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

Explanation

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

### Step 1: Assume the Square Root
Let \( z = x + iy \) be the square root of \(-i\). Therefore, we have:
\[ z^2 = -i \]

### Step 2: Expand the Square
Now, squaring \( z \):
\[ (x + iy)^2 = x^2 + 2xyi - y^2 = (x^2 - y^2) + (2xy)i \]
Setting this equal to \(-i\), we have:
\[ (x^2 - y^2) + (2xy)i = 0 - 1i \]

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

### Step 4: Solve the First Equation
From the first equation \( x^2 - y^2 = 0 \), we can write:
\[ x^2 = y^2 \]
This implies:
\[ y = \pm x \]

### Step 5: Substitute into the Second Equation
Now, substitute \( y = x \) and \( y = -x \) into the second equation \( 2xy = -1 \).

#### Case 1: \( y = x \)
Substituting gives:
\[ 2x^2 = -1 \quad \Rightarrow \quad x^2 = -\frac{1}{2} \]
This case does not yield real solutions.

#### Case 2: \( y = -x \)
Substituting gives:
\[ 2x(-x) = -1 \quad \Rightarrow \quad -2x^2 = -1 \quad \Rightarrow \quad 2x^2 = 1 \quad \Rightarrow \quad x^2 = \frac{1}{2} \]
Thus, we find:
\[ x = \pm \frac{1}{\sqrt{2}} \]

### Step 6: Find Corresponding \( y \) Values
Since \( y = -x \):
- If \( x = \frac{1}{\sqrt{2}} \), then \( y = -\frac{1}{\sqrt{2}} \).
- If \( x = -\frac{1}{\sqrt{2}} \), then \( y = \frac{1}{\sqrt{2}} \).

### Step 7: Write the Square Roots
Thus, the square roots of \(-i\) are:
\[ z_1 = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i \]
\[ z_2 = -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i \]

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

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