Find the square root of the following :
(i) `3-4i`
(ii) `4+6isqrt(5)`
(iii) `-i`
(iv) `8i`
(v) `-7+24i`
(vi) `-24- 10i`

Explanation

To find the square roots of the given complex numbers, we will use the method of expressing the square root in the form \( x + iy \) and then squaring both sides to compare real and imaginary parts.

### (i) Find the square root of \( 3 - 4i \)

1. Let \( \sqrt{3 - 4i} = x + iy \).
2. Squaring both sides gives:
\[ 3 - 4i = (x + iy)^2 = x^2 - y^2 + 2xyi \]
3. From this, we can equate the real and imaginary parts:
- Real part: \( x^2 - y^2 = 3 \) (1)
- Imaginary part: \( 2xy = -4 \) (2)
4. From (2), we can express \( xy \):
\[ xy = -2 \quad \Rightarrow \quad y = \frac{-2}{x} \quad (3) \]
5. Substitute (3) into (1):
\[ x^2 - \left(\frac{-2}{x}\right)^2 = 3 \]
\[ x^2 - \frac{4}{x^2} = 3 \]
\[ x^4 - 3x^2 - 4 = 0 \]
6. Let \( z = x^2 \). Then the equation becomes:
\[ z^2 - 3z - 4 = 0 \]
7. Solving this using the quadratic formula:
\[ z = \frac{3 \pm \sqrt{(3)^2 + 4 \cdot 4}}{2} = \frac{3 \pm 5}{2} \]
\[ z = 4 \quad \text{or} \quad z = -1 \]
Thus, \( x^2 = 4 \) gives \( x = \pm 2 \) (since \( z = -1 \) is not valid).
8. Substitute back to find \( y \):
- If \( x = 2 \):
\[ y = \frac{-2}{2} = -1 \]
- If \( x = -2 \):
\[ y = \frac{-2}{-2} = 1 \]
9. Therefore, the square roots are:
\[ \sqrt{3 - 4i} = \pm (2 - i) \]

### (ii) Find the square root of \( 4 + 6i\sqrt{5} \)

1. Let \( \sqrt{4 + 6i\sqrt{5}} = x + iy \).
2. Squaring gives:
\[ 4 + 6i\sqrt{5} = x^2 - y^2 + 2xyi \]
3. Equating parts:
- \( x^2 - y^2 = 4 \) (1)
- \( 2xy = 6\sqrt{5} \) (2)
4. From (2):
\[ xy = 3\sqrt{5} \quad \Rightarrow \quad y = \frac{3\sqrt{5}}{x} \quad (3) \]
5. Substitute (3) into (1):
\[ x^2 - \left(\frac{3\sqrt{5}}{x}\right)^2 = 4 \]
\[ x^2 - \frac{45}{x^2} = 4 \]
\[ x^4 - 4x^2 - 45 = 0 \]
6. Let \( z = x^2 \):
\[ z^2 - 4z - 45 = 0 \]
7. Solving:
\[ z = \frac{4 \pm \sqrt{16 + 180}}{2} = \frac{4 \pm 14}{2} \]
\[ z = 9 \quad \text{or} \quad z = -5 \]
Thus, \( x^2 = 9 \) gives \( x = \pm 3 \).
8. Substitute back to find \( y \):
- If \( x = 3 \):
\[ y = \frac{3\sqrt{5}}{3} = \sqrt{5} \]
- If \( x = -3 \):
\[ y = \frac{3\sqrt{5}}{-3} = -\sqrt{5} \]
9. Therefore, the square roots are:
\[ \sqrt{4 + 6i\sqrt{5}} = \pm (3 + i\sqrt{5}) \]

### (iii) Find the square root of \( -i \)

1. Let \( \sqrt{-i} = x + iy \).
2. Squaring gives:
\[ -i = x^2 - y^2 + 2xyi \]
3. Equating parts:
- \( x^2 - y^2 = 0 \) (1)
- \( 2xy = -1 \) (2)
4. From (1):
\[ x^2 = y^2 \quad \Rightarrow \quad y = \pm x \quad (3) \]
5. Substitute (3) into (2):
- If \( y = x \):
\[ 2x^2 = -1 \quad \text{(not valid)} \]
- If \( y = -x \):
\[ -2x^2 = -1 \quad \Rightarrow \quad x^2 = \frac{1}{2} \quad \Rightarrow \quad x = \pm \frac{1}{\sqrt{2}} \]
6. Thus, \( y = \mp \frac{1}{\sqrt{2}} \).
7. Therefore, the square roots are:
\[ \sqrt{-i} = \pm \left(\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}\right) \]

