α ∈ [ 0 ; 2 π ] \alpha \in [0; 2\pi] α ∈ [ 0 ; 2 π ]
( E α ) ; z 2 − 2 α e i α ( 1 + 2 i ) z + i 2 2 α + 1 e i 2 α = 0 (E_\alpha);~z^2 - 2^\alpha e^{i\alpha} (1 + 2i) z + i2^{2\alpha + 1} e^{i2\alpha} = 0 ( E α ) ; z 2 − 2 α e i α ( 1 + 2 i ) z + i 2 2 α + 1 e i 2 α = 0 Partie I
1/a
On note :
a = 1 a = 1 a = 1 b = − 2 α e i α ( 1 + 2 i ) b = -2^\alpha e^{i\alpha}(1 + 2i) b = − 2 α e i α ( 1 + 2 i ) c = i ⋅ 2 2 α + 1 e i 2 α c = i \cdot 2^{2\alpha + 1} e^{i2\alpha} c = i ⋅ 2 2 α + 1 e i 2 α
On cherche à calculer le discriminant Δ α = b 2 − 4 a c \Delta_\alpha = b^2 - 4ac Δ α = b 2 − 4 a c
b 2 = ( − 2 α e i α ( 1 + 2 i ) ) 2 = 2 2 α e i 2 α ( 1 + 2 i ) 2 = 2 2 α e i 2 α ( − 3 + 4 i ) \begin{align*}
b^2 &= \left(-2^\alpha e^{i\alpha}(1 + 2i)\right)^2 \\
&= 2^{2\alpha} e^{i2\alpha}(1 + 2i)^2 \\
&= 2^{2\alpha} e^{i2\alpha}(-3 + 4i)
\end{align*} b 2 = ( − 2 α e i α ( 1 + 2 i ) ) 2 = 2 2 α e i 2 α ( 1 + 2 i ) 2 = 2 2 α e i 2 α ( − 3 + 4 i ) 4 a c = 4 ⋅ 1 ⋅ i ⋅ 2 2 α + 1 e i 2 α = i ⋅ 2 2 α + 3 e i 2 α \begin{align*}
4ac &= 4 \cdot 1 \cdot i \cdot 2^{2\alpha + 1} e^{i2\alpha} \\
&= i \cdot 2^{2\alpha + 3} e^{i2\alpha}
\end{align*} 4 a c = 4 ⋅ 1 ⋅ i ⋅ 2 2 α + 1 e i 2 α = i ⋅ 2 2 α + 3 e i 2 α Alors :
Δ α = b 2 − 4 a c = 2 2 α e i 2 α ( − 3 + 4 i ) − i ⋅ 2 2 α + 3 e i 2 α = 2 2 α e i 2 α [ ( − 3 + 4 i ) − 8 i ] = 2 2 α e i 2 α ( − 3 − 4 i ) = ( 2 α e i α ( 1 − 2 i ) ) 2 \begin{align*}
\Delta_\alpha &= b^2 - 4ac \\
&= 2^{2\alpha} e^{i2\alpha}(-3 + 4i) - i \cdot 2^{2\alpha + 3} e^{i2\alpha} \\
&= 2^{2\alpha} e^{i2\alpha} \left[ (-3 + 4i) - 8i \right] \\
&= 2^{2\alpha} e^{i2\alpha}(-3 - 4i) \\
&= \left(2^\alpha e^{i\alpha}(1 - 2i)\right)^2
\end{align*} Δ α = b 2 − 4 a c = 2 2 α e i 2 α ( − 3 + 4 i ) − i ⋅ 2 2 α + 3 e i 2 α = 2 2 α e i 2 α [ ( − 3 + 4 i ) − 8 i ] = 2 2 α e i 2 α ( − 3 − 4 i ) = ( 2 α e i α ( 1 − 2 i ) ) 2 car :
( 1 − 2 i ) 2 = 1 − 4 i + 4 i 2 = − 3 − 4 i (1 - 2i)^2 = 1 - 4i + 4i^2 = -3 - 4i ( 1 − 2 i ) 2 = 1 − 4 i + 4 i 2 = − 3 − 4 i Donc :
Δ α = ( 2 α e i α ( 1 − 2 i ) ) 2 \boxed{\Delta_\alpha = \left(2^\alpha e^{i\alpha}(1 - 2i)\right)^2} Δ α = ( 2 α e i α ( 1 − 2 i ) ) 2 1/b
Soient a a a b b b ( E α ) (E_\alpha) ( E α ) ∣ a ∣ < ∣ b ∣ |a| < |b| ∣ a ∣ < ∣ b ∣
