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

Explanation

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

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

### Step 2: Square Both Sides
Squaring both sides gives us:
\[ (x + iy)^2 = 1 + i \]
Expanding the left-hand side:
\[ x^2 + 2xyi - y^2 = 1 + i \]
This can be rewritten 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 \(2xy = 1\), 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:
\[ x^2 - \frac{1}{4x^2} = 1 \]
Multiplying through by \(4x^2\) to eliminate the fraction:
\[ 4x^4 - 1 = 4x^2 \]
Rearranging gives:
\[ 4x^4 - 4x^2 - 1 = 0 \]

### Step 6: Let \(u = x^2\)
Let \(u = x^2\), then we have:
\[ 4u^2 - 4u - 1 = 0 \]
This is a quadratic equation in \(u\).

### Step 7: Use the Quadratic Formula
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 two possible values for \(u\):
1. \(u_1 = \frac{1 + \sqrt{2}}{2}\)
2. \(u_2 = \frac{1 - \sqrt{2}}{2}\) (this will be negative, so we discard it)

Taking the positive square root:
\[ x = \sqrt{\frac{1 + \sqrt{2}}{2}} \]

### Step 9: Find \(y\)
Using \(y = \frac{1}{2x}\):
\[ y = \frac{1}{2\sqrt{\frac{1 + \sqrt{2}}{2}}} = \frac{1}{\sqrt{2(1 + \sqrt{2})}} = \frac{\sqrt{2}}{2 + 2\sqrt{2}} = \frac{\sqrt{2}(2 - 2\sqrt{2})}{-2} = \frac{2\sqrt{2} - 4}{-2} = \sqrt{2} - 2 \]

### Step 10: Final Complex Numbers
Thus, the two square roots of \(1 + i\) are:
\[ \sqrt{\frac{1 + \sqrt{2}}{2}} + i\left(\sqrt{2} - 2\right) \quad \text{and} \quad -\sqrt{\frac{1 + \sqrt{2}}{2}} - i\left(\sqrt{2} - 2\right) \]