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
`3i, 1-i`

Answer

To solve the problem, we need to find the values of \( r+s \), \( r-s \), \( rs \), and \( \frac{r}{s} \) for the complex numbers \( r = 3i \) and \( s = 1 - i \).

### Step 1: Find \( r + s \)
Given:
- \( r = 3i \)
- \( s = 1 - i \)

We calculate:
\[ r + s = 3i + (1 - i) = 1 + 3i - i = 1 + 2i \]

### Step 2: Find \( r - s \)
We calculate:
\[ r - s = 3i - (1 - i) = 3i - 1 + i = -1 + 4i \]

### Step 3: Find \( rs \)
We calculate:
\[ rs = (3i)(1 - i) = 3i - 3i^2 \]
Since \( i^2 = -1 \), we have:
\[ rs = 3i - 3(-1) = 3i + 3 = 3 + 3i \]

### Step 4: Find \( \frac{r}{s} \)
We calculate:
\[ \frac{r}{s} = \frac{3i}{1 - i} \]
To simplify this, we multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{3i}{1 - i} \cdot \frac{1 + i}{1 + i} = \frac{3i(1 + i)}{(1 - i)(1 + i)} = \frac{3i + 3i^2}{1^2 - i^2} \]
Again, since \( i^2 = -1 \):
\[ = \frac{3i - 3}{1 - (-1)} = \frac{3i - 3}{2} = \frac{-3}{2} + \frac{3}{2}i \]

### Final Results
- \( r + s = 1 + 2i \)
- \( r - s = -1 + 4i \)
- \( rs = 3 + 3i \)
- \( \frac{r}{s} = -\frac{3}{2} + \frac{3}{2}i \)