Nombres complexes : point de vue géométrique
Déterminer des modules à l'aide des propriétés Exercice 1 1
z 1 = ( 1 + 2 i ) ( 3 − i ) z_{1} =\left(1+2i\right)\left(3-i\right) z 1 = ( 1 + 2 i ) ( 3 − i ) Il n'est pas utile de déterminer la forme algébrique de
z 1 z_1 z 1 .
Pour tous complexes z 1 z_1 z 1 z 2 z_2 z 2 ∣ z 1 z 2 ∣ = ∣ z 1 ∣ × ∣ z 2 ∣ \left|z_{1} z_{2} \right|=\left|z_{1} \right|\times \left|z_{2} \right| ∣ z 1 z 2 ∣ = ∣ z 1 ∣ × ∣ z 2 ∣ ∣ z 1 ∣ = ∣ ( 1 + 2 i ) ( 3 − i ) ∣ \left|z_{1} \right|=\left|\left(1+2i\right)\left(3-i\right)\right| ∣ z 1 ∣ = ∣ ( 1 + 2 i ) ( 3 − i ) ∣ ∣ z 1 ∣ = ∣ 1 + 2 i ∣ × ∣ 3 − i ∣ \left|z_{1} \right|=\left|1+2i\right|\times \left|3-i\right| ∣ z 1 ∣ = ∣ 1 + 2 i ∣ × ∣ 3 − i ∣ ∣ z 1 ∣ = ( 1 2 + 2 2 ) × ( 3 2 + ( − 1 ) 2 ) \left|z_{1} \right|=\left(\sqrt{1^{2} +2^{2} } \right)\times \left(\sqrt{3^{2} +\left(-1\right)^{2} } \right) ∣ z 1 ∣ = ( 1 2 + 2 2 ) × ( 3 2 + ( − 1 ) 2 ) ∣ z 1 ∣ = 5 × 10 \left|z_{1} \right|=\sqrt{5} \times \sqrt{10} ∣ z 1 ∣ = 5 × 1 0 ∣ z 1 ∣ = 50 \left|z_{1} \right|=\sqrt{50} ∣ z 1 ∣ = 5 0 ∣ z 1 ∣ = 25 × 2 \left|z_{1} \right|=\sqrt{25\times 2} ∣ z 1 ∣ = 2 5 × 2 Ainsi :
∣ z 1 ∣ = 5 2 \left|z_{1} \right|=5\sqrt{2} ∣ z 1 ∣ = 5 2 Exercice 2 1
z 1 = 2 + i 3 + 4 i z_{1} =\frac{2+i}{3+4i} z 1 = 3 + 4 i 2 + i Pour tous complexes z 1 z_1 z 1 z 2 z_2 z 2 z 2 ≠ 0 z_2\ne 0 z 2 = 0 ∣ z 1 z 2 ∣ = ∣ z 1 ∣ ∣ z 2 ∣ \left|\frac{z_{1} }{z_{2} } \right|=\frac{\left|z_{1} \right|}{\left|z_{2} \right|} ∣ ∣ ∣ ∣ z 2 z 1 ∣ ∣ ∣ ∣ = ∣ z 2 ∣ ∣ z 1 ∣ ∣ z 1 ∣ = ∣ 2 + i 3 + 4 i ∣ \left|z_{1} \right|=\left|\frac{2+i}{3+4i} \right| ∣ z 1 ∣ = ∣ ∣ ∣ ∣ 3 + 4 i 2 + i ∣ ∣ ∣ ∣ ∣ z 1 ∣ = ∣ 2 + i ∣ ∣ 3 + 4 i ∣ \left|z_{1} \right|=\frac{\left|2+i\right|}{\left|3+4i\right|} ∣ z 1 ∣ = ∣ 3 + 4 i ∣ ∣ 2 + i ∣ ∣ z 1 ∣ = 2 2 + 1 2 3 2 + 4 2 \left|z_{1} \right|=\frac{\sqrt{2^{2} +1^{2} } }{\sqrt{3^{2} +4^{2} } } ∣ z 1 ∣ = 3 2 + 4 2 2 2 + 1 2 ∣ z 1 ∣ = 5 25 \left|z_{1} \right|=\frac{\sqrt{5} }{\sqrt{25} } ∣ z 1 ∣ = 2 5 5 Ainsi :
∣ z 1 ∣ = 5 5 \left|z_{1} \right|=\frac{\sqrt{5} }{5} ∣ z 1 ∣ = 5 5 2
z 2 = 3 1 + i z_{2} =\frac{3}{1+i} z 2 = 1 + i 3 3
z 3 = 2 − 2 i 3 − i z_{3} =\frac{2-2i}{\sqrt{3} -i} z 3 = 3 − i 2 − 2 i Exercice 3 1
z 1 = ( 1 + i ) 8 z_{1} =\left(1+i\right)^{8} z 1 = ( 1 + i ) 8 Soit z 1 z_1 z 1 n n n ∣ ( z 1 ) n ∣ = ∣ z 1 ∣ n \left|\left(z_{1} \right)^{n} \right|=\left|z_{1} \right|^{n} ∣ ( z 1 ) n ∣ = ∣ z 1 ∣ n ∣ z 1 ∣ = ∣ ( 1 + i ) 8 ∣ \left|z_{1} \right|=\left|\left(1+i\right)^{8} \right| ∣ z 1 ∣ = ∣ ∣ ∣ ( 1 + i ) 8 ∣ ∣ ∣ ∣ z 1 ∣ = ∣ ( 1 + i ) ∣ 8 \left|z_{1} \right|=\left|\left(1+i\right)\right|^{8} ∣ z 1 ∣ = ∣ ( 1 + i ) ∣ 8 ∣ z 1 ∣ = ( 1 2 + 1 2 ) 8 \left|z_{1} \right|=\left(\sqrt{1^{2} +1^{2} } \right)^{8} ∣ z 1 ∣ = ( 1 2 + 1 2 ) 8 ∣ z 1 ∣ = ( 2 ) 8 \left|z_{1} \right|=\left(\sqrt{2} \right)^{8} ∣ z 1 ∣ = ( 2 ) 8 Ainsi :
∣ z 1 ∣ = 16 \left|z_{1} \right|=16 ∣ z 1 ∣ = 1 6 2
z 2 = ( 5 + 4 i ) 6 z_{2} =\left(\sqrt{5} +4i\right)^{6} z 2 = ( 5 + 4 i ) 6 3
z 3 = ( 1 − 3 ) 12 z_{3} =\left(1-\sqrt{3} \right)^{12} z 3 = ( 1 − 3 ) 1 2 Exercice 4 1
Z A = z 1 z 2 Z_{A} =z_{1} z_{2} Z A = z 1 z 2 2
Z B = z 1 z 2 Z_{B} =\frac{z_{1} }{z_{2} } Z B = z 2 z 1 3
Z C = ( z 1 ) 3 ( z 2 ) 4 Z_{C} =\left(z_{1} \right)^{3} \left(z_{2} \right)^{4} Z C = ( z 1 ) 3 ( z 2 ) 4