Salut,
Fie z=a+bi,
|z|=z+3+i, sau [tex]\sqrt{a^2+b^2}=a+bi+3+i,\;sau\;\sqrt{a^2+b^2}=a+3+(b+1)\cdot i[/tex]
De aici rezultă că: b+1=0, deci b = - 1 (minus 1) => b²=1.
[tex]\sqrt{a^2+1}=a+3,\;sau\;a^2+1=a^2+6a+9\Rightarrow a=-\dfrac{4}{3}.\\\\Deci\;z=-\dfrac{4}{3}-i.[/tex]
Green eyes.
