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

Answer: (D)

$\frac{2\sqrt{6}}{5}$

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: $\frac{2 \sqrt{6}}{5}$.