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 root of the given complex numbers, we will use the method of expressing the square root in the form \( x + iy \) and then equate the real and imaginary parts.

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

1. Let \( z = x + iy \).
2. Then, \( z^2 = 3 - 4i \).
3. Expanding \( z^2 \):
\[ z^2 = (x + iy)^2 = x^2 - y^2 + 2xyi \]
4. Equate the real and imaginary parts:
\[ x^2 - y^2 = 3 \quad (1) \]
\[ 2xy = -4 \quad (2) \]
5. From (2), we can express \( y \):
\[ y = \frac{-4}{2x} = \frac{-2}{x} \]
6. Substitute \( y \) in (1):
\[ x^2 - \left(\frac{-2}{x}\right)^2 = 3 \]
\[ x^2 - \frac{4}{x^2} = 3 \]
7. Multiply through by \( x^2 \) to eliminate the fraction:
\[ x^4 - 3x^2 - 4 = 0 \]
8. Let \( u = x^2 \):
\[ u^2 - 3u - 4 = 0 \]
9. Factor or use the quadratic formula:
\[ u = \frac{3 \pm \sqrt{(3)^2 + 4 \cdot 4}}{2} = \frac{3 \pm 5}{2} \]
\[ u = 4 \quad \text{or} \quad u = -1 \]
(Only \( u = 4 \) is valid since \( u = x^2 \geq 0 \))
10. Thus, \( x^2 = 4 \) gives \( x = \pm 2 \).
11. Substitute \( x = 2 \) into (2):
\[ 2(2)y = -4 \Rightarrow y = -1 \]
12. Therefore, one square root is \( 2 - i \). The other is \( -2 + i \).

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

1. Let \( z = x + iy \).
2. Then, \( z^2 = 4 + 6i\sqrt{5} \).
3. Expanding \( z^2 \):
\[ z^2 = x^2 - y^2 + 2xyi \]
4. Equate the real and imaginary parts:
\[ x^2 - y^2 = 4 \quad (1) \]
\[ 2xy = 6\sqrt{5} \quad (2) \]
5. From (2):
\[ y = \frac{3\sqrt{5}}{x} \]
6. Substitute \( y \) in (1):
\[ x^2 - \left(\frac{3\sqrt{5}}{x}\right)^2 = 4 \]
\[ x^2 - \frac{45}{x^2} = 4 \]
7. Multiply through by \( x^2 \):
\[ x^4 - 4x^2 - 45 = 0 \]
8. Let \( u = x^2 \):
\[ u^2 - 4u - 45 = 0 \]
9. Using the quadratic formula:
\[ u = \frac{4 \pm \sqrt{16 + 180}}{2} = \frac{4 \pm 14}{2} \]
\[ u = 9 \quad \text{or} \quad u = -5 \]
(Only \( u = 9 \) is valid)
10. Thus, \( x^2 = 9 \) gives \( x = \pm 3 \).
11. Substitute \( x = 3 \) into (2):
\[ 2(3)y = 6\sqrt{5} \Rightarrow y = \sqrt{5} \]
12. Therefore, one square root is \( 3 + i\sqrt{5} \). The other is \( -3 - i\sqrt{5} \).

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

1. Let \( z = x + iy \).
2. Then, \( z^2 = -i \).
3. Expanding \( z^2 \):
\[ z^2 = x^2 - y^2 + 2xyi \]
4. Equate the real and imaginary parts:
\[ x^2 - y^2 = 0 \quad (1) \]
\[ 2xy = -1 \quad (2) \]
5. From (1), \( x^2 = y^2 \) implies \( y = \pm x \).
6. Substitute \( y = x \) into (2):
\[ 2x^2 = -1 \quad \text{(not possible)} \]
Substitute \( y = -x \):
\[ 2x(-x) = -1 \Rightarrow -2x^2 = -1 \Rightarrow x^2 = \frac{1}{2} \]
7. Thus, \( x = \pm \frac{1}{\sqrt{2}} \) and \( y = \mp \frac{1}{\sqrt{2}} \).
8. Therefore, the square roots are \( \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i \) and \( -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i \).

