Réponse :
Bonsoir,
Explications étape par étape
Soit z=a+ib
z²=a²-b²+2iab
[tex]z^2-3\overline{z}+2=0\\\\\left\{\begin{array}{ccc}a^2-b^2-3a+2&=&0\\b(2a+3)&=&0\\\end{array} \right. \\\\1)\\\left\{\begin{array}{ccc}b&=&0\\a^2-3a+2&=&0\\\end{array} \right. \\\\\\\left\{\begin{array}{ccc}b&=&0\\(a-1)(a-2)&=&0\\\end{array} \right. \\\\z_1=1,z_2=2\\\\2)\\\left\{\begin{array}{ccc}a&=&-\dfrac{2}{3} \\\dfrac{35}{4} -b^2&=&0\\\end{array} \right. \\\\\left\{\begin{array}{ccc}a&=&-\dfrac{2}{3} \\(\dfrac{\sqrt{35}}{2} +b)(\dfrac{\sqrt{35}}{2} -b)&=&0\\\end{array} \right. \\\\[/tex]
[tex]z_3=\dfrac{-3}{2} -i\dfrac{\sqrt{35} }{2} \\\\z_4=\dfrac{-3}{2} +i\dfrac{\sqrt{35} }{2}[/tex]
