Solve z

z^2+ 2z̅+ 2 = 0 for z, where z ∈ C.

To solve \( z^2 + 2\overline{z} + 2 = 0 \) where \( z \in \mathbb{C} \), we proceed with the following steps:

1. **Express \( z \) and \( \overline{z} \):**
   Let \( z = x + yi \) where \( x, y \in \mathbb{R} \). Then, \( \overline{z} = x - yi \).

2. **Substitute \( z \) and \( \overline{z} \) into the equation:**
   \[
   (x + yi)^2 + 2(x - yi) + 2 = 0
   \]

3. **Expand and simplify:**
   \[
   (x + yi)^2 = x^2 + 2xyi - y^2
   \]
   So the equation becomes:
   \[
   x^2 + 2xyi - y^2 + 2x - 2yi + 2 = 0
   \]

4. **Separate the real and imaginary parts:**
   \[
   (x^2 - y^2 + 2x + 2) + (2xy - 2y)i = 0
   \]

   This gives us two equations:
   \[
   x^2 - y^2 + 2x + 2 = 0 \quad \text{(real part)}
   \]
   \[
   2xy - 2y = 0 \quad \text{(imaginary part)}
   \]

5. **Solve the imaginary part equation first:**
   \[
   2y(x - 1) = 0
   \]

   This implies:
   \[
   y = 0 \quad \text{or} \quad x = 1
   \]

6. **Consider each case:**

   - **Case 1: \( y = 0 \)**
     \[
     x^2 + 2x + 2 = 0
     \]
     Solve the quadratic equation:
     \[
     x = \frac{-2 \pm \sqrt{4 - 8}}{2} = \frac{-2 \pm \sqrt{-4}}{2} = \frac{-2 \pm 2i}{2} = -1 \pm i
     \]
     So, \( z = -1 + i \) and \( z = -1 - i \).

   - **Case 2: \( x = 1 \)**
     \[
     1 - y^2 + 2 + 2 = 0
     \]
     \[
     3 - y^2 = 0
     \]
     \[
     y^2 = 3
     \]
     \[
     y = \pm \sqrt{3}
     \]
     So, \( z = 1 + \sqrt{3}i \) and \( z = 1 - \sqrt{3}i \).

7. **Combine the solutions:**
   \[
   z = -1 + i, \quad z = -1 - i, \quad z = 1 + \sqrt{3}i, \quad z = 1 - \sqrt{3}i
   \]

Thus, the solutions to \( z^2 + 2\overline{z} + 2 = 0 \) are:
\[
z = -1 + i, \quad z = -1 - i, \quad z = 1 + \sqrt{3}i, \quad z = 1 - \sqrt{3}i
\]

Would you like more details or have any questions?

Here are 5 additional questions you might be interested in:
1. How do you solve higher-degree polynomial equations in complex numbers?
2. What is the significance of the roots of complex equations in the complex plane?
3. Can you explain the concept of the complex conjugate root theorem?
4. How do you determine the nature of the roots of a complex quadratic equation?
5. What are some applications of complex numbers in engineering and physics?

**Tip:** When solving complex equations, always consider both the real and imaginary parts separately. This often simplifies the problem into manageable real-number equations.

Learn how to solve the complex quadratic equation z^2 + 2z̅ + 2 = 0 step-by-step. This problem involves complex numbers and the quadratic equation formula, exploring the complex conjugate root theorem. Suitable for advanced high school students.

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