What are the square roots of `-2i` ?
`(i=sqrt(-1))`

Options

A. `+-(1+i)`
B. `+-(1-i)`
C. `+-i`
D. `+-1`

Explanation

To find the square roots of \(-2i\), we can represent the square root in the form \(z = x + iy\), where \(x\) and \(y\) are real numbers.

### Step-by-step Solution:

1. **Set up the equation**:
\[ z^2 = -2i \]
Let \(z = x + iy\).

2. **Square both sides**:
\[ (x + iy)^2 = -2i \]
Expanding the left side gives:
\[ x^2 + 2xyi - y^2 = -2i \]
This can be rearranged to:
\[ (x^2 - y^2) + (2xy)i = 0 - 2i \]

3. **Compare real and imaginary parts**:
From the equation, we can separate the real and imaginary parts:
- Real part: \(x^2 - y^2 = 0\)
- Imaginary part: \(2xy = -2\)

4. **Solve the equations**:
From the real part equation \(x^2 - y^2 = 0\), we can express \(y\) in terms of \(x\):
\[ y^2 = x^2 \implies y = \pm x \]
Substitute \(y = x\) into the imaginary part equation:
\[ 2x(x) = -2 \implies 2x^2 = -2 \implies x^2 = -1 \]
This is not possible for real \(x\). Now substitute \(y = -x\):
\[ 2x(-x) = -2 \implies -2x^2 = -2 \implies x^2 = 1 \implies x = \pm 1 \]

5. **Find corresponding \(y\) values**:
- If \(x = 1\), then \(y = -1\).
- If \(x = -1\), then \(y = 1\).

6. **Write the square roots**:
Thus, the two square roots of \(-2i\) are:
\[ z_1 = 1 - i \quad \text{and} \quad z_2 = -1 + i \]

### Final Answer:
The square roots of \(-2i\) are \(1 - i\) and \(-1 + i\).

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