Chapitre 1 — Nombres complexes : algèbre

1. \(\text{Re}(z_1)=5\), \(\text{Im}(z_1)=-3\)
2. \(\text{Re}(z_2)=0\), \(\text{Im}(z_2)=-2\)
3. \(\text{Re}(z_3)=\sqrt{3}\), \(\text{Im}(z_3)=1\)
4. \((2+i)^2 = 4 + 4i + i^2 = 3 + 4i\), donc \(\text{Re}(z_4)=3\), \(\text{Im}(z_4)=4\)
5. \(\dfrac{1}{i} = \dfrac{-i}{i \cdot (-i)} = \dfrac{-i}{1} = -i\), donc \(\text{Re}(z_5)=0\), \(\text{Im}(z_5)=-1\)

1. \(z+w = 2+2i\)
2. \(z-w = 4-6i\)
3. \(z \times w = (3)(-1) + (3)(4i) + (-2i)(-1) + (-2i)(4i) = -3 + 12i + 2i - 8i^2 = -3 + 14i + 8 = 5 + 14i\)
4. \(z^2 = (3-2i)^2 = 9 - 12i + 4i^2 = 9 - 12i - 4 = 5 - 12i\)
5. \(z\bar{z} = 3^2 + (-2)^2 = 9 + 4 = 13\)

\(z \times w = (1+2i)(3-i) = 3 - i + 6i - 2i^2 = 3 + 5i + 2 = 5 + 5i\)

Donc \(\overline{z \times w} = 5 - 5i\).

D’autre part : \(\bar{z} = 1-2i\) et \(\bar{w} = 3+i\).

\(\bar{z} \times \bar{w} = (1-2i)(3+i) = 3 + i - 6i - 2i^2 = 3 - 5i + 2 = 5 - 5i\). ✓

1. \(\dfrac{1}{3+4i} = \dfrac{3-4i}{9+16} = \dfrac{3-4i}{25} = \dfrac{3}{25} - \dfrac{4}{25}i\)
2. \(\dfrac{2-i}{1+i} = \dfrac{(2-i)(1-i)}{(1+i)(1-i)} = \dfrac{2 - 2i - i + i^2}{2} = \dfrac{1 - 3i}{2} = \dfrac{1}{2} - \dfrac{3}{2}i\)
3. \((1+i)^2 = 2i\), donc \(\dfrac{2i}{2-i} = \dfrac{2i(2+i)}{5} = \dfrac{4i + 2i^2}{5} = \dfrac{-2 + 4i}{5} = -\dfrac{2}{5} + \dfrac{4}{5}i\)
4. \(\dfrac{3+2i}{3-2i} = \dfrac{(3+2i)^2}{(3-2i)(3+2i)} = \dfrac{9+12i-4}{13} = \dfrac{5+12i}{13} = \dfrac{5}{13} + \dfrac{12}{13}i\)

1. \(23 = 4 \times 5 + 3\), donc \(i^{23} = i^3 = -i\)
2. \(100 = 4 \times 25\), donc \(i^{100} = 1\)
3. \(i^{-5} = i^{-5+8} = i^3 = -i\) (ou \(\frac{1}{i^5} = \frac{1}{i} = -i\))
4. \((-i)^{17} = (-1)^{17} \cdot i^{17} = -1 \cdot i^1 = -i\)

1. \((1-i)^2 = 1 - 2i + i^2 = -2i\), donc \((1-i)^4 = (-2i)^2 = 4i^2 = -4\).
2. \((1+i\sqrt{3})^3 = 1 + 3 \cdot i\sqrt{3} + 3(i\sqrt{3})^2 + (i\sqrt{3})^3 = 1 + 3i\sqrt{3} - 9 - 3i\sqrt{3} = -8\). Résultat réel !
3. D’un côté : \((1+i)^2 = 2i\) donc \((1+i)^6 = (2i)^3 = 8i^3 = -8i\).
   De l’autre, par le binôme :

\(\sum_{k=0}^6 \binom{6}{k} i^k = \binom{6}{0} + \binom{6}{1}i + \binom{6}{2}i^2 + \binom{6}{3}i^3 + \binom{6}{4}i^4 + \binom{6}{5}i^5 + \binom{6}{6}i^6\)

\(= 1 + 6i - 15 - 20i + 15 + 6i - 1 = (1-15+15-1) + (6-20+6)i = 0 - 8i\)

La partie imaginaire donne : \(6 - 20 + 6 = -8\), soit \(\binom{6}{1} - \binom{6}{3} + \binom{6}{5} = -8\). ✓

1. \(z = \dfrac{3+i}{1+2i} = \dfrac{(3+i)(1-2i)}{5} = \dfrac{3 - 6i + i - 2i^2}{5} = \dfrac{5 - 5i}{5} = 1 - i\)
2. \(z^2 = -9\) : on cherche \(z = bi\) avec \(b \in \mathbb{R}\). \((bi)^2 = -b^2 = -9\) donc \(b = \pm 3\). Solutions : \(z = 3i\) ou \(z = -3i\).
3. \(z + \bar{z} = 2a = 4\) donc \(a = 2\). \(z - \bar{z} = 2bi = 6i\) donc \(b = 3\). Donc \(z = 2 + 3i\).
4. Avec \(z = a+bi\) : \(z^2 = a^2 - b^2 + 2abi\). Condition : \(\text{Re}(z^2) = \text{Im}(z)\) donne \(a^2 - b^2 = b\). Pas de contrainte sur \(a\) : famille de solutions \(\{a + bi \mid a^2 - b^2 - b = 0\}\).