Bonjour,
[tex]a)[/tex]
[tex]P(i\sqrt{2}) = (i\sqrt{2})^3 -(2 + i\sqrt{2})(i\sqrt{2})^2 + 2(1 + i\sqrt{2})\times i\sqrt{2} - 2i\sqrt{2}[/tex]
[tex]= -2\sqrt{2}i - (2 + i \sqrt{2}) \times (-2) + 2(i\sqrt{2} -2) - 2i\sqrt{2}[/tex]
[tex]= -2\sqrt{2}i + 4 + 2i \sqrt{2} + 2i\sqrt{2} -4 - 2i\sqrt{2}[/tex]
[tex]= -2\sqrt{2}i + 2\sqrt{2}i + 2\sqrt{2}i - 2\sqrt{2}i = 0[/tex]
[tex]\textnormal{On peut donc en d\'eduire que $z_0$ est une racine de P}[/tex]
[tex]b)[/tex]
[tex]P(z) = (z - i\sqrt{2})(z^2 + az + b)[/tex]
[tex]= z^3 + az^2 + bz - i\sqrt{2}z^2 - i\sqrt{2}az - i\sqrt{2}b[/tex]
[tex]= z^3 + (a -i\sqrt{2})z^2 + (b -i\sqrt{2}a) z - i\sqrt{2}b[/tex]
[tex]a - i\sqrt{2} = -2 -i\sqrt{2} \iff a = -2 -i\sqrt{2} + i\sqrt{2} \iff a = -2[/tex]
[tex]b - i\sqrt{2 }\times (-2) = 2 + 2\sqrt{2}i \iff b + 2\sqrt{2}i = 2 + 2\sqrt{2}i \iff b = 2 + 2\sqrt{2}i - 2\sqrt{2}i \iff b = 2[/tex]
[tex]\textnormal{Les r\'eels a et b sont -2 et 2}[/tex]
[tex]c)[/tex]
[tex]P(z) = 0 \iff (z - i\sqrt{2})(z^2 -2z + 2) = 0[/tex]
[tex]z - i\sqrt{2} = 0 \iff z = i\sqrt{2}[/tex]
[tex]z^2 -2z + 2 =0[/tex]
[tex]\Delta = (-2)^2 - 4 \times 1 \times 2 = 4 - 8 = -4 < 0[/tex]
[tex]\textnormal{Le polyn\^ome a deux racines dans $\mathbb{C}$ :}[/tex]
[tex]z_1 = \frac{-b-i\sqrt{-\Delta} }{2a} = \frac{2 - i\sqrt{4} }{2} = 1- i[/tex]
[tex]z_2 = \overline{z_1} = 1 + i[/tex]
[tex]S = \{i\sqrt{2} \ ;1-i \ ; \ 1+i\}[/tex]
