Key Concepts and Formulas

• Purely Imaginary Number: A complex number $z$ is purely imaginary if $z + \bar{z} = 0$.
• Purely Real Number: A complex number $z$ is purely real if $z - \bar{z} = 0$.
• Trigonometric Identities: $\sin^2 \theta + \cos^2 \theta = 1$, $\sin(2\theta) = 2\sin\theta\cos\theta$, $\cos(2\theta) = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$.

Step 1: Determine $\alpha$ from the purely imaginary condition

We are given that $z_1 = \frac{1-i \sin \alpha}{1+2 i \sin \alpha}$ is purely imaginary. This means $z_1 + \bar{z_1} = 0$. We first find the conjugate of $z_1$: $\bar{z_1} = \frac{1+i \sin \alpha}{1-2 i \sin \alpha}$ Now, we set $z_1 + \bar{z_1} = 0$: $\frac{1-i \sin \alpha}{1+2 i \sin \alpha} + \frac{1+i \sin \alpha}{1-2 i \sin \alpha} = 0$ Multiplying both sides by $(1+2i\sin\alpha)(1-2i\sin\alpha) = 1 + 4\sin^2\alpha$, we get $(1-i \sin \alpha)(1-2 i \sin \alpha) + (1+i \sin \alpha)(1+2 i \sin \alpha) = 0$ Expanding, we have $1 - 3i\sin\alpha - 2\sin^2\alpha + 1 + 3i\sin\alpha - 2\sin^2\alpha = 0$ $2 - 4\sin^2\alpha = 0$ $\sin^2\alpha = \frac{1}{2}$ $\sin\alpha = \pm\frac{1}{\sqrt{2}}$ Since $\pi < \alpha < 2\pi$, $\alpha$ is in the third or fourth quadrant. Thus, $\sin\alpha$ must be negative. Therefore, $\sin\alpha = -\frac{1}{\sqrt{2}}$. The solutions in the given range are $\alpha = \frac{5\pi}{4}$ and $\alpha = \frac{7\pi}{4}$.

Step 2: Determine $\beta$ from the purely real condition

We are given that $z_2 = \frac{1+i \cos \beta}{1-2 i \cos \beta}$ is purely real. This means $z_2 - \bar{z_2} = 0$. We first find the conjugate of $z_2$: $\bar{z_2} = \frac{1-i \cos \beta}{1+2 i \cos \beta}$ Now, we set $z_2 - \bar{z_2} = 0$: $\frac{1+i \cos \beta}{1-2 i \cos \beta} - \frac{1-i \cos \beta}{1+2 i \cos \beta} = 0$ Multiplying both sides by $(1-2i\cos\beta)(1+2i\cos\beta) = 1 + 4\cos^2\beta$, we get $(1+i \cos \beta)(1+2 i \cos \beta) - (1-i \cos \beta)(1-2 i \cos \beta) = 0$ Expanding, we have $1 + 3i\cos\beta - 2\cos^2\beta - (1 - 3i\cos\beta - 2\cos^2\beta) = 0$ $6i\cos\beta = 0$ This implies $\cos\beta = 0$. Since $\pi < \beta < 2\pi$, the only solution is $\beta = \frac{3\pi}{2}$.

Step 3: Determine the Set $S$

From Step 1, the possible values for $\alpha$ are $\frac{5\pi}{4}$ and $\frac{7\pi}{4}$. From Step 2, the only possible value for $\beta$ is $\frac{3\pi}{2}$. Therefore, the set $S$ of all pairs $(\alpha, \beta)$ is: $S = \left\{ \left(\frac{5\pi}{4}, \frac{3\pi}{2}\right), \left(\frac{7\pi}{4}, \frac{3\pi}{2}\right) \right\}$

Step 4: Calculate $Z_{\alpha \beta}$ for each pair in $S$

We are given $Z_{\alpha \beta}=\sin 2 \alpha+i \cos 2 \beta$.

For $(\alpha, \beta) = \left(\frac{5\pi}{4}, \frac{3\pi}{2}\right)$: $Z_1 = \sin\left(2 \cdot \frac{5\pi}{4}\right) + i \cos\left(2 \cdot \frac{3\pi}{2}\right) = \sin\left(\frac{5\pi}{2}\right) + i \cos(3\pi) = \sin\left(\frac{\pi}{2}\right) + i \cos(\pi) = 1 - i$

For $(\alpha, \beta) = \left(\frac{7\pi}{4}, \frac{3\pi}{2}\right)$: $Z_2 = \sin\left(2 \cdot \frac{7\pi}{4}\right) + i \cos\left(2 \cdot \frac{3\pi}{2}\right) = \sin\left(\frac{7\pi}{2}\right) + i \cos(3\pi) = \sin\left(\frac{3\pi}{2}\right) + i \cos(\pi) = -1 - i$

Step 5: Evaluate the Sum $\sum\limits_{(\alpha, \beta) \in S}\left(i Z_{\alpha \beta}+\frac{1}{i \bar{Z}_{\alpha \beta}}\right)$

We need to calculate $\left(i Z_1 + \frac{1}{i \bar{Z_1}}\right) + \left(i Z_2 + \frac{1}{i \bar{Z_2}}\right)$, where $Z_1 = 1-i$ and $Z_2 = -1-i$.

For $Z_1 = 1-i$: $i Z_1 = i(1-i) = i - i^2 = i + 1 = 1 + i$ $\bar{Z_1} = 1+i$ $\frac{1}{i \bar{Z_1}} = \frac{1}{i(1+i)} = \frac{1}{i - 1} = \frac{1}{-1+i} = \frac{-1-i}{(-1+i)(-1-i)} = \frac{-1-i}{1+1} = \frac{-1-i}{2} = -\frac{1}{2} - \frac{1}{2}i$ $i Z_1 + \frac{1}{i \bar{Z_1}} = (1+i) + \left(-\frac{1}{2} - \frac{1}{2}i\right) = \frac{1}{2} + \frac{1}{2}i$

For $Z_2 = -1-i$: $i Z_2 = i(-1-i) = -i - i^2 = -i + 1 = 1 - i$ $\bar{Z_2} = -1+i$ $\frac{1}{i \bar{Z_2}} = \frac{1}{i(-1+i)} = \frac{1}{-i - 1} = \frac{1}{-1-i} = \frac{-1+i}{(-1-i)(-1+i)} = \frac{-1+i}{1+1} = \frac{-1+i}{2} = -\frac{1}{2} + \frac{1}{2}i$ $i Z_2 + \frac{1}{i \bar{Z_2}} = (1-i) + \left(-\frac{1}{2} + \frac{1}{2}i\right) = \frac{1}{2} - \frac{1}{2}i$

The sum is: $\left(\frac{1}{2} + \frac{1}{2}i\right) + \left(\frac{1}{2} - \frac{1}{2}i\right) = 1$

Common Mistakes & Tips

• Remember to check the range of $\alpha$ and $\beta$ when solving trigonometric equations.
• Be careful with complex conjugates and $i^2 = -1$.
• Rationalize denominators to simplify complex fractions.

Summary

We found the values of $\alpha$ and $\beta$ that satisfy the given conditions, determined the set $S$, calculated $Z_{\alpha\beta}$ for each pair in $S$, and finally evaluated the given sum. The final result is 1.

Final Answer

The final answer is \boxed{1}, which corresponds to option (C).