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

Answer

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

### Step 1: Set up the equation
Let \( z = x + yi \) be the square root of \(-8 - 6i\). Then we have:
\[ z^2 = -8 - 6i \]
This implies:
\[ (x + yi)^2 = -8 - 6i \]

### Step 2: Expand the left side
Expanding the left side, we get:
\[ x^2 + 2xyi - y^2 = -8 - 6i \]
This can be separated into real and imaginary parts:
\[ (x^2 - y^2) + (2xy)i = -8 - 6i \]

### Step 3: Set up the system of equations
From the above equation, we can equate the real and imaginary parts:
1. \( x^2 - y^2 = -8 \) (1)
2. \( 2xy = -6 \) (2)

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

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

### Step 6: Multiply through by \(x^2\)
To eliminate the fraction, multiply through by \(x^2\):
\[ x^4 + 8x^2 - 9 = 0 \]

### Step 7: Let \(u = x^2\)
Let \(u = x^2\). Then, we have a quadratic equation:
\[ u^2 + 8u - 9 = 0 \]

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

Calculating the two possible values:
1. \(u = \frac{2}{2} = 1\)
2. \(u = \frac{-18}{2} = -9\) (not valid since \(u = x^2\) must be non-negative)

Thus, \(x^2 = 1\) implies:
\[ x = \pm 1 \]

### Step 9: Find corresponding \(y\) values
Using \(y = \frac{-3}{x}\):
1. If \(x = 1\), then \(y = -3\).
2. If \(x = -1\), then \(y = 3\).

### Step 10: Write the square roots
Thus, the two square roots of \(-8 - 6i\) are:
\[ z = 1 - 3i \quad \text{and} \quad z = -1 + 3i \]

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

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