The correct option is A 12+i√32
Let z=x+iy
Given, arg(z)=π3
⇒yx=tan(argz)=√3
⇒y=x√3
Now, z−1=(x+iy)−1=(x−1)+iy
Since, arg(z−1)=2π3, therefore z−1 lies in 2nd quadrant in argand plane.
∴x−1<0 and y>0
Now π−α=arg(z−1)
⇒α=π3
tanα=∣∣∣yx−1∣∣∣
⇒√3=∣∣∣yx−1∣∣∣
⇒y1−x=√3
⇒x√3=√3(1−x)
⇒2x=1
⇒x=12 and y=√32
Thus, the complex number is 12+i√32