Answer
$a.\quad (2\sqrt{2},-3\pi/4)$
$b.\quad (3,\pi/2)$
$c.\quad (2,5\pi/6)$
$d.\displaystyle \quad (13,-\arctan(\frac{12}{5}))$
Work Step by Step
Cartesian: $(x,y)=(r\cos\theta,r\sin\theta)$
$r^{2}=x^{2}+y^{2},\displaystyle \qquad\tan\theta=\frac{y}{x}$
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.\quad$
$\left[\begin{array}{lll}
r^{2}=4+4 & & \tan\theta=\frac{-2}{-2}\\
r=2\sqrt{2} & & \theta=\pi/4+k\pi
\end{array}\right], \quad$
$(-2,-2)$ is in quadrant III; we take $\theta=-3\pi/4$
Polar coordinates:$\quad (2\sqrt{2},-3\pi/4)$
$ b.\quad$
$\left[\begin{array}{lll}
r^{2}=0+9 & & \tan\theta=undef.\\
r=3 & & \theta=\pi/2+k\pi
\end{array}\right], \quad$
$(0,3)$ is on the +y axis; we take $\theta=\pi/2$
Polar coordinates:$\quad (3,\pi/2)$
$ c.\quad$
$\left[\begin{array}{lll}
r^{2}=3+1 & & \tan\theta=\frac{1}{-\sqrt{3}}\\
r=2 & & \theta=\pi/6+k\pi
\end{array}\right],$
$(-\sqrt{3},1)$ is in quadrant II; we take $\theta=5\pi/6$
Polar coordinates:$\quad (2,5\pi/6)$
$ d.\quad$
$\left[\begin{array}{lll}
r^{2}=25+144 & & \tan\theta=\frac{-12}{5}\\
r=13 & & \theta=\arctan(\frac{-12}{5})+k\pi
\end{array}\right]$
The point is in quadrant IV; we take $\displaystyle \theta=-\arctan(\frac{12}{5})$
Polar coordinates: $\displaystyle \quad (13,-\arctan(\frac{12}{5}))$
