Determine whether z=2ei5π/6z = 2e^{i5\pi/6}z=2ei5π/6 lies on the circle ∣z∣=2|z| = 2∣z∣=2.
Yes, and arg(z)=5π/6\text{arg}(z) = 5\pi/6arg(z)=5π/6
Yes, but arg(z)=−7π/6\text{arg}(z) = -7\pi/6arg(z)=−7π/6
No, because ∣z∣≠2|z| \neq 2∣z∣=2
Cannot determine without more info