Answer
$(a)$
$(2,\ \pi/2+2k\pi),\ k\in \mathbb{Z}$ or $(-2,\ \pi/2+(2k+1)\pi),\ k\in \mathbb{Z}$
$(b)$
$(2,\ 0+2k\pi),\ k\in \mathbb{Z}$ or $(-2,\ 0+(2k+1)\pi),\ k\in \mathbb{Z}$
$(c)$
$(-2,\ \pi/2+2k\pi),\ k\in \mathbb{Z}$ or $(+2,\ \pi/2+(2k+1)\pi),\ k\in \mathbb{Z}$
$(d)$
$(-2,\ 0+2k\pi),\ k\in \mathbb{Z}$ or $(+2,\ 0+(2k+1)\pi),\ k\in \mathbb{Z}$
.
Work Step by Step
Plotting $(r,\theta):$
- if $r$ is positive, then the point lies on the terminal side of $\theta+2k\pi,\ k\in \mathbb{Z}$
- if $r$ is negative, then the point lies opposite the terminal side of $\theta$; it lies on the terminal side of $\theta\pm\pi+2k\pi=\theta+ (2k+1)\pi,\ k\in \mathbb{Z}$
$(a)$
$(2,\ \pi/2+2k\pi),\ k\in \mathbb{Z}$ or $(-2,\ \pi/2+(2k+1)\pi),\ k\in \mathbb{Z}$
$(b)$
$(2,\ 0+2k\pi),\ k\in \mathbb{Z}$ or $(-2,\ 0+(2k+1)\pi),\ k\in \mathbb{Z}$
$(c)$
$(-2,\ \pi/2+2k\pi),\ k\in \mathbb{Z}$ or $(+2,\ \pi/2+(2k+1)\pi),\ k\in \mathbb{Z}$
$(d)$
$(-2,\ 0+2k\pi),\ k\in \mathbb{Z}$ or $(+2,\ 0+(2k+1)\pi),\ k\in \mathbb{Z}$
