Question

Let $z$ be a complex number such that $|z + 2| = 1$ and $\text{Im}\left(\frac{z+1}{z+2}\right) = \frac{1}{5}$. Then the value of $|\text{Re}(z+2)|$ is:

• A. $\frac{\sqrt{6}}{5}$
• B. $\frac{1+\sqrt{6}}{5}$
• C. $\frac{24}{5}$
• D. $\frac{2\sqrt{6}}{5}$

Answer 1

The Correct Option is D

To solve this problem, let's first understand the given conditions about the complex number \( z \).

1. It is given that \(|z + 2| = 1\). This indicates that the point \((z+2)\) lies on a circle centered at \((-2, 0)\) with a radius of 1 in the complex plane.
2. We are also given that \(\text{Im}\left(\frac{z+1}{z+2}\right) = \frac{1}{5}\). This focuses on the imaginary part of the complex fraction.

Let's set \( z = x + yi \), where \( x, y \) are real numbers. The equation \(|z+2| = 1\) can be rewritten as:

\(|(x+2) + yi| = 1\)

Squaring both sides, we have:

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

Next, for the condition \(\text{Im}\left(\frac{z+1}{z+2}\right) = \frac{1}{5}\):

Using the expression for division of complex numbers, we have:

\(\frac{z+1}{z+2} = \frac{(x+1) + yi}{(x+2) + yi}\)

Multiply numerator and denominator by the conjugate of the denominator:

\(\frac{[(x+1) + yi]\cdot[(x+2) - yi]}{((x+2)^2 + y^2)}\)

This will resolve to:

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

The imaginary part here simplifies to:

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

Setting this equal to \(\frac{1}{5}\), we get:

\(\frac{y}{(x+2)^2 + y^2} = \frac{1}{5}\)

Therefore, \(5y = (x+2)^2 + y^2\). Using \((x+2)^2 + y^2 = 1\):

Substituting the constraint, we get:

\(5y = 1\)

Thus, the solution is \(y = \frac{1}{5}\).

Remember, we need \(|\mathrm{Re}(z+2)|\), which simplifies to \(|x+2|\). From \( (x+2)^2 + y^2 = 1 \) and \( y = \frac{1}{5} \):

\((x+2)^2 + \left(\frac{1}{5}\right)^2 = 1\)

Substitute \( \left(\frac{1}{5}\right)^2 = \frac{1}{25} \) leading to:

\((x+2)^2 = 1 - \frac{1}{25} = \frac{24}{25}\)

Therefore, \(|x+2| = \sqrt{\frac{24}{25}} = \frac{2\sqrt{6}}{5}\).

Thus, the value of \(|\mathrm{Re}(z+2)|\) is \(\frac{2\sqrt{6}}{5}\). The correct answer is therefore \(\frac{2\sqrt{6}}{5}\).

Answer 2

Let:

\[ z + 2 = \cos \theta + i \sin \theta \implies \frac{1}{z + 2} = \cos \theta - i \sin \theta. \]

Now:

\[ \frac{z + 1}{z + 2} = 1 - \frac{1}{z + 2} = 1 - (\cos \theta - i \sin \theta). \]

Simplify:

\[ \frac{z + 1}{z + 2} = (1 - \cos \theta) + i \sin \theta. \]

The imaginary part is:

\(\ Im \left( \frac{z + 1}{z + 2} \right) = \sin \theta = \frac{1}{5}.\)

Using \(\sin^2 \theta + \cos^2 \theta = 1\):

\[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \left( \frac{1}{5} \right)^2 = 1 - \frac{1}{25} = \frac{24}{25}. \]

\[ \cos \theta = \pm \sqrt{\frac{24}{25}} = \pm \frac{2 \sqrt{6}}{5}. \]

Now, the real part of \(z + 2\) is:

\(\ Re(z + 2) = \cos \theta.\)

The magnitude of \(\Re(z + 2)\) is:

\[ |\ Re(z + 2)| = \frac{2 \sqrt{6}}{5}. \]

Final Answer: \(\frac{2 \sqrt{6}}{5}\).