Finding the Square Root of i using Complex Numbers

The question asks us to find the square root of the imaginary unit \( i \), where \( i = \sqrt{-1} \). We are looking for a complex number \( z \) such that \( z^2 = i \).

Let the square root of \( i \) be a complex number \( x + yi \), where \( x \) and \( y \) are real numbers. So, we have:

\( \sqrt{i} = x + yi \)

To find \( x \) and \( y \), we can square both sides of the equation:

\( (x + yi)^2 = i \)

Expand the left side:

\( x^2 + 2xyi + (yi)^2 = i \)

Since \( i^2 = -1 \), we substitute this into the equation:

\( x^2 + 2xyi - y^2 = i \)

Rearrange the terms to group the real and imaginary parts:

\( (x^2 - y^2) + (2xy)i = 0 + 1i \)

For two complex numbers to be equal, their real parts must be equal, and their imaginary parts must be equal. Equating the real and imaginary parts, we get a system of two equations:

Equation 1 (Real parts): \( x^2 - y^2 = 0 \)

Equation 2 (Imaginary parts): \( 2xy = 1 \)

From Equation 1, we have \( x^2 = y^2 \), which means \( x = y \) or \( x = -y \).

Let's consider these two cases:

Case 1: \( x = y \)

Substitute \( y = x \) into Equation 2:

\( 2x(x) = 1 \)

\( 2x^2 = 1 \)

\( x^2 = \frac{1}{2} \)

Taking the square root of both sides gives:

\( x = \pm \sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}} \)

• If \( x = \frac{1}{\sqrt{2}} \), then \( y = x = \frac{1}{\sqrt{2}} \). This gives the square root \( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i = \frac{1 + i}{\sqrt{2}} \).
• If \( x = -\frac{1}{\sqrt{2}} \), then \( y = x = -\frac{1}{\sqrt{2}} \). This gives the square root \( -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i = -\frac{1 + i}{\sqrt{2}} \).

Case 2: \( x = -y \)

Substitute \( y = -x \) into Equation 2:

\( 2x(-x) = 1 \)

\( -2x^2 = 1 \)

\( x^2 = -\frac{1}{2} \)

Since \( x \) is a real number, \( x^2 \) cannot be negative. Therefore, there are no real solutions for \( x \) and \( y \) in this case.

Thus, the two square roots of \( i \) are \( \frac{1 + i}{\sqrt{2}} \) and \( -\frac{1 + i}{\sqrt{2}} \). We are looking for one of these values among the options.

Let's check the options provided:

• Option 1: \( \frac{1 + i}{2} \)
• Option 2: \( \frac{1 - i}{2} \)
• Option 3: \( \frac{1 + i}{\sqrt{2}} \)

Comparing our results with the options, we see that \( \frac{1 + i}{\sqrt{2}} \) matches Option 3.

We can verify this by squaring the expression from Option 3:

\( \left( \frac{1 + i}{\sqrt{2}} \right)^2 = \frac{(1 + i)^2}{(\sqrt{2})^2} = \frac{1^2 + 2(1)(i) + i^2}{2} = \frac{1 + 2i - 1}{2} = \frac{2i}{2} = i \)

Since squaring \( \frac{1 + i}{\sqrt{2}} \) results in \( i \), this is indeed a square root of \( i \).

Therefore, the square root of \( i \) among the given options is \( \frac{1 + i}{\sqrt{2}} \).