The Cartesian coordinates of a point are given. a) (2,-2)

Consider the Cartesian coordinates be (x,y).
The polar coordinates ${\displaystyle (r,\theta )}$ is defined as,
${\displaystyle r^2=x^2+y^2}$
${\displaystyle x=r\cos \theta }$
${\displaystyle y=r\sin \theta }$
This implies that, ${\displaystyle \tan \theta =\frac{y}{x}}$
a) The Cartesian coordinates be ${\displaystyle (2,-2)}$.
Calculate r and ${\displaystyle \theta }$ as follows:
${\displaystyle r^2=2^2+{(-2)}^2}$
${\displaystyle =8}$
${\displaystyle r=\pm 2\sqrt{2}}$
${\displaystyle \tan \theta =\frac{y}{x}}$
${\displaystyle =\frac{-2}{2}}$
${\displaystyle =-1}$
Thus, the angle ${\displaystyle \theta }$ is ${\displaystyle \frac{3\pi }{4},\frac{7\pi }{4}}$ as the angle ${\displaystyle 0\le \theta 2\pi }$
The polar coordinates of (2,-2) are,
${\displaystyle (2\sqrt{2},\frac{7\pi }{4})}$, where r>0
${\displaystyle (-2\sqrt{2},\frac{3\pi }{4})}$, where r<0
b) The Cartesian coordinates be ${\displaystyle (-1,\sqrt{3})}$.
Calculate r and ${\displaystyle \theta }$ as follows:
${\displaystyle r^2={(-1)}^2+{\left(\sqrt{3}\right)}^2}$
${\displaystyle =4}$
${\displaystyle r=\pm \sqrt{2}}$
${\displaystyle \tan \theta =\frac{y}{x}}$
${\displaystyle =\frac{\sqrt{3}}{-1}}$
${\displaystyle =-\sqrt{3}}$
Thus, the angle ${\displaystyle \theta }$ is ${\displaystyle \frac{2\pi }{3},\frac{5\pi }{3}}$ as the angle ${\displaystyle 0\le \theta \le 2\pi }$
The polar coordinates of ${\displaystyle (-1,\sqrt{3})}$ are,
${\displaystyle (2,\frac{2\pi }{3})}$, where r>0
${\displaystyle (-2,\frac{5\pi }{3})}$, where r<0