In each of the following find `r+s, r- s, rs, (r )/(s)` if r denotes the first complex number and s denotes the second complex number
`7+ 3i, 3i-7`

Answer

To solve the problem, we need to find \( r+s \), \( r-s \), \( rs \), and \( \frac{r}{s} \) for the complex numbers \( r = 7 + 3i \) and \( s = 3i - 7 \).

### Step 1: Find \( r + s \)

\[ r + s = (7 + 3i) + (3i - 7) \]
Combine like terms:
\[ = 7 - 7 + 3i + 3i = 0 + 6i = 6i \]

### Step 2: Find \( r - s \)

\[ r - s = (7 + 3i) - (3i - 7) \]
Distribute the negative sign:
\[ = 7 + 3i - 3i + 7 = 7 + 7 = 14 \]

### Step 3: Find \( rs \)

\[ rs = (7 + 3i)(3i - 7) \]
Use the distributive property (FOIL):
\[ = 7 \cdot 3i + 7 \cdot (-7) + 3i \cdot 3i + 3i \cdot (-7) \]
Calculating each term:
\[ = 21i - 49 + 9i^2 - 21i \]
Since \( i^2 = -1 \):
\[ = 21i - 49 - 9 - 21i = -49 - 9 = -58 \]

### Step 4: Find \( \frac{r}{s} \)

\[ \frac{r}{s} = \frac{7 + 3i}{3i - 7} \]
To simplify, multiply the numerator and denominator by the conjugate of the denominator:
\[ = \frac{(7 + 3i)(3i + 7)}{(3i - 7)(3i + 7)} \]
Calculating the denominator:
\[ (3i - 7)(3i + 7) = (3i)^2 - 7^2 = -9 - 49 = -58 \]
Calculating the numerator:
\[ (7 + 3i)(3i + 7) = 7 \cdot 3i + 7 \cdot 7 + 3i \cdot 3i + 3i \cdot 7 \]
Calculating each term:
\[ = 21i + 49 + 9i^2 + 21i = 21i + 49 - 9 + 21i = 49 - 9 + 42i = 40 + 42i \]
Thus:
\[ \frac{r}{s} = \frac{40 + 42i}{-58} \]
This can be simplified:
\[ = -\frac{40}{58} - \frac{42}{58}i = -\frac{20}{29} - \frac{21}{29}i \]

### Final Results

1. \( r + s = 6i \)
2. \( r - s = 14 \)
3. \( rs = -58 \)
4. \( \frac{r}{s} = -\frac{20}{29} - \frac{21}{29}i \)