Convert the following Cartesian coordinates to the cooresponding polar coordinates using positive r.
(i) `(1,-1)`
(ii) `(-3,-4)`

Explanation

To convert Cartesian coordinates to polar coordinates, we use the formulas:

1. \( r = \sqrt{x^2 + y^2} \)
2. \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \)

We also need to consider the quadrant in which the point lies to determine the correct angle \( \theta \).

### Step-by-Step Solution

#### (i) For the point `(1, -1)`

1. **Calculate \( r \)**:
\[ r = \sqrt{x^2 + y^2} = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

2. **Calculate \( \theta \)**:
\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{-1}{1}\right) = \tan^{-1}(-1) \]
The angle \( \tan^{-1}(-1) \) corresponds to \( -\frac{\pi}{4} \). Since we want the angle in the range \( [0, 2\pi) \), we can convert this to:
\[ \theta = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4} \]

3. **Final Polar Coordinates**:
\[ \text{Polar Coordinates} = \left(\sqrt{2}, \frac{7\pi}{4}\right) \]

#### (ii) For the point `(-3, -4)`

1. **Calculate \( r \)**:
\[ r = \sqrt{x^2 + y^2} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

2. **Calculate \( \theta \)**:
\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{-4}{-3}\right) = \tan^{-1}\left(\frac{4}{3}\right) \]
Since the point is in the third quadrant, we need to add \( \pi \) to the angle:
\[ \theta = \pi + \tan^{-1}\left(\frac{4}{3}\right) \]

3. **Final Polar Coordinates**:
\[ \text{Polar Coordinates} = \left(5, \pi + \tan^{-1}\left(\frac{4}{3}\right)\right) \]

### Summary of Polar Coordinates
- For the point `(1, -1)`: \( \left(\sqrt{2}, \frac{7\pi}{4}\right) \)
- For the point `(-3, -4)`: \( \left(5, \pi + \tan^{-1}\left(\frac{4}{3}\right)\right) \)