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:

Updated On: Nov 27, 2024
  • $\frac{\sqrt{6}}{5}$
  • $\frac{1+\sqrt{6}}{5}$
  • $\frac{24}{5}$
  • $\frac{2\sqrt{6}}{5}$
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The Correct Option is D

Solution and Explanation

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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