We depict these five vectors in Figure 5.

(a) Here v is in the first quadrant, and since \tan \theta=2 / 2=1, \theta=\pi / 4.
(b) Here \theta=\tan ^{-1} 2 \sqrt{3} / 2=\tan ^{-1} \sqrt{3}=\pi / 3.
(c) We see that \mathbf{v} is in the second quadrant, and since \tan ^{-3} 2 /(2 \sqrt{3})=\tan ^{-1} 1 / \sqrt{3}=\pi / 6, we see from the figure that

\theta=\pi-(\pi / 6)=5 \pi / 6
(d) Here v is in the third quadrant, and since \tan ^{-1} 1=\pi / 4,  \theta=\pi+(\pi / 4)=5 \pi / 4
(e) Since \mathbf{v} is in the fourth quadrant, and since \tan ^{-1}(-1)=-\pi / 4, \theta=2 \pi-(\pi / 4)=7 \pi / 4