z 1 = − b + Δ α 2 et z 2 = − b − Δ α 2 z_1 = \frac{-b + \sqrt{\Delta_\alpha}}{2} \text{ et } z_2 = \frac{-b - \sqrt{\Delta_\alpha}}{2} z 1 = 2 − b + Δ α et z 2 = 2 − b − Δ α Δ α = 2 α e i α ( 1 − 2 i ) \sqrt{\Delta_\alpha} = 2^\alpha e^{i\alpha}(1 - 2i) Δ α = 2 α e i α ( 1 − 2 i ) z 1 = 2 α e i α ( 1 + 2 i ) + 2 α e i α ( 1 − 2 i ) 2 = 2 α e i α ( 1 + 2 i + 1 − 2 i ) 2 = 2 α e i α \begin{align*}
z_1 &= \frac{2^\alpha e^{i\alpha}(1 + 2i) + 2^\alpha e^{i\alpha}(1 - 2i)}{2} \\
&= \frac{2^\alpha e^{i\alpha}(1 + 2i + 1 - 2i)}{2} \\
&= 2^\alpha e^{i\alpha}
\end{align*} z 1 = 2 2 α e i α ( 1 + 2 i ) + 2 α e i α ( 1 − 2 i ) = 2 2 α e i α ( 1 + 2 i + 1 − 2 i ) = 2 α e i α z 2 = 2 α e i α ( 1 + 2 i ) − 2 α e i α ( 1 − 2 i ) 2 = 2 α e i α ( 1 + 2 i − 1 + 2 i ) 2 = 2 α + 1 i e i α \begin{align*}
z_2 &= \frac{2^\alpha e^{i\alpha}(1 + 2i) - 2^\alpha e^{i\alpha}(1 - 2i)}{2} \\
&= \frac{2^\alpha e^{i\alpha}(1 + 2i - 1 + 2i)}{2} \\
&= 2^{\alpha+1} i e^{i\alpha}
\end{align*} z 2 = 2 2 α e i α ( 1 + 2 i ) − 2 α e i α ( 1 − 2 i ) = 2 2 α e i α ( 1 + 2 i − 1 + 2 i ) = 2 α + 1 i e i α Donc :
z 1 = 2 α e i α ; z 2 = 2 α + 1 i e i α z_1 = 2^\alpha e^{i\alpha} \quad ; \quad z_2 = 2^{\alpha+1} i e^{i\alpha} z 1 = 2 α e i α ; z 2 = 2 α + 1 i e i α ∣ z 1 ∣ = 2 α ; ∣ z 2 ∣ = 2 α + 1 |z_1| = 2^\alpha \quad ; \quad |z_2| = 2^{\alpha+1} ∣ z 1 ∣ = 2 α ; ∣ z 2 ∣ = 2 α + 1 puisque ∣ a ∣ < ∣ b ∣ |a| < |b| ∣ a ∣ < ∣ b ∣
a = z 1 = 2 α e i α ; b = z 2 = 2 α + 1 i e i α a = z_1 = 2^\alpha e^{i\alpha} ~; \quad b = z_2 = 2^{\alpha+1} i e^{i\alpha} a = z 1 = 2 α e i α ; b = z 2 = 2 α + 1 i e i α 2
b a = 2 α + 1 i e i α 2 α e i α = 2 i \frac{b}{a} = \frac{2^{\alpha+1} i e^{i\alpha}}{2^\alpha e^{i\alpha}} = 2i a b = 2 α e i α 2 α + 1 i e i α = 2 i donc b a \frac{b}{a} a b
Partie II
b a = λ i avec λ = Im ( b a ) \frac{b}{a} = \lambda i \quad \text{avec} \quad \lambda = \text{Im} \left( \frac{b}{a} \right) a b = λi avec λ = Im ( a b ) 1 A ( a ) A(a) A ( a ) B ( b ) B(b) B ( b ) H ( h ) H(h) H ( h ) 1 h = 1 a + 1 b \frac{1}{h} = \frac{1}{a} + \frac{1}{b} h 1 = a 1 + b 1
1/a
on a h = a b a + b h = \dfrac{ab}{a + b} h = a + b ab
h b − a = a b b 2 − a 2 \frac{h}{b - a} = \dfrac{ab}{b^2 - a^2} b − a h = b 2 − a 2 ab et b a = λ i ⟹ b = λ i a \frac{b}{a} = \lambda i \implies b = \lambda i a a b = λi ⟹ b = λia
∙ b 2 − a 2 = − λ 2 a 2 − a 2 = − ( λ 2 + 1 ) a 2 ∙ a b = a λ i a = λ i a 2 \begin{align*}
&\bullet~ b^2 - a^2 = -\lambda^2 a^2 - a^2 = -(\lambda^2 + 1) a^2 \\
&\bullet~ ab = a \lambda i a = \lambda i a^2
