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:
- $\frac{\sqrt{6}}{5}$
- $\frac{1+\sqrt{6}}{5}$
- $\frac{24}{5}$
- $\frac{2\sqrt{6}}{5}$
The Correct Option is D
Approach Solution - 1
To solve this problem, let's first understand the given conditions about the complex number \( z \).
- 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.
- 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 \(\frac{2\sqrt{6}}{5}\).
Approach Solution -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: \(\frac{2 \sqrt{6}}{5}\).
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