Calculer un angle avec les complexes Exercice

\dfrac{z_C-z_A}{z_B-z_A}=\dfrac{-\sqrt{2}+i\sqrt{2}-\left(\sqrt{2}+i\sqrt{2}\right)}{\sqrt{2}-i\sqrt{2}-\left(\sqrt{2}+i\sqrt{2}\right)}

\dfrac{z_C-z_A}{z_B-z_A}=\dfrac{-\sqrt{2}+i\sqrt{2}-\sqrt{2}-i\sqrt{2}}{\sqrt{2}-i\sqrt{2}-\sqrt{2}-i\sqrt{2}}

\dfrac{z_C-z_A}{z_B-z_A}=\dfrac{-2\sqrt{2}}{-2i\sqrt{2}}

\dfrac{z_C-z_A}{z_B-z_A}=\dfrac{1}{i}

On multiplie le dénominateur par son complexe conjugué :

\dfrac{z_C-z_A}{z_B-z_A}=\dfrac{-i}{i\times\left(-i\right)}

\dfrac{z_C-z_A}{z_B-z_A}=-i

On sait directement que arg\left(-i\right)=-\dfrac{\pi}{2}+2k\pi, k\in\mathbb{Z}

\dfrac{z_C-z_A}{z_B-z_A}=\dfrac{-9+5i-\left(3+5i\right)}{9+5i-\left(3+5i\right)}

\dfrac{z_C-z_A}{z_B-z_A}=\dfrac{-9+5i-3-5i}{9+5i-3-5i}

\dfrac{z_C-z_A}{z_B-z_A}=\dfrac{-12}{6}

\dfrac{z_C-z_A}{z_B-z_A}=-2

On sait directement que arg\left(-2\right)= arg\left(-1\right) =\pi+2k\pi, k\in\mathbb{Z}

\dfrac{z_C-z_A}{z_B-z_A}=\dfrac{3-i\sqrt{3}-\left(2-2i\sqrt{3}\right)}{4-2i\sqrt{3}-\left(2-2i\sqrt{3}\right)}

\dfrac{z_C-z_A}{z_B-z_A}=\dfrac{3-i\sqrt{3}-2+2i\sqrt{3}}{4-2i\sqrt{3}-2+2i\sqrt{3}}

\dfrac{z_C-z_A}{z_B-z_A}=\dfrac{1+i\sqrt{3}}{2}

\left| \dfrac{1+i\sqrt{3}}{2} \right|=\sqrt{\left(\dfrac{1}{2}\right)^2+\left(\dfrac{\sqrt{3}}{2}\right)^2}

\left| \dfrac{1+i\sqrt{3}}{2} \right|=\sqrt{\dfrac{1}{4}+\dfrac{3}{4}}

\left| \dfrac{1+i\sqrt{3}}{2} \right|=1

On note \theta=arg\left(\dfrac{1+i\sqrt{3}}{2}\right)

On a :

• \cos\left(\theta\right)=\dfrac{\dfrac{1}{2}}{1}=\dfrac{1}{2}
• \sin\left(\theta\right)=\dfrac{\dfrac{\sqrt{3}}{2}}{1}=\dfrac{\sqrt{3}}{2}

Ainsi, on obtient \theta=\dfrac{\pi}{3}+2k\pi, k\in\mathbb{Z}

\dfrac{z_C-z_A}{z_B-z_A}=\dfrac{5-i-\left(1+3i\right)}{5+3i-\left(1+3i\right)}

\dfrac{z_C-z_A}{z_B-z_A}=\dfrac{5-i-1-3i}{5+3i-1-3i}

\dfrac{z_C-z_A}{z_B-z_A}=\dfrac{4-4i}{4}

\dfrac{z_C-z_A}{z_B-z_A}=1-i

\left| 1-i \right|=\sqrt{1^2+\left(-1\right)^2}

\left| 1-i \right|=\sqrt{2}

On note \theta=arg\left(1-i\right)

On a :

• \cos\left(\theta\right)=\dfrac{{1}}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}
• \sin\left(\theta\right)=\dfrac{-1}{\sqrt2}=-\dfrac{\sqrt{2}}{2}

Ainsi, on obtient \theta=-\dfrac{\pi}{4}+2k\pi, k\in\mathbb{Z}

\dfrac{z_C-z_A}{z_B-z_A}=\dfrac{\sqrt3-3+i-\left(-3+2i\right)}{-5+2i-\left(-3+2i\right)}

\dfrac{z_C-z_A}{z_B-z_A}=\dfrac{\sqrt3-3+i+3-2i}{-5+2i+3-2i}

\dfrac{z_C-z_A}{z_B-z_A}=\dfrac{\sqrt3-i}{-2}

\dfrac{z_C-z_A}{z_B-z_A}=\dfrac{-\sqrt3+i}{2}

\left| \dfrac{-\sqrt3+i}{2} \right|=\sqrt{\left(\dfrac{-\sqrt3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2}

\left| \dfrac{-\sqrt3+i}{2} \right|=\sqrt{\dfrac{3}{4}+\dfrac{1}{4}}

\left| \dfrac{-\sqrt3+i}{2} \right|=\sqrt{1}

\left| \dfrac{-\sqrt3+i}{2} \right|=1

On note \theta=arg\left(\dfrac{-\sqrt3+i}{2}\right)

On a :

• \cos\left(\theta\right)=\dfrac{\dfrac{-\sqrt3}{2}}{1}=-\dfrac{\sqrt3}{2}
• \sin\left(\theta\right)=\dfrac{\dfrac{1}{2}}{1}=\dfrac{1}{2}

Ainsi, on obtient \theta=\dfrac{5\pi}{6}+2k\pi, k\in\mathbb{Z}

\dfrac{z_D-z_B}{z_C-z_A}=\dfrac{-3+2\sqrt{3}i-\left(-4+3\sqrt{3}i\right)}{-6+3i-\left(-6+i\right)}

\dfrac{z_D-z_B}{z_C-z_A}=\dfrac{-3+2\sqrt{3}i+4-3\sqrt{3}i}{-6+3i+6-i}

\dfrac{z_D-z_B}{z_C-z_A}=\dfrac{1-\sqrt{3}i}{2i}

On multiplie le numérateur et le dénominateur par i :

\dfrac{z_D-z_B}{z_C-z_A}=\dfrac{i-\sqrt{3}i^2}{2i^2}

\dfrac{z_D-z_B}{z_C-z_A}=\dfrac{-\sqrt{3}-i}{2}

\left| \dfrac{-\sqrt{3}-i}{2}\right|=\sqrt{\left(\dfrac{-\sqrt{3}}{2}\right)^2+\left(\dfrac{-1}{2}\right)^2}

\left| \dfrac{-\sqrt{3}-i}{2}\right|=\sqrt{\dfrac{3}{4}+\dfrac{1}{4}}

\left| \dfrac{-\sqrt{3}-i}{2}\right|=1

On note \theta=arg\left(\dfrac{-\sqrt{3}-i}{2}\right)

On a :

• \cos\left(\theta\right)=\dfrac{\dfrac{-\sqrt{3}}{2}}{1}=-\dfrac{\sqrt{3}}{2}
• \sin\left(\theta\right)=\dfrac{-\dfrac{1}{2}}{1}=-\dfrac{1}{2}

Ainsi, on obtient \theta=\dfrac{7\pi}{6}+2k\pi, k\in\mathbb{Z}