Find the square roots of the following:`-8\- 6i`

Explanation

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

### Step 1: Assume the square root
Let the square root of \(-8 - 6i\) 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:
\[ z^2 = (a^2 - b^2) + (2ab)i \]
We want this to equal \(-8 - 6i\).

### Step 3: Set up equations
From the above expression, we can equate the real and imaginary parts:
1. \(a^2 - b^2 = -8\) (real part)
2. \(2ab = -6\) (imaginary part)

### Step 4: Solve for \(b\)
From the second equation \(2ab = -6\), we can express \(b\) in terms of \(a\):
\[ b = \frac{-6}{2a} = \frac{-3}{a} \]

### Step 5: Substitute \(b\) into the first equation
Substituting \(b\) into the first equation gives:
\[ a^2 - \left(\frac{-3}{a}\right)^2 = -8 \]
This simplifies to:
\[ a^2 - \frac{9}{a^2} = -8 \]
Multiplying through by \(a^2\) to eliminate the fraction:
\[ a^4 + 8a^2 - 9 = 0 \]

### Step 6: Let \(x = a^2\)
Let \(x = a^2\). Then we have:
\[ x^2 + 8x - 9 = 0 \]
We can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot (-9)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ x = \frac{-8 \pm \sqrt{64 + 36}}{2} = \frac{-8 \pm \sqrt{100}}{2} = \frac{-8 \pm 10}{2} \]
This gives us two solutions:
1. \(x = 1\)
2. \(x = -9\) (not valid since \(x = a^2\) must be non-negative)

### Step 7: Find \(a\)
Since \(x = a^2 = 1\), we have:
\[ a = \pm 1 \]

### Step 8: Find \(b\)
Now, substituting \(a\) back to find \(b\):
- If \(a = 1\):
\[ b = \frac{-3}{1} = -3 \]
- If \(a = -1\):
\[ b = \frac{-3}{-1} = 3 \]

### Step 9: Write the square roots
Thus, the square roots of \(-8 - 6i\) are:
\[ 1 - 3i \quad \text{and} \quad -1 + 3i \]
We can express this as:
\[ \sqrt{-8 - 6i} = \pm(1 - 3i) \]

### Final Answer
The square roots of \(-8 - 6i\) are:
\[ 1 - 3i \quad \text{and} \quad -1 + 3i \]

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