Find the multiplicative inverse of the following complex numbers:
(i) `4-3i`
(ii) `3+2i`

Answer

To find the multiplicative inverse of the complex numbers \(4 - 3i\) and \(3 + 2i\), we will use the formula for the multiplicative inverse of a complex number \(z\), which is given by:

\[ z^{-1} = \frac{1}{z} \]

### Part (i): Finding the multiplicative inverse of \(4 - 3i\)

1. **Write the expression for the multiplicative inverse:**
\[ z^{-1} = \frac{1}{4 - 3i} \]

2. **Rationalize the denominator:**
To eliminate the imaginary part from the denominator, we multiply the numerator and denominator by the conjugate of the denominator, which is \(4 + 3i\):
\[ z^{-1} = \frac{1 \cdot (4 + 3i)}{(4 - 3i)(4 + 3i)} \]

3. **Calculate the denominator:**
Using the formula \( (a - b)(a + b) = a^2 - b^2 \):
\[ (4 - 3i)(4 + 3i) = 4^2 - (3i)^2 = 16 - 9(-1) = 16 + 9 = 25 \]

4. **Calculate the numerator:**
\[ 1 \cdot (4 + 3i) = 4 + 3i \]

5. **Combine the results:**
\[ z^{-1} = \frac{4 + 3i}{25} = \frac{4}{25} + \frac{3}{25}i \]

### Part (ii): Finding the multiplicative inverse of \(3 + 2i\)

1. **Write the expression for the multiplicative inverse:**
\[ z^{-1} = \frac{1}{3 + 2i} \]

2. **Rationalize the denominator:**
Multiply the numerator and denominator by the conjugate of the denominator, which is \(3 - 2i\):
\[ z^{-1} = \frac{1 \cdot (3 - 2i)}{(3 + 2i)(3 - 2i)} \]

3. **Calculate the denominator:**
\[ (3 + 2i)(3 - 2i) = 3^2 - (2i)^2 = 9 - 4(-1) = 9 + 4 = 13 \]

4. **Calculate the numerator:**
\[ 1 \cdot (3 - 2i) = 3 - 2i \]

5. **Combine the results:**
\[ z^{-1} = \frac{3 - 2i}{13} = \frac{3}{13} - \frac{2}{13}i \]

### Final Answers:
- The multiplicative inverse of \(4 - 3i\) is \(\frac{4}{25} + \frac{3}{25}i\).
- The multiplicative inverse of \(3 + 2i\) is \(\frac{3}{13} - \frac{2}{13}i\).