Let z∈C , if A=z:arg(z)=π4 and B=z:arg(z-3-3i)=2π3 then n(A∩B) is equal to

Argz=π4⇒Tan−1yx=π4⇒y=x     (x>0,y>0)arg  (z−3−3i)=2π3 Tan−1y−3x−3=2π/3y−3=−3x−3y−3>0,x−3<0We can observe that 3+3i∈A but ∉B∴nA∩B=0