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 \(x + iy\), where \(x\) and \(y\) are real numbers.

### Step 2: Square both sides
Squaring both sides gives us:
\[ (x + iy)^2 = 1 - i \]
Expanding the left side:
\[ x^2 + 2xyi - y^2 = 1 - i \]
This can be rearranged as:
\[ (x^2 - y^2) + (2xy)i = 1 - i \]

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

### Step 4: Solve for \(y\)
From the imaginary part equation, we can express \(y\) in terms of \(x\):
\[ y = \frac{-1}{2x} \]

### Step 5: Substitute \(y\) into the real part equation
Substituting \(y\) into the real part equation:
\[ x^2 - \left(\frac{-1}{2x}\right)^2 = 1 \]
This simplifies to:
\[ x^2 - \frac{1}{4x^2} = 1 \]

### Step 6: Clear the fraction
Multiply through by \(4x^2\) to eliminate the fraction:
\[ 4x^4 - 1 = 4x^2 \]
Rearranging gives us:
\[ 4x^4 - 4x^2 - 1 = 0 \]

### Step 7: Use the quadratic formula
Let \(u = x^2\). We can rewrite the equation as:
\[ 4u^2 - 4u - 1 = 0 \]
Using the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ u = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4} \]
Calculating the discriminant:
\[ u = \frac{4 \pm \sqrt{16 + 16}}{8} = \frac{4 \pm \sqrt{32}}{8} = \frac{4 \pm 4\sqrt{2}}{8} = \frac{1 \pm \sqrt{2}}{2} \]

### Step 8: Find \(x\)
Thus, we have:
\[ x^2 = \frac{1 + \sqrt{2}}{2} \quad \text{or} \quad x^2 = \frac{1 - \sqrt{2}}{2} \]
Since \(x^2\) must be non-negative, we only consider:
\[ x^2 = \frac{1 + \sqrt{2}}{2} \]
So,
\[ x = \sqrt{\frac{1 + \sqrt{2}}{2}} \quad \text{or} \quad x = -\sqrt{\frac{1 + \sqrt{2}}{2}} \]

### Step 9: Find \(y\)
Using \(y = \frac{-1}{2x}\):
1. For \(x = \sqrt{\frac{1 + \sqrt{2}}{2}}\):
\[ y = \frac{-1}{2\sqrt{\frac{1 + \sqrt{2}}{2}}} \]
2. For \(x = -\sqrt{\frac{1 + \sqrt{2}}{2}}\):
\[ y = \frac{1}{2\sqrt{\frac{1 + \sqrt{2}}{2}}} \]

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