If `a=sqrt(2i)`, then which of the following is correct?

Options

A. `a=1+i`
B. `a=1-i`
C. `a=-2(sqrt(2))i`
D. none of these

Answer

To solve the problem where \( a = \sqrt{2i} \), we will follow these steps:

### Step 1: Express \( a \) in terms of real and imaginary parts
Assume \( a = x + iy \), where \( x \) and \( y \) are real numbers. We know from the problem that:
\[ a = \sqrt{2i} \]

### Step 2: Square both sides
Squaring both sides gives us:
\[ (x + iy)^2 = 2i \]
Expanding the left side:
\[ x^2 + 2xyi - y^2 = 2i \]
This can be rewritten as:
\[ (x^2 - y^2) + (2xy)i = 0 + 2i \]

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

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

From the imaginary part equation \( 2xy = 2 \), we can simplify to:
\[ xy = 1 \]

### Step 5: Substitute \( x \) in terms of \( y \)
1. If \( x = y \):
\[ y^2 = 1 \implies y = 1 \text{ or } y = -1 \]
Thus, \( x = 1 \) or \( x = -1 \).

2. If \( x = -y \):
\[ -y^2 = 1 \implies y^2 = -1 \]
This does not yield real solutions.

### Step 6: Conclusion
Thus, the possible values for \( (x, y) \) are:
- \( (1, 1) \) leading to \( a = 1 + i \)
- \( (-1, -1) \) leading to \( a = -1 - i \)

Therefore, the values of \( a \) are \( a = 1 + i \) or \( a = -1 - i \).

### Final Answer
The correct option is \( a = \pm(1 + i) \).

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