Bonjour Eazzy78
[tex]\dfrac{z+2}{z-1}=zi\\\\Cond.:z\neq1\\\\z+2=zi(z-1)\\z+2=z^2i-zi\\z^2i-zi-z-2=0\\iz^2-(i+1)z-2=0\\-i[iz^2-(i+1)z-2]=0\\-i^2z^2+i(i+1)z+2i=0\\\\z^2+(i-1)z+2i=0\\\Delta=(i-1)^2-4\times1\times2i\\\Delta=-1-2i+1-8\\\Delta=-10i=10e^{i(\frac{-\pi}{2})}[/tex]
Les racines carrées de Δ sont [tex]\pm\sqrt{10}e^{i(\frac{-\pi}{4})}[/tex], soit
[tex]\pm\sqrt{10}\times[\cos(\dfrac{-\pi}{4})+i\sin(\dfrac{-\pi}{4})]=\pm\sqrt{10}\times(\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})\\\\=\pm(\dfrac{\sqrt{20}}{2}-i\dfrac{\sqrt{20}}{2})=\pm(\dfrac{2\sqrt{5}}{2}-i\dfrac{2\sqrt{5}}{2})=\pm(\sqrt{5}-i\sqrt{5})[/tex]
D'où
[tex]z_1=\dfrac{-(i-1)+(\sqrt{5}-i\sqrt{5})}{2}=\dfrac{-i+1+\sqrt{5}-i\sqrt{5}}{2}\\\\z_1=\dfrac{1+\sqrt{5}-i(1+\sqrt{5})}{2}\\\\\boxed{z_1=\dfrac{1+\sqrt{5}}{2}-i(\dfrac{1+\sqrt{5}}{2})}[/tex]
[tex]z_2=\dfrac{-(i-1)-(\sqrt{5}-i\sqrt{5})}{2}=\dfrac{-i+1-\sqrt{5}+i\sqrt{5}}{2}\\\\z_2=\dfrac{1-\sqrt{5}-i(1-\sqrt{5})}{2}\\\\\boxed{z_2=\dfrac{1-\sqrt{5}}{2}-i(\dfrac{1-\sqrt{5}}{2})}[/tex]
