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
`-i, 5+2i`

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 \( r = -i \) and \( s = 5 + 2i \).

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

\[ r + s = -i + (5 + 2i) \]
Combine the real and imaginary parts:
\[ = 5 + (-i + 2i) \]
\[ = 5 + i \]

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

\[ r - s = -i - (5 + 2i) \]
Distributing the negative sign:
\[ = -i - 5 - 2i \]
Combine the real and imaginary parts:
\[ = -5 + (-i - 2i) \]
\[ = -5 - 3i \]

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

\[ rs = (-i)(5 + 2i) \]
Distributing \( -i \):
\[ = -5i - 2i^2 \]
Since \( i^2 = -1 \):
\[ = -5i + 2 \]
Rearranging gives:
\[ = 2 - 5i \]

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

\[ \frac{r}{s} = \frac{-i}{5 + 2i} \]
To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator:
\[ = \frac{-i(5 - 2i)}{(5 + 2i)(5 - 2i)} \]
Calculating the denominator:
\[ = 5^2 - (2i)^2 = 25 - 4(-1) = 25 + 4 = 29 \]
Calculating the numerator:
\[ = -5i + 2i^2 = -5i - 2 \]
So we have:
\[ = \frac{-2 - 5i}{29} \]
Which can be separated into real and imaginary parts:
\[ = -\frac{2}{29} - \frac{5}{29}i \]

### Final Results:

1. \( r + s = 5 + i \)
2. \( r - s = -5 - 3i \)
3. \( rs = 2 - 5i \)
4. \( \frac{r}{s} = -\frac{2}{29} - \frac{5}{29}i \)