Answer
$a.\quad (-2,0)$
$b.\quad (-1,\pi)$
$c.\quad (-3,\pi/2)$
$d.\quad (-1, 7\pi/6)$
Work Step by Step
$(x,y)=(r\cos\theta,r\sin\theta)$
$r^{2}=x^{2}+y^{2},\displaystyle \qquad\tan\theta=\frac{y}{x}$
$ a.\quad$
$\left[\begin{array}{lll}
r^{2}=4+0 & & \tan\theta=\frac{0}{-2}\\
r=-2 & & \theta=0+k\pi
\end{array}\right], \quad$
$(-2,0)$ is on the -x axis, and, since we chose a negative $r$, we take the angle for the opposite axis, the +x axis.
We take $\theta=0$.
Polar coordinates:$\quad (-2,0)$
$ b.\quad$
$\left[\begin{array}{lll}
r^{2}=1+0 & & \tan\theta=\frac{0}{1}\\
r=-1 & & \theta=0+k\pi
\end{array}\right], \quad$
$(1,0)$ is on the +x axis, and, since we chose a negative $r$, we take the angle for the opposite axis, the -x axis.
We take $\theta=\pi$.
Polar coordinates:$\quad (-1,\pi)$
$ c.\quad$
$\left[\begin{array}{lll}
r^{2}=0+9 & & \tan\theta=undef.\\
r=-3 & & \theta=\pi/2+k\pi
\end{array}\right], $
$(0,-3)$ is on the -y axis, and, since we chose a negative $r$, we take the angle for the opposite axis, the +y axis.
We take $\theta=\pi/2$.
Polar coordinates:$\quad (-3,\pi/2)$
$ d.\quad$
$\left[\begin{array}{lll}
r^{2}=\frac{3}{4}+\frac{1}{4} & & \tan\theta=\frac{1}{\sqrt{3}}\\
r=-1 & & \theta=\pi/6+k\pi
\end{array}\right]$
$(\sqrt{3}/2, 1/2)$ is in quadrant I, and, since we chose a negative $r$, we take the angle in the opposite quadrant, quadrant III.
We take $\theta=7\pi/6$
Polar coordinates:$\quad (-1, 7\pi/6)$
