If `i^(2)=-1`, which of the following is a square root of `8 - 6i` ?

Options

A. `3-i`
B. `3+i`
C. `3-4i`
D. `4-3i`

Explanation

To find the square root of the complex number \(8 - 6i\), we will assume that it can be expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers. We will then square this expression and equate it to \(8 - 6i\).

### Step 1: Set up the equation
Let \(z = a + bi\). Then, we have:
\[ z^2 = (a + bi)^2 \]
Expanding this, we get:
\[ z^2 = a^2 + 2abi + (bi)^2 = a^2 + 2abi - b^2 \]
Thus, we can write:
\[ z^2 = (a^2 - b^2) + (2ab)i \]

### Step 2: Equate the real and imaginary parts
Now, we equate the real and imaginary parts of \(z^2\) with \(8 - 6i\):
\[ a^2 - b^2 = 8 \quad \text{(1)} \]
\[ 2ab = -6 \quad \text{(2)} \]

### Step 3: Solve for \(b\) in terms of \(a\)
From equation (2), we can express \(b\) in terms of \(a\):
\[ b = \frac{-6}{2a} = \frac{-3}{a} \]

### Step 4: Substitute \(b\) into equation (1)
Substituting \(b = \frac{-3}{a}\) into equation (1):
\[ a^2 - \left(\frac{-3}{a}\right)^2 = 8 \]
This simplifies to:
\[ a^2 - \frac{9}{a^2} = 8 \]

### Step 5: Clear the fraction
Multiply through by \(a^2\) to eliminate the fraction:
\[ a^4 - 9 = 8a^2 \]
Rearranging gives:
\[ a^4 - 8a^2 - 9 = 0 \]

### Step 6: Let \(x = a^2\)
Let \(x = a^2\). Then we have:
\[ x^2 - 8x - 9 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ x = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot (-9)}}{2 \cdot 1} \]
\[ x = \frac{8 \pm \sqrt{64 + 36}}{2} \]
\[ x = \frac{8 \pm \sqrt{100}}{2} \]
\[ x = \frac{8 \pm 10}{2} \]
This gives us:
\[ x = 9 \quad \text{or} \quad x = -1 \]

### Step 8: Find \(a\)
Since \(x = a^2\), we take \(x = 9\) (as \(x = -1\) is not valid for \(a^2\)):
\[ a^2 = 9 \implies a = 3 \quad \text{or} \quad a = -3 \]

### Step 9: Find \(b\)
Using \(b = \frac{-3}{a}\):
- If \(a = 3\), then \(b = \frac{-3}{3} = -1\).
- If \(a = -3\), then \(b = \frac{-3}{-3} = 1\).

### Step 10: Write the square roots
Thus, the possible square roots of \(8 - 6i\) are:
\[ 3 - i \quad \text{and} \quad -3 + i \]

Since the problem asks for a square root, we can conclude that one of the square roots is:
\[ \boxed{3 - i} \]