Réponse : 3)[tex]z'=\frac{x+iy-1+2i}{x+iy-i} =\frac{x-1+(y+2)i}{x+(y-1)i} =\frac{((x-1)+(y+2)i)(x-(y-1)i)}{(x+(y-1)i)(x-(y-1)i)} =\frac{x(x-1)-(x-1)(y-1)i+x(y+2)i-(y+2)(y-1)i^{2}}{x^{2}-((y-1)i)^{2}} \\=\frac{x(x-1)+(y+2)(y-1)+(-(x-1)(y-1)+x(y+2))i}{x^{2}+(y-1)^{2}} =\frac{x^{2}-x+y^{2}-y+2y-2}{x^{2}+(y-1)^{2}} +\frac{-(xy-x-y+1)+xy+2x}{x^{2}+(y-1)^{2}} i\\[/tex]
(suite)[tex]=\frac{x^{2}+y^{2}-x+y-2}{x^{2}+(y-1)^{2}} +\frac{-xy+x+y-1+xy+2x}{x^{2}+(y-1)^{2}} i=\frac{x^{2}+y^{2}-x+y-2}{x^{2}+(y-1)^{2}} +\frac{3x+y-1}{x^{2}+(y-1)^{2}} i[/tex]
