Answer
(a) (i) $(2\sqrt 2, \frac{7\pi}{4})$; (ii) $ (-2\sqrt 2, \frac{3\pi}{4})$
(b) (i) $(2, \frac{2\pi}{3})$; (ii)$ (-2,\frac{5\pi}{3})$
Work Step by Step
(a)
$r =\sqrt{(x)^2+(y)^2} = \sqrt {(2)^2 + (-2)^2} = \sqrt {4 + 4} = \sqrt 8 = 2\sqrt 2$
$tan(\theta) = \frac{y}{x} = \frac{-2}{2} = -1$
$\theta = tan^{-1}(-1)$
$\theta = \frac{3\pi}{4}$ or $\theta = \frac{7\pi}{4}$
But, the given point (2,-2) lies in the fourth quadrant, therefore, the correct angle is $\theta = \frac{7\pi}{4}$.
(i) $(2\sqrt 2, \frac{7\pi}{4})$
To get the same coordinate, but with r < 0, we should multiply the "r" by $-1$, and add* or subtract* "$\pi$" from the angle:
* The angle must be between: 0 $\leq\theta \lt 2\pi$
(ii) $(-2\sqrt 2, \frac{7\pi}{4} - \pi) = (-2\sqrt 2, \frac{3\pi}{4})$
(b)
$r = \sqrt{(x)^2+(y)^2} = \sqrt{(-1)^2 + (\sqrt 3)^2} = \sqrt{1 + 3} = 2$
$tan(\theta) = \frac{y}{x} = \frac{\sqrt 3}{-1} = -\sqrt 3$
$\theta = tan^{-1}(-\sqrt 3)$
$\theta = \frac{2\pi}{3}$ or $\theta = \frac{5\pi}{3}$
But, the given point $(-1, \sqrt 3)$ lies in the second quadrant, therefore, the correct angle is $\theta = \frac{2\pi}{3}$.
(i) $(2, 2\pi/3)$
To get the same coordinate, but with r < 0, we should multiply the "r" by $-1$, and add* or subtract* "$\pi$" from the angle:
* The angle must be between: 0 $\leq\theta \lt 2\pi$
(ii)$ (-2, 2\pi/3 + \pi) = (-2,5\pi/3)$
