Find the square roots of the following: `15-8i`

Explanation

To find the square roots of the complex number \( 15 - 8i \), we can follow these steps:

### Step 1: Assume the square root
Let \( z_1 \) be the square root of \( 15 - 8i \). We can express \( z_1 \) in the form \( z_1 = x + yi \), where \( x \) and \( y \) are real numbers.

### Step 2: Set up the equation
Since \( z_1^2 = 15 - 8i \), we can expand \( z_1^2 \):
\[ z_1^2 = (x + yi)^2 = x^2 + 2xyi - y^2 \]
This gives us:
\[ z_1^2 = (x^2 - y^2) + (2xy)i \]

### Step 3: Equate real and imaginary parts
From the equation \( z_1^2 = 15 - 8i \), we can equate the real and imaginary parts:
1. \( x^2 - y^2 = 15 \) (1)
2. \( 2xy = -8 \) (2)

### Step 4: Solve for \( y \)
From equation (2), we can express \( y \) in terms of \( x \):
\[ y = \frac{-8}{2x} = \frac{-4}{x} \]

### Step 5: Substitute \( y \) into equation (1)
Substituting \( y \) into equation (1):
\[ x^2 - \left(\frac{-4}{x}\right)^2 = 15 \]
This simplifies to:
\[ x^2 - \frac{16}{x^2} = 15 \]

### Step 6: Multiply through by \( x^2 \)
To eliminate the fraction, multiply through by \( x^2 \):
\[ x^4 - 16 = 15x^2 \]
Rearranging gives us:
\[ x^4 - 15x^2 - 16 = 0 \]

### Step 7: Let \( t = x^2 \)
Let \( t = x^2 \). Then the equation becomes:
\[ t^2 - 15t - 16 = 0 \]

### Step 8: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{15 \pm \sqrt{(-15)^2 - 4 \cdot 1 \cdot (-16)}}{2 \cdot 1} \]
\[ t = \frac{15 \pm \sqrt{225 + 64}}{2} \]
\[ t = \frac{15 \pm \sqrt{289}}{2} \]
\[ t = \frac{15 \pm 17}{2} \]
This gives us two solutions:
1. \( t = \frac{32}{2} = 16 \)
2. \( t = \frac{-2}{2} = -1 \) (not valid since \( t = x^2 \))

### Step 9: Find \( x \)
From \( t = 16 \):
\[ x^2 = 16 \implies x = 4 \text{ or } x = -4 \]

### Step 10: Find \( y \)
Using \( y = \frac{-4}{x} \):
1. If \( x = 4 \):
\[ y = \frac{-4}{4} = -1 \]
2. If \( x = -4 \):
\[ y = \frac{-4}{-4} = 1 \]

### Step 11: Write the square roots
Thus, the two square roots are:
1. \( z_1 = 4 - i \)
2. \( z_2 = -4 + i \)

### Final Answer
The square roots of \( 15 - 8i \) are \( 4 - i \) and \( -4 + i \).

---