What is the square root of `i, where, ``i = ``sqrt( -1)` ?

Options

A. `( 1+i)/(2)`
B. `(1-i)/(2)`
C. `(1+i)/( sqrt(2))`
D. None of these

Answer

To find the square root of \( i \) (where \( i = \sqrt{-1} \)), we can express the square root in terms of complex numbers. Let's denote the square root of \( i \) as \( A + Bi \), where \( A \) and \( B \) are real numbers.

### Step-by-Step Solution:

1. **Set Up the Equation**:
\[ \sqrt{i} = A + Bi \]
Squaring both sides gives:
\[ i = (A + Bi)^2 \]

2. **Expand the Right Side**:
Using the formula \( (x + y)^2 = x^2 + 2xy + y^2 \):
\[ i = A^2 + 2ABi + (Bi)^2 \]
Since \( (Bi)^2 = B^2(-1) = -B^2 \), we can rewrite the equation as:
\[ i = A^2 - B^2 + 2ABi \]

3. **Separate Real and Imaginary Parts**:
From the equation \( i = A^2 - B^2 + 2ABi \), we can equate the real and imaginary parts:
- Real part: \( A^2 - B^2 = 0 \)
- Imaginary part: \( 2AB = 1 \)

4. **Solve the Real Part**:
From \( A^2 - B^2 = 0 \), we can factor this as:
\[ (A - B)(A + B) = 0 \]
This gives us two cases:
- Case 1: \( A = B \)
- Case 2: \( A = -B \)

5. **Substituting Case 1**:
If \( A = B \), substitute into the imaginary part:
\[ 2A^2 = 1 \implies A^2 = \frac{1}{2} \implies A = \pm \frac{1}{\sqrt{2}} \]
Thus, \( B = A = \pm \frac{1}{\sqrt{2}} \).

6. **Substituting Case 2**:
If \( A = -B \), substitute into the imaginary part:
\[ 2A(-A) = 1 \implies -2A^2 = 1 \implies A^2 = -\frac{1}{2} \]
This case does not yield real solutions since \( A^2 \) cannot be negative.

7. **Final Results**:
From Case 1, we have two possible values:
\[ \sqrt{i} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i \quad \text{or} \quad \sqrt{i} = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i \]
We can express these in a more standard form:
\[ \sqrt{i} = \frac{1}{\sqrt{2}}(1 + i) \quad \text{or} \quad \sqrt{i} = \frac{1}{\sqrt{2}}(-1 - i) \]

### Conclusion:
Thus, the square root of \( i \) can be expressed as:
\[ \sqrt{i} = \frac{1}{\sqrt{2}}(1 + i) \quad \text{or} \quad \sqrt{i} = \frac{1}{\sqrt{2}}(-1 - i) \]