Răspuns:
z=a+bi
z`=a-bi
z``=a+bi=z
f(z)+2f(z`)=2z+3z`
f(z`)+2f(z``)=2z`+3z``<=>
f(z`)+2f(z)=2z`+3z Inmultesti aceasta ecuatie cu -2 si o adaogi la prima
-2f(z`)-4f(z)+f(z)+2f(z`)=-4z`-6z+2z--3z`
-3f(z)=-4z-5z`
f(z)=4z/3+5z`/3
Injectivitate
f(z1)=f(z2)
4z1/3+5z`1/3=4z2/3+5z`2/3
4(a1+b1i)/3+5(a1-b1i)/3=4(a2+b2i)/3+5(a2-b2i)/3║×3
4a1+4b1i+5a1-5b1i=4a2+4b2i+5a2-5b2i
9a1-b1i=9a2-b2i=>
a1=a2 si b1= b2 adica z1=z2=> f(z) injectiva
surjectiva
fie y∈C a.i f(z)=y y=m+ni
4z/3+5z`/3 =m+ni
4(a+bi)/3+5(a-bi)/3=m+ni║×3
4a+4bi+5a-5bi=m+ni
9a-bi=m+ni
m=9a Admite solutii reale ∀m
-b=n admite solutii reale pt ∀n=>
∀z∈C E y∈C a.i f(z)=C
functia este surjectiva
Explicație pas cu pas:
