Find the value of `sqrt(3-4i)`.

Answer

To find the value of \( \sqrt{3 - 4i} \), we will express it in the form \( z = a + bi \), where \( a \) and \( b \) are real numbers.

### Step-by-Step Solution:

1. **Assume the form of the square root**:
Let \( z = a + bi \), where \( a \) and \( b \) are real numbers. Therefore, we have:
\[ z^2 = 3 - 4i \]

2. **Square both sides**:
Squaring \( z \):
\[ (a + bi)^2 = a^2 + 2abi + (bi)^2 = a^2 + 2abi - b^2 \]
This can be rewritten as:
\[ (a^2 - b^2) + (2ab)i = 3 - 4i \]

3. **Equate real and imaginary parts**:
From the equation above, we can equate the real and imaginary parts:
\[ a^2 - b^2 = 3 \quad \text{(1)} \]
\[ 2ab = -4 \quad \text{(2)} \]

4. **Solve for \( b \)**:
From equation (2), we can express \( b \) in terms of \( a \):
\[ b = \frac{-4}{2a} = \frac{-2}{a} \]

5. **Substitute \( b \) into equation (1)**:
Substitute \( b = \frac{-2}{a} \) into equation (1):
\[ a^2 - \left(\frac{-2}{a}\right)^2 = 3 \]
Simplifying this gives:
\[ a^2 - \frac{4}{a^2} = 3 \]

6. **Multiply through by \( a^2 \)**:
To eliminate the fraction, multiply through by \( a^2 \):
\[ a^4 - 4 = 3a^2 \]
Rearranging gives:
\[ a^4 - 3a^2 - 4 = 0 \]

7. **Let \( x = a^2 \)**:
Substitute \( x = a^2 \):
\[ x^2 - 3x - 4 = 0 \]

8. **Factor the quadratic**:
Factoring the quadratic:
\[ (x - 4)(x + 1) = 0 \]
Thus, \( x = 4 \) or \( x = -1 \). Since \( x = a^2 \) must be non-negative, we have:
\[ a^2 = 4 \implies a = \pm 2 \]

9. **Find corresponding \( b \)**:
Using \( a = 2 \):
\[ b = \frac{-2}{2} = -1 \]
Using \( a = -2 \):
\[ b = \frac{-2}{-2} = 1 \]

10. **Write the solutions for \( z \)**:
Therefore, the two possible values for \( z \) are:
\[ z = 2 - i \quad \text{and} \quad z = -2 + i \]

11. **Final Result**:
Thus, the value of \( \sqrt{3 - 4i} \) is:
\[ \sqrt{3 - 4i} = \pm (2 - i) \]