Find the multiplicative inverse of each of the following :
`(i)" "(1-sqrt(3)i)" "(ii)" "(2+5i)" "(iii)" "((2+3i))/((1+i))" "(iv)" "((1+i)(1+2i))/((1+3i))`

Explanation

To find the multiplicative inverse of the given complex numbers, we will follow a systematic approach for each case. The multiplicative inverse of a complex number \( z \) is given by \( \frac{1}{z} \). To simplify this, we will rationalize the denominator.

### (i) Find the multiplicative inverse of \( 1 - \sqrt{3}i \)

1. **Write the inverse**:
\[ \frac{1}{1 - \sqrt{3}i} \]

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

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

4. **Final result**:
\[ \frac{1 + \sqrt{3}i}{4} = \frac{1}{4} + \frac{\sqrt{3}}{4}i \]

### (ii) Find the multiplicative inverse of \( 2 + 5i \)

1. **Write the inverse**:
\[ \frac{1}{2 + 5i} \]

2. **Rationalize the denominator**:
\[ \frac{1 \cdot (2 - 5i)}{(2 + 5i)(2 - 5i)} = \frac{2 - 5i}{2^2 - (5i)^2} \]

3. **Calculate the denominator**:
\[ 2^2 - (5i)^2 = 4 - (-25) = 4 + 25 = 29 \]

4. **Final result**:
\[ \frac{2 - 5i}{29} = \frac{2}{29} - \frac{5}{29}i \]

### (iii) Find the multiplicative inverse of \( \frac{2 + 3i}{1 + i} \)

1. **Write the inverse**:
\[ \frac{1 + i}{2 + 3i} \]

2. **Rationalize the denominator**:
\[ \frac{(1 + i)(2 - 3i)}{(2 + 3i)(2 - 3i)} = \frac{(1 + i)(2 - 3i)}{2^2 - (3i)^2} \]

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

4. **Expand the numerator**:
\[ (1 + i)(2 - 3i) = 2 - 3i + 2i - 3i^2 = 2 - i + 3 = 5 - i \]

5. **Final result**:
\[ \frac{5 - i}{13} = \frac{5}{13} - \frac{1}{13}i \]

### (iv) Find the multiplicative inverse of \( \frac{(1 + i)(1 + 2i)}{1 + 3i} \)

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

2. **Write the inverse**:
\[ \frac{1 + 3i}{-1 + 3i} \]

3. **Rationalize the denominator**:
\[ \frac{(1 + 3i)(-1 - 3i)}{(-1 + 3i)(-1 - 3i)} = \frac{(-1 - 3i + 3i - 9i^2)}{1 - 9} = \frac{-1 + 9}{1 - 9} = \frac{8 - 0i}{-8} = -1 \]

4. **Final result**:
\[ -1 \]

### Summary of Results
1. \( \frac{1}{4} + \frac{\sqrt{3}}{4}i \)
2. \( \frac{2}{29} - \frac{5}{29}i \)
3. \( \frac{5}{13} - \frac{1}{13}i \)
4. \( -1 \)