Find the square root of `(-8 -6i)`.

Options

A. `pm(1-3i)`
B. `pm(2-3i).`
C. `pm(1-2i).`
D. `pm(2-2i).`

Explanation

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

### Step 1: Assume the square root
Let \( z = x + iy \) be the square root of \(-8 - 6i\). Therefore, we have:
\[ z^2 = -8 - 6i \]

### Step 2: Expand \(z^2\)
Expanding \(z^2\):
\[ (x + iy)^2 = x^2 + 2xyi - y^2 = (x^2 - y^2) + (2xy)i \]
So, we can equate:
\[ x^2 - y^2 = -8 \quad \text{(1)} \]
\[ 2xy = -6 \quad \text{(2)} \]

### Step 3: Solve for \(x^2 + y^2\)
We can find \(x^2 + y^2\) using the identity:
\[ (x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2 \]
Substituting the values from equations (1) and (2):
\[ (x^2 + y^2)^2 = (-8)^2 + (-6)^2 \]
Calculating:
\[ (x^2 + y^2)^2 = 64 + 36 = 100 \]
Thus:
\[ x^2 + y^2 = \sqrt{100} = 10 \quad \text{(3)} \]

### Step 4: Solve the system of equations
Now we have two equations:
1. \(x^2 - y^2 = -8\) (from equation (1))
2. \(x^2 + y^2 = 10\) (from equation (3))

Adding these two equations:
\[ (x^2 - y^2) + (x^2 + y^2) = -8 + 10 \]
This simplifies to:
\[ 2x^2 = 2 \implies x^2 = 1 \implies x = \pm 1 \]

### Step 5: Find \(y^2\)
Using \(x^2 = 1\) in equation (3):
\[ 1 + y^2 = 10 \implies y^2 = 9 \implies y = \pm 3 \]

### Step 6: Determine the signs of \(x\) and \(y\)
From equation (2), \(2xy = -6\), which implies \(xy < 0\). Therefore, \(x\) and \(y\) must have opposite signs.

- If \(x = 1\), then \(y = -3\).
- If \(x = -1\), then \(y = 3\).

### Step 7: Write the solutions
Thus, the two possible values for \(z\) are:
1. \(z = 1 - 3i\)
2. \(z = -1 + 3i\)

### Final Result
The square root of \(-8 - 6i\) is:
\[ \sqrt{-8 - 6i} = \pm (1 - 3i) \quad \text{or} \quad \pm (-1 + 3i) \]