The question asks to solve the equation

$(z + 2 + i)^* + (2 + i)z = 0$

where $*$ represents the complex conjugate, and the answer should be in the form $x + iy$.

Step 1: Understanding the terms

1. Let $z = x + iy$, where $x$ and $y$ are real numbers.
2. The conjugate $z^* = x - iy$.
3. Rewrite the expression $(z + 2 + i)^*$: $(z + 2 + i)^* = (x + iy + 2 + i)^* = x + 2 - i(y + 1).$

Step 2: Substituting into the equation

Substitute into the given equation: $(z + 2 + i)^* + (2 + i)z = 0.$ Replace the terms: $(x + 2 - i(y + 1)) + (2 + i)(x + iy) = 0.$

Simplify $(2 + i)(x + iy)$: $(2 + i)(x + iy) = 2x + 2iy + ix - y.$

Combine real and imaginary terms: $(2 + i)(x + iy) = (2x - y) + i(2y + x).$

Thus, the equation becomes: $(x + 2 - i(y + 1)) + (2x - y + i(2y + x)) = 0.$

Step 3: Separating real and imaginary parts

Combine the real and imaginary parts of the equation:

• Real part: $x + 2 + 2x - y$,
• Imaginary part: $-(y + 1) + (2y + x)$.

This gives two separate equations:

1. Real part: x + 2 + 2x - y = 0 \quad \Rightarrow \quad 3x - y + 2 = 0. \tag{1}
2. Imaginary part: -(y + 1) + (2y + x) = 0 \quad \Rightarrow \quad x + y - 1 = 0. \tag{2}

Step 4: Solving the system of equations

From equation $(2)$: $x + y = 1 \quad \Rightarrow \quad y = 1 - x.$

Substitute $y = 1 - x$ into equation $(1)$: $3x - (1 - x) + 2 = 0.$

Simplify: $3x - 1 + x + 2 = 0 \quad \Rightarrow \quad 4x + 1 = 0.$

Solve for $x$: $x = -\frac{1}{4}.$

Substitute $x = -\frac{1}{4}$ into $y = 1 - x$: $y = 1 - \left(-\frac{1}{4}\right) = 1 + \frac{1}{4} = \frac{5}{4}.$

Step 5: Final solution

The solution is: $z = x + iy = -\frac{1}{4} + i\frac{5}{4}.$

Final Answer:

$z = -\frac{1}{4} + \frac{5}{4}i.$

Do you want further clarification on any part of this solution? Let me know!

Related Questions:

1. What is a complex conjugate, and how is it used in solving equations?
2. How do you separate real and imaginary parts of a complex equation?
3. Solve $z + z^* = 4$, where $z$ is a complex number.
4. Solve $(z - 3 + 2i)^* + z = 0$ for $z$.
5. How do you represent a complex number in polar form?

Tip: Always isolate real and imaginary components when solving complex number equations.