Let z be a complex number then `|z + 2| = 1` and imaginary part of `(z + 1)/(z + 2) = 1/6` then find the value real part of z

Options

A. `-2 pm (sqrt (35))/6`
B. `-2 pm (sqrt (37))/6`
C. `-3 pm (sqrt (39))/6`
D. `3 pm (sqrt (33))/6`

Answer

To solve the problem, we need to find the real part of the complex number \( z \) given the conditions \( |z + 2| = 1 \) and the imaginary part of \( \frac{z + 1}{z + 2} = \frac{1}{6} \).

Let's denote \( z = x + iy \), where \( x \) and \( y \) are the real and imaginary parts of \( z \), respectively.

### Step 1: Analyze the first condition

The first condition is given by:

\[ |z + 2| = 1 \]

This can be rewritten as:

\[ |x + 2 + iy| = 1 \]

Using the formula for the modulus of a complex number, we have:

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

Squaring both sides gives:

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

This is our first equation.

### Step 2: Analyze the second condition

The second condition is:

\[ \text{Im}\left(\frac{z + 1}{z + 2}\right) = \frac{1}{6} \]

Substituting \( z = x + iy \):

\[ \frac{(x + 1) + iy}{(x + 2) + iy} \]

To find the imaginary part, we multiply the numerator and denominator by the conjugate of the denominator:

\[ \frac{((x + 1) + iy)((x + 2) - iy)}{((x + 2) + iy)((x + 2) - iy)} \]

Calculating the denominator:

\[ (x + 2)^2 + y^2 \]

Calculating the numerator:

\[ (x + 1)(x + 2) - y^2 + i(y(x + 2) - y(x + 1)) \]

This simplifies to:

\[ (x + 1)(x + 2) - y^2 + i(y(x + 2 - x - 1)) = (x + 1)(x + 2) - y^2 + i(y) \]

Thus, the imaginary part is:

\[ \frac{y}{(x + 2)^2 + y^2} \]

Setting this equal to \( \frac{1}{6} \):

\[ \frac{y}{(x + 2)^2 + y^2} = \frac{1}{6} \]

Cross-multiplying gives:

\[ 6y = (x + 2)^2 + y^2 \]

### Step 3: Solve the equations

Now we have two equations:

1. \( (x + 2)^2 + y^2 = 1 \) (Equation 1)
2. \( 6y = (x + 2)^2 + y^2 \) (Equation 2)

From Equation 1, we can express \( y^2 \):

\[ y^2 = 1 - (x + 2)^2 \]

Substituting this into Equation 2:

\[ 6y = (x + 2)^2 + (1 - (x + 2)^2) \]

This simplifies to:

\[ 6y = 1 \]

Thus, we find:

\[ y = \frac{1}{6} \]

### Step 4: Substitute \( y \) back to find \( x \)

Substituting \( y = \frac{1}{6} \) back into Equation 1:

\[ (x + 2)^2 + \left(\frac{1}{6}\right)^2 = 1 \]

Calculating \( \left(\frac{1}{6}\right)^2 = \frac{1}{36} \):

\[ (x + 2)^2 + \frac{1}{36} = 1 \]

Subtracting \( \frac{1}{36} \) from both sides:

\[ (x + 2)^2 = 1 - \frac{1}{36} = \frac{36}{36} - \frac{1}{36} = \frac{35}{36} \]

Taking the square root:

\[ x + 2 = \pm \sqrt{\frac{35}{36}} = \pm \frac{\sqrt{35}}{6} \]

Thus:

\[ x = -2 \pm \frac{\sqrt{35}}{6} \]

### Final Answer

The real part of \( z \) can take two values:

\[ x = -2 + \frac{\sqrt{35}}{6} \quad \text{or} \quad x = -2 - \frac{\sqrt{35}}{6} \]