z=x+iy
x=2+sqrt(3)
y=1
|z|=sqrt(x^2+y^2)=sqrt((2+sqrt(3))^2+1)=sqrt(4+4sqrt(3)+3+1)=sqrt(8+4sqrt(3))=2sqrt(2+sqrt(3))
cos φ =x/|z|=(2+sqrt(3))/2sqrt(2+sqrt(3))=sqrt(2+sqrt(3))/2
sin φ =y/|z|=1/2sqrt(2+sqrt(3))=(2-sqrt(3))/2
⇒
tg φ =y/x=1/(2+sqrt(3)) =2-sqrt(3) ⇒ φ =[b]π/12[/b]
О т в е т. [b]В)[/b]