\end{align*} ∙ b 2 − a 2 = − λ 2 a 2 − a 2 = − ( λ 2 + 1 ) a 2 ∙ ab = aλia = λi a 2 h b − a = λ i a 2 − ( λ 2 + 1 ) a 2 = − ( λ λ 2 + 1 ) i \begin{align*}
\frac{h}{b - a} &= \dfrac{\lambda i a^2}{-(\lambda^2 + 1)a^2} \\
&= - \left( \frac{\lambda}{\lambda^2 + 1} \right)i
\end{align*} b − a h = − ( λ 2 + 1 ) a 2 λi a 2 = − ( λ 2 + 1 λ ) i on a
( A B → , O H → ‾ ) ≡ arg ( h b − a ) [ 2 π ] ≡ ± π 2 [ 2 π ] \begin{align*}
\left(\overline{\overrightarrow{AB},\overrightarrow{OH}}\right) &\equiv \arg\left(\frac{h}{b - a}\right)~[2\pi] \\
&\equiv \pm\frac\pi2~ [2\pi]
\end{align*} ( A B , O H ) ≡ arg ( b − a h ) [ 2 π ] ≡ ± 2 π [ 2 π ] Donc ( A B ) ⊥ ( O H ) (AB)\perp(OH) ( A B ) ⊥ ( O H )
1/b
h − a b − a = h b − a − a b − a \frac{h - a}{b - a} = \frac{h}{b - a} - \frac{a}{b - a} b − a h − a = b − a h − b − a a a b − a = a λ i a − a = 1 λ i − 1 \frac{a}{b - a} = \frac{a}{\lambda i a - a} = \frac{1}{\lambda i - 1} b − a a = λia − a a = λi − 1 1 h − a b − a = − ( λ λ 2 + 1 ) i − 1 λ i − 1 = − ( λ λ 2 + 1 ) i − − λ i − 1 λ 2 + 1 = 1 λ 2 + 1 ∈ R \begin{align*}
\frac{h - a}{b - a} &= - \left( \frac{\lambda}{\lambda^2 + 1} \right)i - \frac{1}{\lambda i - 1} \\
&= - \left( \frac{\lambda}{\lambda^2 + 1} \right)i - \frac{-\lambda i - 1}{\lambda^2 + 1} \\
&= \frac{1}{\lambda^2 + 1} \in \R
\end{align*} b − a h − a = − ( λ 2 + 1 λ ) i − λi − 1 1 = − ( λ 2 + 1 λ ) i − λ 2 + 1 − λi − 1 = λ 2 + 1 1 ∈ R Alors les points H , A H, A H , A B B B
2/a
on a :
m = h 2 et n = h + b 2 m = \frac h2 \quad \text{et} \quad n = \frac{h + b}{2} m = 2 h et n = 2 h + b Donc
m − a = − h − 2 a 2 m - a = -\frac{h - 2a}{2} m − a = − 2 h − 2 a donc :
n m − a = − h + b h − 2 a \frac{n}{m - a} = -\frac{h + b}{h - 2a} m − a n = − h − 2 a h + b et on a : h = a b a + b h = \dfrac{ab}{a + b} h = a + b ab
h + b = 2 a b + b 2 a + b h + b = \frac{2ab + b^2}{a + b} h + b = a + b 2 ab + b 2 h − 2 a = − a b − 2 a 2 a + b h - 2a = \frac{-ab - 2a^2}{a + b} h − 2 a = a + b − ab − 2 a 2 Alors
n m − a = 2 a b + b 2 − a b − 2 a 2 = − b a = − λ i \begin{align*}
\frac{n}{m - a} &= \frac{2ab + b^2}{-ab - 2a^2} = -\frac{b}{a} = -\lambda i
\end{align*} m − a n = − ab − 2 a 2 2 ab + b 2 = − a b = − λi 2/b
on a
( A I → , O J → ‾ ) ≡ arg ( n m − a ) [ 2 π ] ≡ ± π 2 [ 2 π ] \begin{align*}
\left(\overline{\overrightarrow{AI},\overrightarrow{OJ}}\right) &\equiv \arg\left(\frac{n}{m - a}\right)~[2\pi] \\
