Find the square roots of the following: `1- i`

Explanation

To find the square roots of the complex number \(1 - i\), we can follow these steps:

### Step 1: Assume the square root
Let the square root of \(1 - i\) be \(a + ib\), where \(a\) and \(b\) are real numbers.

### Step 2: Square both sides
We square both sides:
\[ (a + ib)^2 = 1 - i \]
Expanding the left side:
\[ a^2 + 2abi + (ib)^2 = 1 - i \]
Since \(i^2 = -1\), this becomes:
\[ a^2 - b^2 + 2abi = 1 - i \]

### Step 3: Equate real and imaginary parts
Now, we can equate the real and imaginary parts:
- Real part: \(a^2 - b^2 = 1\)
- Imaginary part: \(2ab = -1\)

### Step 4: Solve the equations
From the imaginary part, we can express \(b\) in terms of \(a\):
\[ b = \frac{-1}{2a} \]
Substituting this into the equation for the real part:
\[ a^2 - \left(\frac{-1}{2a}\right)^2 = 1 \]
This simplifies to:
\[ a^2 - \frac{1}{4a^2} = 1 \]
Multiplying through by \(4a^2\) to eliminate the fraction:
\[ 4a^4 - 1 = 4a^2 \]
Rearranging gives us a polynomial:
\[ 4a^4 - 4a^2 - 1 = 0 \]

### Step 5: Let \(x = a^2\)
Let \(x = a^2\). The equation becomes:
\[ 4x^2 - 4x - 1 = 0 \]
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4} \]
\[ x = \frac{4 \pm \sqrt{16 + 16}}{8} \]
\[ x = \frac{4 \pm \sqrt{32}}{8} \]
\[ x = \frac{4 \pm 4\sqrt{2}}{8} \]
\[ x = \frac{1 \pm \sqrt{2}}{2} \]

### Step 6: Find \(a\) and \(b\)
Thus, we have:
\[ a^2 = \frac{1 + \sqrt{2}}{2} \quad \text{or} \quad a^2 = \frac{1 - \sqrt{2}}{2} \]
Since \(a^2\) must be non-negative, we take:
\[ a^2 = \frac{1 + \sqrt{2}}{2} \]
Then:
\[ a = \pm \sqrt{\frac{1 + \sqrt{2}}{2}} \]

Now, substituting \(a\) back to find \(b\):
\[ b = \frac{-1}{2a} = \frac{-1}{2\sqrt{\frac{1 + \sqrt{2}}{2}}} = \frac{-\sqrt{2}}{2\sqrt{1 + \sqrt{2}}} \]

### Step 7: Final results
Thus, the square roots of \(1 - i\) are:
\[ \sqrt{1 - i} = \pm \left( \sqrt{\frac{1 + \sqrt{2}}{2}} - i \frac{\sqrt{2}}{2\sqrt{1 + \sqrt{2}}} \right) \]