The polar coordinates \((r,\theta)\) is defined as,

\(r^2=x^2+y^2\)

\(x=r\cos\theta\)

\(y=r\sin\theta\)

This implies that, \(\tan\theta=\frac{y}{x}\)

a) The Cartesian coordinates be \((2,-2)\).

Calculate r and \(\theta\) as follows:

\(r^2=2^2+(-2)^2\)

\(=8\)

\(r=\pm2\sqrt{2}\)

\(\tan\theta=\frac{y}{x}\)

\(=\frac{-2}{2}\)

\(=-1\)

Thus, the angle \(\theta\) is \(\frac{3\pi}{4},\frac{7\pi}{4}\) as the angle \(0\leq\theta2\pi\)

The polar coordinates of (2,-2) are,

\((2\sqrt{2},\frac{7\pi}{4})\), where r>0

\((-2\sqrt{2},\frac{3\pi}{4})\), where r

b) The Cartesian coordinates be \((-1,\sqrt{3})\).

Calculate r and \(\theta\) as follows:

\(r^2=(-1)^2+(\sqrt{3})^2\)

\(=4\)

\(r=\pm\sqrt{2}\)

\(\tan\theta=\frac{y}{x}\)

\(=\frac{\sqrt{3}}{-1}\)

\(=-\sqrt{3}\)

Thus, the angle \(\theta\) is \(\frac{2\pi}{3},\frac{5\pi}{3}\) as the angle \(0\leq\theta\leq2\pi\)

The polar coordinates of \((-1,\sqrt{3})\) are,

\((2,\frac{2\pi}{3})\), where r>0

\((-2,\frac{5\pi}{3})\), where r<0