Find the square root of complex number `-i`.

Answer

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

### Step 1: Assume the square root
Let us assume that the square root of \(-i\) can be expressed in the form:
\[ \sqrt{-i} = x + iy \]
where \(x\) and \(y\) are real numbers.

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

### Step 3: Compare real and imaginary parts
Now we can compare the real and imaginary parts from both sides of the equation:
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:
\[ x^2 = y^2 \implies x = y \quad \text{or} \quad x = -y \]

#### Case 1: \(x = y\)
Substituting \(y = x\) into the second equation:
\[ 2x^2 = -1 \implies x^2 = -\frac{1}{2} \implies x = \pm \frac{\sqrt{2}}{2}i \]
Thus, \(y = \pm \frac{\sqrt{2}}{2}i\).

#### Case 2: \(x = -y\)
Substituting \(y = -x\) into the second equation:
\[ 2x(-x) = -1 \implies -2x^2 = -1 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2} \]
Thus, \(y = \mp \frac{\sqrt{2}}{2}\).

### Step 5: Write the final results
Combining both cases, we find:
\[ \sqrt{-i} = \pm \left(\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}\right) \quad \text{or} \quad \pm \left(\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}\right) \]

Thus, the square roots of \(-i\) are:
\[ \sqrt{-i} = \frac{\sqrt{2}}{2}(1 + i) \quad \text{and} \quad \sqrt{-i} = \frac{\sqrt{2}}{2}(1 - i) \]

### Final Answer:
\[ \sqrt{-i} = \pm \frac{\sqrt{2}}{2}(1 + i) \quad \text{or} \quad \pm \frac{\sqrt{2}}{2}(1 - i) \]