Find the square roots of the following:` -i`

Explanation

To find the square roots of the complex number \(-i\), we can follow these steps:

### Step 1: Assume the square root
Let the square root of \(-i\) be \(x + iy\), where \(x\) is the real part and \(y\) is the imaginary part.

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

### 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: \(2xy = -1\)

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

### Step 5: Substitute into the second equation
Let's first consider \(x = y\):
Substituting into the second equation \(2xy = -1\):
\[ 2x^2 = -1 \quad \Rightarrow \quad x^2 = -\frac{1}{2} \]
This does not yield a real solution.

Now consider \(x = -y\):
Substituting into the second equation:
\[ 2(-y)y = -1 \quad \Rightarrow \quad -2y^2 = -1 \quad \Rightarrow \quad 2y^2 = 1 \quad \Rightarrow \quad y^2 = \frac{1}{2} \]
Thus, we find:
\[ y = \pm \frac{1}{\sqrt{2}} \]

### Step 6: Find corresponding \(x\) values
Since \(x = -y\), we have:
1. If \(y = \frac{1}{\sqrt{2}}\), then \(x = -\frac{1}{\sqrt{2}}\).
2. If \(y = -\frac{1}{\sqrt{2}}\), then \(x = \frac{1}{\sqrt{2}}\).

### Step 7: Write the final answers
Thus, the two square roots of \(-i\) are:
\[ -\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} \quad \text{and} \quad \frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}} \]

### Final Answer
The square roots of \(-i\) are:
\[ -\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} \quad \text{and} \quad \frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}} \]

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