Let z∈C , if A=z:arg(z)=π4 and B=z:arg(z-3-3i)=2π3 then n(A∩B) is equal to
1
2
3
0
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<0
We can observe that
3+3i∈A but ∉B
∴nA∩B=0