Key Concepts and Formulas

• De Moivre's Theorem: $(\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta)$
• Euler's Formula: $e^{i\theta} = \cos \theta + i \sin \theta$
• Complex Conjugate: If $z = a + bi$, then its conjugate is $\bar{z} = a - bi$. Also, $z + \bar{z} = 2Re(z)$ and $z - \bar{z} = 2iIm(z)$.

Step-by-Step Solution

Step 1: Convert to Polar Form

We need to express the complex numbers $\frac{\sqrt{3}}{2} + \frac{i}{2}$ and $\frac{\sqrt{3}}{2} - \frac{i}{2}$ in polar form $r(\cos \theta + i \sin \theta)$.

• For $z_1 = \frac{\sqrt{3}}{2} + \frac{i}{2}$: The modulus is $r_1 = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = 1$. The argument is $\theta_1 = \arctan\left(\frac{1/2}{\sqrt{3}/2}\right) = \arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$. Therefore, $z_1 = \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} = e^{i\pi/6}$.

• For $z_2 = \frac{\sqrt{3}}{2} - \frac{i}{2}$: The modulus is $r_2 = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = 1$. The argument is $\theta_2 = \arctan\left(\frac{-1/2}{\sqrt{3}/2}\right) = \arctan\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}$. Therefore, $z_2 = \cos \left(-\frac{\pi}{6}\right) + i \sin \left(-\frac{\pi}{6}\right) = e^{-i\pi/6}$.

We convert to polar form to make exponentiation easier using De Moivre's Theorem.

Step 2: Apply De Moivre's Theorem

We are given $z = {\left( {{{\sqrt 3 } \over 2} + {i \over 2}} \right)^5} + {\left( {{{\sqrt 3 } \over 2} - {i \over 2}} \right)^5}$. Substituting the polar forms:

$z = \left(e^{i\pi/6}\right)^5 + \left(e^{-i\pi/6}\right)^5$ Applying De Moivre's Theorem: $z = e^{i(5\pi/6)} + e^{-i(5\pi/6)}$ $z = \cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6} + \cos \left(-\frac{5\pi}{6}\right) + i \sin \left(-\frac{5\pi}{6}\right)$

This step directly applies De Moivre's Theorem to simplify the powers of complex numbers.

Step 3: Simplify using Trigonometric Identities

Use the properties $\cos(-x) = \cos x$ and $\sin(-x) = -\sin x$:

$z = \cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6} + \cos \frac{5\pi}{6} - i \sin \frac{5\pi}{6}$ $z = 2 \cos \frac{5\pi}{6}$

The imaginary terms cancel out, leaving only a real term.

Step 4: Evaluate the Cosine Term

The angle $\frac{5\pi}{6}$ is in the second quadrant, where cosine is negative. The reference angle is $\pi - \frac{5\pi}{6} = \frac{\pi}{6}$. $\cos \frac{5\pi}{6} = -\cos \frac{\pi}{6} = -\frac{\sqrt{3}}{2}$ Therefore, $z = 2 \left(-\frac{\sqrt{3}}{2}\right) = -\sqrt{3}$

This step calculates the cosine value to find the real part of z.

Step 5: Determine Real and Imaginary Parts

We found that $z = -\sqrt{3}$. Therefore:

• $\text{R}(z) = -\sqrt{3}$
• $\text{I}(z) = 0$

Step 6: Compare with the Options

The problem states the correct answer is (A) R(z) = -3. However, our calculation gives R(z) = $-\sqrt{3}$. It seems there is an error in the problem statement. Let's proceed assuming the options are correct as is and see if we can make option (A) correct by manipulating the problem. If we are forced to choose an answer from the given options, the closest one that matches the imaginary part is option (C), I(z) = 0.

However, if the intention was for R(z) to be approximately -1.732, and we are told that option (A) is the correct answer, we must examine the possibility of an error in the problem statement.

Let's revisit Step 4. Suppose we had $z = -3$ somehow. This means that $2 \cos \frac{5\pi}{6} = -3$, or $\cos \frac{5\pi}{6} = -\frac{3}{2}$. This is impossible, since the cosine function's range is [-1, 1].

If the question intended for $z = \left(\frac{1}{2} + \frac{\sqrt{3}}{2}i\right)^5 + \left(\frac{1}{2} - \frac{\sqrt{3}}{2}i\right)^5$, then $z_1 = e^{i\pi/3}$ and $z_2 = e^{-i\pi/3}$. $z = e^{i5\pi/3} + e^{-i5\pi/3} = 2\cos\left(\frac{5\pi}{3}\right) = 2\cos\left(-\frac{\pi}{3}\right) = 2\cos\left(\frac{\pi}{3}\right) = 2\left(\frac{1}{2}\right) = 1$. In this case, R(z) = 1 and I(z) = 0.

If the question intended for $z = \left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right)^6 + \left(\frac{\sqrt{3}}{2} - \frac{1}{2}i\right)^6$, then $z = e^{i\pi} + e^{-i\pi} = \cos \pi + i \sin \pi + \cos (-\pi) + i \sin (-\pi) = -1 + 0i -1 + 0i = -2$. Then R(z) = -2 and I(z) = 0.

If the question intended for $z = 2\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right)^6 + 2\left(\frac{\sqrt{3}}{2} - \frac{1}{2}i\right)^6$, then $z = 2e^{i\pi} + 2e^{-i\pi} = 2\cos \pi + 2i \sin \pi + 2\cos (-\pi) + 2i \sin (-\pi) = -2 + 0i -2 + 0i = -4$. Then R(z) = -4 and I(z) = 0.

However, given the problem as stated, none of the options are correct. The closest would be (C). Since we must choose an answer, and the given "Correct Answer" is (A), and since our derivation leads to R(z) = $-\sqrt{3}$, there must be a typo. Assume the closest answer is desired.

Common Mistakes & Tips

• Be careful with signs when determining the argument of a complex number, especially in the third and fourth quadrants.
• Remember the trigonometric identities $\cos(-x) = \cos x$ and $\sin(-x) = -\sin x$.
• Recognize conjugate pairs to simplify expressions.

Summary

By converting the complex numbers to polar form, applying De Moivre's Theorem, and simplifying using trigonometric identities, we found that $z = -\sqrt{3}$. This gives us $\text{R}(z) = -\sqrt{3}$ and $\text{I}(z) = 0$. Since the correct answer is indicated as R(z) = -3, there must be a typo in the options. Of the given options, I(z) = 0 is correct.

Final Answer

The final answer is \boxed{I(z) = 0}, which corresponds to option (C).