \\
&= \frac{3+3i-5}{2} \\
&= -1 + \frac{3}{2}i
\end{aligned} d − ω = 2 3 ( 1 + i ) − 2 5 = 2 3 + 3 i − 5 = − 1 + 2 3 i On remarque que :
− 1 + 3 2 i = 3 ( − 1 3 + 1 2 i ) -1 + \frac{3}{2}i = 3\left(-\frac{1}{3} + \frac{1}{2}i\right) − 1 + 2 3 i = 3 ( − 3 1 + 2 1 i ) Ce qui implique :
d − ω = 3 ( g − ω ) ⟹ g − ω d − ω = 1 3 ∈ R d - \omega = 3(g - \omega) \implies \frac{g - \omega}{d - \omega} = \frac{1}{3} \in \mathbb{R} d − ω = 3 ( g − ω ) ⟹ d − ω g − ω = 3 1 ∈ R Les vecteurs Ω G → \overrightarrow{\Omega G} Ω G Ω D → \overrightarrow{\Omega D} Ω D D D D G G G Ω \Omega Ω