Find the conjugate of each of the following :
`{:((i),(-5-2i),(ii),(1)/((4+3i)),(iii),((1+i)^(2))/((3-i)),(iv),((1+i)(2+i))/((3+i))),((v),sqrt(-3),(vi),sqrt(2),(vii),-sqrt(-1),(viii),(2-5i)^(2)):}`

Answer

To find the conjugate of each of the given complex numbers, we will follow the definition of the conjugate of a complex number. The conjugate of a complex number \( z = x + iy \) is given by \( \overline{z} = x - iy \).

Let’s solve each part step by step.

### Step 1: Find the conjugate of \( i \)
- Given: \( z = i \)
- Here, \( x = 0 \) and \( y = 1 \).
- Conjugate: \( \overline{z} = 0 - 1i = -i \)

### Step 2: Find the conjugate of \( -5 - 2i \)
- Given: \( z = -5 - 2i \)
- Here, \( x = -5 \) and \( y = -2 \).
- Conjugate: \( \overline{z} = -5 + 2i \)

### Step 3: Find the conjugate of \( \frac{1}{4 + 3i} \)
- Given: \( z = \frac{1}{4 + 3i} \)
- To eliminate \( i \) from the denominator, multiply by the conjugate of the denominator:
\[ \overline{4 + 3i} = 4 - 3i \]
- Thus,
\[ z = \frac{1 \cdot (4 - 3i)}{(4 + 3i)(4 - 3i)} = \frac{4 - 3i}{16 + 9} = \frac{4 - 3i}{25} \]
- Conjugate: \( \overline{z} = \frac{4}{25} + \frac{3}{25}i \)

### Step 4: Find the conjugate of \( \frac{(1+i)^2}{3-i} \)
- First, calculate \( (1+i)^2 \):
\[ (1+i)^2 = 1^2 + 2(1)(i) + i^2 = 1 + 2i - 1 = 2i \]
- Now, \( z = \frac{2i}{3-i} \).
- Multiply by the conjugate of the denominator:
\[ \overline{3-i} = 3 + i \]
- Thus,
\[ z = \frac{2i(3+i)}{(3-i)(3+i)} = \frac{6i + 2i^2}{9 + 1} = \frac{6i - 2}{10} = \frac{-2}{10} + \frac{6}{10}i = -\frac{1}{5} + \frac{3}{5}i \]
- Conjugate: \( \overline{z} = -\frac{1}{5} - \frac{3}{5}i \)

### Step 5: Find the conjugate of \( \frac{(1+i)(2+i)}{3+i} \)
- First, calculate \( (1+i)(2+i) \):
\[ (1+i)(2+i) = 2 + i + 2i + i^2 = 2 + 3i - 1 = 1 + 3i \]
- Now, \( z = \frac{1 + 3i}{3+i} \).
- Multiply by the conjugate of the denominator:
\[ \overline{3+i} = 3 - i \]
- Thus,
\[ z = \frac{(1 + 3i)(3 - i)}{(3+i)(3-i)} = \frac{3 - i + 9i - 3i^2}{9 + 1} = \frac{3 - i + 9i + 3}{10} = \frac{6 + 8i}{10} = \frac{3}{5} + \frac{4}{5}i \]
- Conjugate: \( \overline{z} = \frac{3}{5} - \frac{4}{5}i \)

### Step 6: Find the conjugate of \( \sqrt{-3} \)
- Given: \( z = \sqrt{-3} = i\sqrt{3} \)
- Conjugate: \( \overline{z} = -i\sqrt{3} \)

### Step 7: Find the conjugate of \( \sqrt{2} \)
- Given: \( z = \sqrt{2} \)
- Conjugate: \( \overline{z} = \sqrt{2} \)

### Step 8: Find the conjugate of \( -\sqrt{-1} \)
- Given: \( z = -\sqrt{-1} = -i \)
- Conjugate: \( \overline{z} = i \)

### Step 9: Find the conjugate of \( (2-5i)^2 \)
- Calculate \( (2-5i)^2 \):
\[ (2-5i)^2 = 4 - 20i + 25i^2 = 4 - 20i - 25 = -21 - 20i \]
- Conjugate: \( \overline{z} = -21 + 20i \)

### Summary of Conjugates:
1. \( i \) → \( -i \)
2. \( -5 - 2i \) → \( -5 + 2i \)
3. \( \frac{1}{4 + 3i} \) → \( \frac{4}{25} + \frac{3}{25}i \)
4. \( \frac{(1+i)^2}{3-i} \) → \( -\frac{1}{5} - \frac{3}{5}i \)
5. \( \frac{(1+i)(2+i)}{3+i} \) → \( \frac{3}{5} - \frac{4}{5}i \)
6. \( \sqrt{-3} \) → \( -i\sqrt{3} \)
7. \( \sqrt{2} \) → \( \sqrt{2} \)
8. \( -\sqrt{-1} \) → \( i \)
9. \( (2-5i)^2 \) → \( -21 + 20i \)