notez z=a+bi
(1+i)(a+bi)=3-i
a+bi+ai+b*i²=3-i
a+bi+ai-b-3+i=0
(a-b-3)+i(a+b+1)=0 <-- am separat partea reala de cea imaginara
egalez fiecare paranteza cu 0 si rezolv sistemul care se formeaza
[tex] \left \{ {{a-b-3=0} \atop {a+b+1=0}} \right. \\ \left \{ {{a-b=3} \atop {a+b=-1}} \right. \\ 2a=2 =\ \textgreater \ a=1 [/tex]
1-b=3 => -b=2 => b=-2
solutie:
z=1-2i
