Answer
$a.\quad(\sqrt{2},\pi/4)$
$b.\quad(3,\pi)$
$c.\quad(2,11\pi/6)$
$d.\displaystyle \quad(5, \arctan(-\frac{4}{3})+\pi)$
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$
$\left[\begin{array}{lll}
r^{2}=1+1 & & \tan\theta=\frac{1}{1}\\
r=\sqrt{2} & & \theta=\pi/4+k\pi
\end{array}\right]$
The point $(1,1)$ is quadrant I, we keep $\theta=\pi/4$
$\qquad (\sqrt{2},\pi/4)$
$ b.\quad$
$\left[\begin{array}{lll}
r^{2}=9+0 & & \tan\theta=0\\
r=3 & & \theta=0+k\pi
\end{array}\right]\qquad$
The point $($-3,0$)$ is on the -x side of the x-axis; we take $\theta=\pi$
$(3,\pi)$
$ c.\quad$
$\left[\begin{array}{lll}
r^{2}=3+1 & & \tan\theta=-\frac{1}{\sqrt{3}}\\
r=2 & & \theta=-\pi/6\frac{1}{\sqrt{3}}+k\pi
\end{array}\right]\qquad$
The point $(\sqrt{3},-1)$ is quadrant I; we take $\theta=11\pi/6$
$\qquad (2,11\pi/6)$
$ d.\quad$
$\left[\begin{array}{lll}
r^{2}=9+16 & & \tan\theta=-\frac{4}{3}\\
r=5 & & \theta=\arctan(-\frac{4}{3})+k\pi
\end{array}\right]$
The point $(-3,4)$ is quadrant II; we take $\displaystyle \theta=\arctan(-\frac{4}{3})+\pi$
$(5, \displaystyle \arctan(-\frac{4}{3})+\pi)$