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

1. Let \( z = x + iy \).
2. Then, \( z^2 = 8i \).
3. Expanding \( z^2 \):
\[ z^2 = x^2 - y^2 + 2xyi \]
4. Equate the real and imaginary parts:
\[ x^2 - y^2 = 0 \quad (1) \]
\[ 2xy = 8 \quad (2) \]
5. From (1), \( x^2 = y^2 \) implies \( y = \pm x \).
6. Substitute \( y = x \) into (2):
\[ 2x^2 = 8 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2 \]
Substitute \( y = -x \):
\[ 2x(-x) = 8 \quad \text{(not possible)} \]
7. Therefore, the square roots are \( 2 + 2i \) and \( -2 - 2i \).

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

1. Let \( z = x + iy \).
2. Then, \( z^2 = -7 + 24i \).
3. Expanding \( z^2 \):
\[ z^2 = x^2 - y^2 + 2xyi \]
4. Equate the real and imaginary parts:
\[ x^2 - y^2 = -7 \quad (1) \]
\[ 2xy = 24 \quad (2) \]
5. From (2):
\[ y = \frac{12}{x} \]
6. Substitute \( y \) in (1):
\[ x^2 - \left(\frac{12}{x}\right)^2 = -7 \]
\[ x^2 - \frac{144}{x^2} = -7 \]
7. Multiply through by \( x^2 \):
\[ x^4 + 7x^2 - 144 = 0 \]
8. Let \( u = x^2 \):
\[ u^2 + 7u - 144 = 0 \]
9. Using the quadratic formula:
\[ u = \frac{-7 \pm \sqrt{49 + 576}}{2} = \frac{-7 \pm 25}{2} \]
\[ u = 9 \quad \text{or} \quad u = -16 \quad \text{(only valid } u = 9\text{)} \]
10. Thus, \( x^2 = 9 \) gives \( x = \pm 3 \).
11. Substitute \( x = 3 \) into (2):
\[ 2(3)y = 24 \Rightarrow y = 4 \]
12. Therefore, the square roots are \( 3 + 4i \) and \( -3 - 4i \).

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

1. Let \( z = x + iy \).
2. Then, \( z^2 = -24 - 10i \).
3. Expanding \( z^2 \):
\[ z^2 = x^2 - y^2 + 2xyi \]
4. Equate the real and imaginary parts:
\[ x^2 - y^2 = -24 \quad (1) \]
\[ 2xy = -10 \quad (2) \]
5. From (2):
\[ y = \frac{-5}{x} \]
6. Substitute \( y \) in (1):
\[ x^2 - \left(\frac{-5}{x}\right)^2 = -24 \]
\[ x^2 - \frac{25}{x^2} = -24 \]
7. Multiply through by \( x^2 \):
\[ x^4 + 24x^2 - 25 = 0 \]
8. Let \( u = x^2 \):
\[ u^2 + 24u - 25 = 0 \]
9. Using the quadratic formula:
\[ u = \frac{-24 \pm \sqrt{576 + 100}}{2} = \frac{-24 \pm 26}{2} \]
\[ u = 1 \quad \text{or} \quad u = -25 \quad \text{(only valid } u = 1\text{)} \]
10. Thus, \( x^2 = 1 \) gives \( x = \pm 1 \).
11. Substitute \( x = 1 \) into (2):
\[ 2(1)y = -10 \Rightarrow y = -5 \]
12. Therefore, the square roots are \( 1 - 5i \) and \( -1 + 5i \).

### Summary of Results:
1. \( \sqrt{3 - 4i} = 2 - i \) or \( -2 + i \)
2. \( \sqrt{4 + 6i\sqrt{5}} = 3 + i\sqrt{5} \) or \( -3 - i\sqrt{5} \)
3. \( \sqrt{-i} = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i \) or \( -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i \)
4. \( \sqrt{8i} = 2 + 2i \) or \( -2 - 2i \)
5. \( \sqrt{-7 + 24i} = 3 + 4i \) or \( -3 - 4i \)
6. \( \sqrt{-24 - 10i} = 1 - 5i \) or \( -1 + 5i \)