Answer
$a.\quad (-3\sqrt{2},5\pi/4)$
$b.\quad (-1,0)$
$c.\quad (-2,5\pi/3)$
$d.\displaystyle \quad (-5,-\arctan(\frac{3}{4})+\pi)$
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$
$\left[\begin{array}{lll}
r^{2}=9+9 & & \tan\theta=\frac{3}{3}\\
r=-3\sqrt{2} & & \theta=\pi/4+k\pi
\end{array}\right], \quad$
$(3,3)$ is in quadrant I, and, since we chose a negative $r$, we take the angle in the opposite quadrant, quadrant III. We take $\theta=5\pi/4$
Polar coordinates:$\quad (-3\sqrt{2},5\pi/4)$
$ 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 $\theta=0$.
Polar coordinates:$\quad (-1,0)$
$ c.\quad$
$\left[\begin{array}{lll}
r^{2}=1+3 & & \tan\theta=\frac{\sqrt{3}}{-1}\\
r=-2 & & \theta=-\pi/3+k\pi
\end{array}\right],$
$(-1,\sqrt{3})$ is in quadrant II, and, since we chose a negative $r$, we take the angle in the opposite quadrant, quadrant IV.
We take $\theta=5\pi/3.$
Polar coordinates:$\quad (-2,5\pi/3)$
$ d.\quad$
$\left[\begin{array}{lll}
r^{2}=16+9 & & \tan\theta=\frac{-3}{4}\\
r=-5 & & \theta=-\arctan(\frac{3}{4})+k\pi
\end{array}\right]$
$(4,-3)$ is in quadrant IV, and, since we chose a negative $r$, we take the angle in the opposite quadrant, quadrant II.
We take $\displaystyle \theta=-\arctan(\frac{3}{4})+\pi$
Polar coordinates:$\displaystyle \quad (-5,-\arctan(\frac{3}{4})+\pi)$
