Find the square root of the following complex numbers
`-8+ 6i`

Answer

To find the square root of the complex number \(-8 + 6i\), we will follow these steps:

### Step 1: Assume the square root
Let the square root of the complex number be \( z = a + bi \), where \( a \) and \( b \) are real numbers.

### Step 2: Square both sides
Squaring both sides gives us:
\[ z^2 = (a + bi)^2 = a^2 + 2abi - b^2 \]
This can be rewritten as:
\[ z^2 = (a^2 - b^2) + (2ab)i \]

### Step 3: Set the equation equal to the complex number
We know that:
\[ z^2 = -8 + 6i \]
Thus, we can equate the real and imaginary parts:
\[ a^2 - b^2 = -8 \quad (1) \]
\[ 2ab = 6 \quad (2) \]

### Step 4: Solve for \(ab\)
From equation (2):
\[ ab = 3 \quad (3) \]
We can express \( b \) in terms of \( a \):
\[ b = \frac{3}{a} \quad (4) \]

### Step 5: Substitute \(b\) into equation (1)
Substituting equation (4) into equation (1):
\[ a^2 - \left(\frac{3}{a}\right)^2 = -8 \]
This simplifies to:
\[ a^2 - \frac{9}{a^2} = -8 \]

### Step 6: Clear the fraction
Multiply through by \( a^2 \) to eliminate the fraction:
\[ a^4 + 8a^2 - 9 = 0 \]

### Step 7: Let \(x = a^2\)
Let \( x = a^2 \). Then the equation becomes:
\[ x^2 + 8x - 9 = 0 \]

### Step 8: 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 two solutions:
\[ x = 1 \quad \text{or} \quad x = -9 \]
Since \( x = a^2 \) cannot be negative, we have:
\[ a^2 = 1 \implies a = \pm 1 \]

### Step 9: Find \(b\) using \(a\)
Using equation (4):
1. If \( a = 1 \):
\[ b = \frac{3}{1} = 3 \]
2. If \( a = -1 \):
\[ b = \frac{3}{-1} = -3 \]

### Step 10: Write the final answers
Thus, the square roots of \(-8 + 6i\) are:
\[ z = 1 + 3i \quad \text{and} \quad z = -1 - 3i \]
So, the final answer is:
\[ \sqrt{-8 + 6i} = \pm (1 + 3i) \]