### (iv) Find the square root of \( 8i \)

1. Let \( \sqrt{8i} = x + iy \).
2. Squaring gives:
\[ 8i = x^2 - y^2 + 2xyi \]
3. Equating parts:
- \( x^2 - y^2 = 0 \) (1)
- \( 2xy = 8 \) (2)
4. From (1):
\[ x^2 = y^2 \quad \Rightarrow \quad y = \pm x \quad (3) \]
5. Substitute (3) into (2):
- If \( y = x \):
\[ 2x^2 = 8 \quad \Rightarrow \quad x^2 = 4 \quad \Rightarrow \quad x = \pm 2 \]
- If \( y = -x \):
\[ -2x^2 = 8 \quad \text{(not valid)} \]
6. Therefore, \( y = 2 \) when \( x = 2 \) or \( y = -2 \) when \( x = -2 \).
7. Thus, the square roots are:
\[ \sqrt{8i} = \pm (2 + 2i) \]

### (v) Find the square root of \( -7 + 24i \)

1. Let \( \sqrt{-7 + 24i} = x + iy \).
2. Squaring gives:
\[ -7 + 24i = x^2 - y^2 + 2xyi \]
3. Equating parts:
- \( x^2 - y^2 = -7 \) (1)
- \( 2xy = 24 \) (2)
4. From (2):
\[ xy = 12 \quad \Rightarrow \quad y = \frac{12}{x} \quad (3) \]
5. Substitute (3) into (1):
\[ x^2 - \left(\frac{12}{x}\right)^2 = -7 \]
\[ x^2 - \frac{144}{x^2} = -7 \]
\[ x^4 + 7x^2 - 144 = 0 \]
6. Let \( z = x^2 \):
\[ z^2 + 7z - 144 = 0 \]
7. Solving:
\[ z = \frac{-7 \pm \sqrt{49 + 576}}{2} = \frac{-7 \pm 25}{2} \]
\[ z = 9 \quad \text{or} \quad z = -16 \]
Thus, \( x^2 = 9 \) gives \( x = \pm 3 \).
8. Substitute back to find \( y \):
- If \( x = 3 \):
\[ y = \frac{12}{3} = 4 \]
- If \( x = -3 \):
\[ y = \frac{12}{-3} = -4 \]
9. Therefore, the square roots are:
\[ \sqrt{-7 + 24i} = \pm (3 + 4i) \]

### (vi) Find the square root of \( -24 - 10i \)

1. Let \( \sqrt{-24 - 10i} = x + iy \).
2. Squaring gives:
\[ -24 - 10i = x^2 - y^2 + 2xyi \]
3. Equating parts:
- \( x^2 - y^2 = -24 \) (1)
- \( 2xy = -10 \) (2)
4. From (2):
\[ xy = -5 \quad \Rightarrow \quad y = \frac{-5}{x} \quad (3) \]
5. Substitute (3) into (1):
\[ x^2 - \left(\frac{-5}{x}\right)^2 = -24 \]
\[ x^2 - \frac{25}{x^2} = -24 \]
\[ x^4 - 24x^2 - 25 = 0 \]
6. Let \( z = x^2 \):
\[ z^2 - 24z - 25 = 0 \]
7. Solving:
\[ z = \frac{24 \pm \sqrt{576 + 100}}{2} = \frac{24 \pm 26}{2} \]
\[ z = 25 \quad \text{or} \quad z = -1 \]
Thus, \( x^2 = 25 \) gives \( x = \pm 5 \).
8. Substitute back to find \( y \):
- If \( x = 5 \):
\[ y = \frac{-5}{5} = -1 \]
- If \( x = -5 \):
\[ y = \frac{-5}{-5} = 1 \]
9. Therefore, the square roots are:
\[ \sqrt{-24 - 10i} = \pm (5 - i) \]

### Summary of Results

1. \( \sqrt{3 - 4i} = \pm (2 - i) \)
2. \( \sqrt{4 + 6i\sqrt{5}} = \pm (3 + i\sqrt{5}) \)
3. \( \sqrt{-i} = \pm \left(\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}\right) \)
4. \( \sqrt{8i} = \pm (2 + 2i) \)
5. \( \sqrt{-7 + 24i} = \pm (3 + 4i) \)
6. \( \sqrt{-24 - 10i} = \pm (5 - i) \)