&\equiv \pm\frac\pi2~ [2\pi]
\end{align*} ( A I , O J ) ≡ arg ( m − a n ) [ 2 π ] ≡ ± 2 π [ 2 π ] Donc ( O J ) ⊥ ( A I ) (OJ)\perp(AI) ( O J ) ⊥ ( A I )
n m − a = − λ i ⟹ ∣ n ∣ ∣ m − a ∣ = ∣ λ ∣ ⟹ O J A I = ∣ λ ∣ ⟹ O J = ∣ λ ∣ A I \begin{align*}
\frac{n}{m - a} = -\lambda i &\implies \frac{|n|}{|m - a|} = |\lambda| \\
&\implies \frac{OJ}{AI} = |\lambda| \\
&\implies OJ = |\lambda| AI
\end{align*} m − a n = − λi ⟹ ∣ m − a ∣ ∣ n ∣ = ∣ λ ∣ ⟹ A I O J = ∣ λ ∣ ⟹ O J = ∣ λ ∣ A I 2/c
on a ( O J ) ⊥ ( A I ) (OJ)\perp(AI) ( O J ) ⊥ ( A I ) K ∈ ( O J ) ∩ ( A I ) K \in (OJ) \cap (AI) K ∈ ( O J ) ∩ ( A I )
Donc, les vecteurs K I → \overrightarrow{KI} K I K J → \overrightarrow{KJ} K J
Si k k k K K K
alors ∃ α ∈ R \exists \alpha \in \R ∃ α ∈ R m − k n − k = α i ( ∗ ) \frac{m - k}{n - k} = \alpha i ~~~~(*) n − k m − k = α i ( ∗ )
on a ( O H ) ⊥ ( A B ) (OH)\perp(AB) ( O H ) ⊥ ( A B )
J ∈ ( H B ) J \in (HB) J ∈ ( H B ) J J J [ H B ] [HB] [ H B ] I ∈ ( O H ) I \in (OH) I ∈ ( O H ) I I I [ O H ] [OH] [ O H ]
Les vecteurs H I → \overrightarrow{HI} H I H J → \overrightarrow{HJ} H J
alors ∃ β ∈ R \exists \beta \in \R ∃ β ∈ R n − h m − h = β i ( ∗ ∗ ) \frac{n - h}{m - h} = \beta i ~~~~(**) m − h n − h = β i ( ∗ ∗ )
de ( ∗ ) (*) ( ∗ ) ( ∗ ∗ ) (**) ( ∗ ∗ )
m − k n − k × n − h m − h = − α β ∈ R \frac{m - k}{n - k} \times \frac{n - h}{m - h} = -\alpha \beta \in \R n − k m − k × m − h n − h = − α β ∈ R Alors les points K , I , H K, I, H K , I , H J J J
Remarque : les points K , I , H K, I, H K , I , H J J J [ I J ] [IJ] [ I J ]
2/d
on a :
n − m = h + b 2 − h 2 = b 2 n - m = \frac{h + b}{2} - \frac{h}{2} = \frac{b}{2} n − m = 2 h + b − 2 h = 2 b donc
n − m a = b 2 a = − 1 2 λ i ∈ i R \frac{n - m}{a} = \frac{b}{2a} = -\frac{1}{2} \lambda i \in i\R a n − m = 2 a b = − 2 1 λi ∈ i R Alors ( I J ) ⊥ ( O A ) (IJ)\perp(OA) ( I J ) ⊥ ( O A )
Exemple on prend a = 2 a=2 a = 2 b = 2 i b=2i b = 2 i
donc b a = i \frac ba=i a b = i λ = 1 \lambda=1 λ = 1
h = a b a + b = 4 i 2 + 2 i = 2 i 1 + i = i ( 1 − i ) = 1 + i h=\frac{ab}{a+b}=\frac{4i}{2+2i}=\frac{2i}{1+i}=i(1-i)=1+i h = a + b ab = 2 + 2 i 4 i = 1 + i 2 i = i ( 1 − i ) = 1 + i
I ( m ) I(m) I ( m ) [ O H ] [OH] [ O H ] m = 1 2 + 1 2 i m=\frac12+\frac12i m = 2 1 + 2 1 i
J ( n ) J(n) J ( n ) [ H B ] [HB] [ H B ] n = 1 + 3 i 2 n=\frac{1+3i}{2} n = 2 1 + 3 i
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