The square root of -8i is

Options

A. ` +-2(1-i)`
B. `2(1+i)`
C. `+-(1-i)`
D. `+-(1+i)`

Answer

To find the square root of \(-8i\), we can follow these steps:

### Step 1: Set up the equation
Let \( z = \sqrt{-8i} \). We can express \( z \) in terms of its real and imaginary parts:
\[ z = x + iy \]
where \( x \) and \( y \) are real numbers.

### Step 2: Square both sides
Squaring both sides gives:
\[ -8i = (x + iy)^2 \]
Expanding the right side:
\[ -8i = x^2 + 2xyi - y^2 \]
This can be rearranged to:
\[ -8i = (x^2 - y^2) + (2xy)i \]

### Step 3: Equate real and imaginary parts
From the equation above, we can separate the real and imaginary parts:
1. Real part: \( x^2 - y^2 = 0 \)
2. Imaginary part: \( 2xy = -8 \)

### Step 4: Solve the equations
From the real part equation \( x^2 - y^2 = 0 \), we can conclude:
\[ x^2 = y^2 \implies x = y \text{ or } x = -y \]

Using the imaginary part \( 2xy = -8 \):
1. If \( x = y \):
\[ 2x^2 = -8 \implies x^2 = -4 \quad (\text{not possible for real } x) \]
2. If \( x = -y \):
\[ 2(-y)y = -8 \implies -2y^2 = -8 \implies y^2 = 4 \implies y = 2 \text{ or } y = -2 \]
Thus, \( x = -2 \) or \( x = 2 \).

### Step 5: Find the values of \( z \)
From the values of \( y \):
1. If \( y = 2 \), then \( x = -2 \):
\[ z = -2 + 2i \]
2. If \( y = -2 \), then \( x = 2 \):
\[ z = 2 - 2i \]

### Conclusion
The square roots of \(-8i\) are:
\[ z = -2 + 2i \quad \text{and} \quad z = 2 - 2i \]

### Final Answer
\[ \sqrt{-8i} = \pm 2(1 - i) \]