If z is a complex number satisfying the relation `|z+ 1|=z+2(1+i)`, then z is

Options

A. `1/2 (1 + 4i)`
B. `1/2(3 + 4 i)`
C. `1/2(1-4i)`
D. `1/2(3-4i)`

Answer

To solve the problem, we need to find the complex number \( z \) that satisfies the equation:

\[ |z + 1| = z + 2(1 + i) \]

### Step 1: Represent \( z \) in terms of real and imaginary parts
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers.

### Step 2: Substitute \( z \) into the equation
Substituting \( z \) into the equation gives:

\[ |x + iy + 1| = (x + iy) + 2(1 + i) \]

This simplifies to:

\[ |x + 1 + iy| = (x + 2) + (y + 2)i \]

### Step 3: Calculate the modulus of the left side
The modulus of the left side can be calculated as:

\[ |x + 1 + iy| = \sqrt{(x + 1)^2 + y^2} \]

### Step 4: Write the equation with the modulus
Now we have:

\[ \sqrt{(x + 1)^2 + y^2} = (x + 2) + (y + 2)i \]

### Step 5: Separate real and imaginary parts
For the equation to hold, the imaginary part must be zero. Therefore, we set:

\[ y + 2 = 0 \implies y = -2 \]

### Step 6: Substitute \( y \) back into the equation
Now substitute \( y = -2 \) back into the modulus equation:

\[ \sqrt{(x + 1)^2 + (-2)^2} = x + 2 \]

This simplifies to:

\[ \sqrt{(x + 1)^2 + 4} = x + 2 \]

### Step 7: Square both sides to eliminate the square root
Squaring both sides gives:

\[ (x + 1)^2 + 4 = (x + 2)^2 \]

### Step 8: Expand both sides
Expanding both sides results in:

\[ x^2 + 2x + 1 + 4 = x^2 + 4x + 4 \]

This simplifies to:

\[ x^2 + 2x + 5 = x^2 + 4x + 4 \]

### Step 9: Rearrange the equation
Subtract \( x^2 \) from both sides:

\[ 2x + 5 = 4x + 4 \]

Rearranging gives:

\[ 5 - 4 = 4x - 2x \implies 1 = 2x \implies x = \frac{1}{2} \]

### Step 10: Write the final answer
Now we have \( x = \frac{1}{2} \) and \( y = -2 \). Therefore, the complex number \( z \) is:

\[ z = \frac{1}{2} - 2i \]

### Final Answer
Thus, the complex number \( z \) is:

\[ \boxed{\frac{1}{2} - 2i} \]