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: \(\frac{2 \sqrt{6}}{5}\).
Let \( z \) satisfy \( |z| = 1, \ z = 1 - \overline{z} \text{ and } \operatorname{Im}(z)>0 \)
Then consider:
Statement-I: \( z \) is a real number
Statement-II: Principal argument of \( z \) is \( \dfrac{\pi}{3} \)
Then:
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: