find the square roots of the following:
(i) 2i
(ii) `-i`

Answer

To find the square roots of the complex numbers \(2i\) and \(-i\), we will follow a systematic approach.

### Part (i): Finding the square roots of \(2i\)

1. **Assume the square root**: Let \(z = a + bi\) be the square root of \(2i\). Therefore, we have:
\[ z^2 = 2i \]
This implies:
\[ (a + bi)^2 = 2i \]

2. **Expand the left side**: Using the formula \((a + bi)^2 = a^2 + 2abi - b^2\), we can rewrite the equation:
\[ a^2 - b^2 + 2abi = 0 + 2i \]

3. **Equate real and imaginary parts**: From the equation, we can separate the real and imaginary parts:
- Real part: \(a^2 - b^2 = 0\)
- Imaginary part: \(2ab = 2\)

4. **Solve the equations**:
- From \(a^2 - b^2 = 0\), we get \(a^2 = b^2\) which gives us \(a = b\) or \(a = -b\).
- From \(2ab = 2\), we simplify to \(ab = 1\).

5. **Case 1: \(a = b\)**:
- Substituting \(b = a\) into \(ab = 1\):
\[ a \cdot a = 1 \implies a^2 = 1 \implies a = 1 \text{ or } a = -1 \]
- Therefore, if \(a = 1\), then \(b = 1\) and if \(a = -1\), then \(b = -1\). This gives us the roots:
\[ z = 1 + i \quad \text{and} \quad z = -1 - i \]

6. **Case 2: \(a = -b\)**:
- Substituting \(b = -a\) into \(ab = 1\):
\[ a \cdot (-a) = 1 \implies -a^2 = 1 \implies a^2 = -1 \]
- This case is not possible since \(a\) and \(b\) are real numbers.

7. **Final roots**: Thus, the square roots of \(2i\) are:
\[ z = 1 + i \quad \text{and} \quad z = -1 - i \]

### Part (ii): Finding the square roots of \(-i\)

1. **Assume the square root**: Let \(z = a + bi\) be the square root of \(-i\). Therefore, we have:
\[ z^2 = -i \]
This implies:
\[ (a + bi)^2 = -i \]

2. **Expand the left side**: Using the formula \((a + bi)^2 = a^2 + 2abi - b^2\), we can rewrite the equation:
\[ a^2 - b^2 + 2abi = 0 - 1i \]

3. **Equate real and imaginary parts**: From the equation, we can separate the real and imaginary parts:
- Real part: \(a^2 - b^2 = 0\)
- Imaginary part: \(2ab = -1\)

4. **Solve the equations**:
- From \(a^2 - b^2 = 0\), we get \(a^2 = b^2\) which gives us \(a = b\) or \(a = -b\).
- From \(2ab = -1\), we simplify to \(ab = -\frac{1}{2}\).

5. **Case 1: \(a = b\)**:
- Substituting \(b = a\) into \(ab = -\frac{1}{2}\):
\[ a \cdot a = -\frac{1}{2} \implies a^2 = -\frac{1}{2} \]
- This case is not possible since \(a\) and \(b\) are real numbers.

6. **Case 2: \(a = -b\)**:
- Substituting \(b = -a\) into \(ab = -\frac{1}{2}\):
\[ a \cdot (-a) = -\frac{1}{2} \implies -a^2 = -\frac{1}{2} \implies a^2 = \frac{1}{2} \]
- Thus, \(a = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}\) and \(b = \mp \frac{\sqrt{2}}{2}\).

7. **Final roots**: Therefore, the square roots of \(-i\) are:
\[ z = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i \quad \text{and} \quad z = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i \]

### Summary of Results:
- The square roots of \(2i\) are \(1 + i\) and \(-1 - i\).
- The square roots of \(-i\) are \(\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\) and \(-\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i\).