Réponse : Bonjour,
En posant [tex]z=x+iy[/tex], on a:
[tex]\displaystyle f(z)=\frac{x+iy-1}{x+iy-i}=\frac{x+iy-1}{x+(y-1)i}=\frac{(x+iy-1)(x-i(y-1))}{(x+(y-1)i)(x-i(y-1))}\\=\frac{x^{2}-ixy+xi+xyi+y^{2}-y-x+iy-i}{x^{2}-i^{2}(y-1)^{2}}=\frac{x^{2}+y^{2}-x-y+i(x+y-1)}{x^{2}+(y-1)^{2}}[/tex]