Convert the following in the form of `(a+ib)` :
(i) `(1+i)^(4)`
(ii) `(-3+(1)/(2)i)^(3)`
(iii) `(1-i)(3+4i)`
(iv) `(1+i)(1+ 2i)(1+ 3i)`
(v) `(3+5i)/(6-i)`
(vi) `((2+3i)^(2))/(2+i)`
(vii) ` ((1+ i)(2+i))/((3+i))`
(viii) ` (2-i)^(-3)`

Explanation

Let's solve each part step by step to convert the given expressions into the form \( a + ib \).

### (i) \( (1+i)^4 \)

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

2. **Now calculate \( (2i)^2 \)**:
\[ (2i)^2 = 4i^2 = 4(-1) = -4 \]

3. **Final result**:
\[ (1+i)^4 = -4 + 0i \]

### (ii) \( (-3 + \frac{1}{2}i)^3 \)

1. **Using the formula \( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \)**:
- Here \( a = -3 \) and \( b = \frac{1}{2}i \).

2. **Calculate \( (-3)^3 \)**:
\[ (-3)^3 = -27 \]

3. **Calculate \( 3(-3)^2(\frac{1}{2}i) \)**:
\[ 3 \cdot 9 \cdot \frac{1}{2}i = \frac{27}{2}i \]

4. **Calculate \( 3(-3)(\frac{1}{2}i)^2 \)**:
\[ 3(-3)(-\frac{1}{4}) = \frac{9}{4} \]

5. **Calculate \( (\frac{1}{2}i)^3 \)**:
\[ (\frac{1}{2}i)^3 = \frac{1}{8}i^3 = -\frac{1}{8}i \]

6. **Combine all parts**:
\[ -27 + \frac{27}{2}i + \frac{9}{4} - \frac{1}{8}i \]

- Convert to a common denominator (8):
\[ -27 = -\frac{216}{8}, \quad \frac{27}{2}i = \frac{108}{8}i, \quad \frac{9}{4} = \frac{18}{8} \]
\[ -\frac{216}{8} + \frac{18}{8} + \left(\frac{108}{8} - \frac{1}{8}\right)i = -\frac{198}{8} + \frac{107}{8}i \]

7. **Final result**:
\[ -\frac{198}{8} + \frac{107}{8}i \]

### (iii) \( (1-i)(3+4i) \)

1. **Use the distributive property**:
\[ 1 \cdot 3 + 1 \cdot 4i - i \cdot 3 - i \cdot 4i = 3 + 4i - 3i - 4i^2 \]

2. **Substituting \( i^2 = -1 \)**:
\[ 3 + (4i - 3i) + 4 = 3 + i + 4 = 7 + i \]

3. **Final result**:
\[ 7 + i \]

### (iv) \( (1+i)(1+2i)(1+3i) \)

1. **First calculate \( (1+i)(1+2i) \)**:
\[ 1 \cdot 1 + 1 \cdot 2i + i \cdot 1 + i \cdot 2i = 1 + 2i + i - 2 = -1 + 3i \]

2. **Now multiply by \( (1+3i) \)**:
\[ (-1 + 3i)(1 + 3i) = -1 \cdot 1 + (-1) \cdot 3i + 3i \cdot 1 + 3i \cdot 3i \]
\[ = -1 - 3i + 3i + 9i^2 = -1 + 9(-1) = -1 - 9 = -10 \]

3. **Final result**:
\[ -10 + 0i \]

### (v) \( \frac{3+5i}{6-i} \)

1. **Multiply numerator and denominator by the conjugate of the denominator**:
\[ \frac{(3+5i)(6+i)}{(6-i)(6+i)} = \frac{18 + 3i + 30i + 5i^2}{36 + 1} \]

2. **Substituting \( i^2 = -1 \)**:
\[ = \frac{18 + 33i - 5}{37} = \frac{13 + 33i}{37} \]

3. **Final result**:
\[ \frac{13}{37} + \frac{33}{37}i \]

### (vi) \( \frac{(2+3i)^2}{2+i} \)

1. **Calculate \( (2+3i)^2 \)**:
\[ = 4 + 12i + 9i^2 = 4 + 12i - 9 = -5 + 12i \]

2. **Now divide by \( (2+i) \)**:
\[ \frac{-5 + 12i}{2+i} \cdot \frac{2-i}{2-i} = \frac{(-5)(2) + (-5)(-i) + (12i)(2) + (12i)(-i)}{(2)^2 + (-1)^2} \]
\[ = \frac{-10 + 5i + 24i - 12}{4 + 1} = \frac{-22 + 29i}{5} \]

3. **Final result**:
\[ -\frac{22}{5} + \frac{29}{5}i \]

### (vii) \( \frac{(1+i)(2+i)}{(3+i)} \)

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

2. **Now divide by \( (3+i) \)**:
\[ \frac{1 + 3i}{3+i} \cdot \frac{3-i}{3-i} = \frac{(1)(3) + (1)(-i) + (3i)(3) + (3i)(-i)}{(3)^2 + (-1)^2} \]
\[ = \frac{3 - i + 9i - 3}{9 + 1} = \frac{8i}{10} = \frac{4i}{5} \]

3. **Final result**:
\[ 0 + \frac{4}{5}i \]

### (viii) \( (2-i)^{-3} \)

1. **Calculate \( (2-i)^3 \)**:
\[ = (2-i)(2-i)(2-i) = (2-i)(4 - 4i + i^2) = (2-i)(3 - 4i) \]
\[ = 6 - 8i - 3i + 4 = 10 - 11i \]

2. **Now take the reciprocal**:
\[ \frac{1}{10 - 11i} \cdot \frac{10 + 11i}{10 + 11i} = \frac{10 + 11i}{100 + 121} = \frac{10 + 11i}{221} \]

3. **Final result**:
\[ \frac{10}{221} + \frac{11}{221}i \]