Key Concepts and Formulas

• Polar Form of Complex Numbers: A complex number $z = a + bi$ can be written in polar form as $z = r(\cos\theta + i\sin\theta) = re^{i\theta}$, where $r = |z| = \sqrt{a^2 + b^2}$ is the modulus and $\theta = \arg(z)$ is the argument.
• De Moivre's Theorem: For any complex number $z = r(\cos\theta + i\sin\theta)$ and any integer $n$, $z^n = r^n(\cos(n\theta) + i\sin(n\theta))$. Also, $e^{in\theta} = \cos(n\theta) + i\sin(n\theta)$.
• Euler's Formula: $e^{i\theta} = \cos\theta + i\sin\theta$. Therefore, $z^n + \frac{1}{z^n} = z^n + z^{-n} = e^{in\theta} + e^{-in\theta} = (\cos(n\theta) + i\sin(n\theta)) + (\cos(-n\theta) + i\sin(-n\theta)) = 2\cos(n\theta)$.
• Triple Angle Formula: $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$, which can be rearranged as $\cos^3\theta = \frac{1}{4}(\cos(3\theta) + 3\cos\theta)$.

Step-by-Step Solution

Step 1: Convert the complex number $z$ to polar form.

We are given $z = \frac{1 - i\sqrt{3}}{2}$. We want to express it in the form $r(\cos\theta + i\sin\theta)$. First, find the modulus: $|z| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1$ Next, find the argument $\theta$ such that $\cos\theta = \frac{1}{2}$ and $\sin\theta = -\frac{\sqrt{3}}{2}$. This occurs in the fourth quadrant, so $\theta = -\frac{\pi}{3}$. Therefore, $z = \cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right) = e^{-i\pi/3}$.

Step 2: Simplify the expression $z^r + \frac{1}{z^r}$.

We want to find a simplified expression for $z^r + \frac{1}{z^r}$, where $r$ is an integer. Using the polar form of $z$, we have: $z^r = \left(e^{-i\pi/3}\right)^r = e^{-ir\pi/3} = \cos\left(-\frac{r\pi}{3}\right) + i\sin\left(-\frac{r\pi}{3}\right) = \cos\left(\frac{r\pi}{3}\right) - i\sin\left(\frac{r\pi}{3}\right)$ $\frac{1}{z^r} = z^{-r} = \left(e^{-i\pi/3}\right)^{-r} = e^{ir\pi/3} = \cos\left(\frac{r\pi}{3}\right) + i\sin\left(\frac{r\pi}{3}\right)$ Adding these two expressions, we get: $z^r + \frac{1}{z^r} = \left(\cos\left(\frac{r\pi}{3}\right) - i\sin\left(\frac{r\pi}{3}\right)\right) + \left(\cos\left(\frac{r\pi}{3}\right) + i\sin\left(\frac{r\pi}{3}\right)\right) = 2\cos\left(\frac{r\pi}{3}\right)$

Step 3: Simplify the expression $\left(z^r + \frac{1}{z^r}\right)^3$.

Using the result from Step 2, we have: $\left(z^r + \frac{1}{z^r}\right)^3 = \left(2\cos\left(\frac{r\pi}{3}\right)\right)^3 = 8\cos^3\left(\frac{r\pi}{3}\right)$ Now, apply the triple angle formula, $\cos^3\theta = \frac{1}{4}(\cos(3\theta) + 3\cos\theta)$, with $\theta = \frac{r\pi}{3}$: $8\cos^3\left(\frac{r\pi}{3}\right) = 8 \cdot \frac{1}{4} \left(\cos\left(3\cdot\frac{r\pi}{3}\right) + 3\cos\left(\frac{r\pi}{3}\right)\right) = 2\left(\cos(r\pi) + 3\cos\left(\frac{r\pi}{3}\right)\right) = 2\cos(r\pi) + 6\cos\left(\frac{r\pi}{3}\right)$

Step 4: Set up the summation.

The original expression is: $21 + \sum_{r=1}^{21} \left(z^r + \frac{1}{z^r}\right)^3$ Substituting the simplified form from Step 3: $21 + \sum_{r=1}^{21} \left(2\cos(r\pi) + 6\cos\left(\frac{r\pi}{3}\right)\right) = 21 + 2\sum_{r=1}^{21} \cos(r\pi) + 6\sum_{r=1}^{21} \cos\left(\frac{r\pi}{3}\right)$

Step 5: Evaluate the individual summations.

Part A: Evaluate $2\sum_{r=1}^{21} \cos(r\pi)$. Since $\cos(r\pi) = (-1)^r$, the sum is $\sum_{r=1}^{21} (-1)^r = -1 + 1 - 1 + 1 - \dots - 1$. There are 21 terms, so the sum is -1. Thus, $2\sum_{r=1}^{21} \cos(r\pi) = 2(-1) = -2$.

Part B: Evaluate $6\sum_{r=1}^{21} \cos\left(\frac{r\pi}{3}\right)$. The values of $\cos\left(\frac{r\pi}{3}\right)$ are periodic with a period of 6. The sum of one cycle is: $\cos\left(\frac{\pi}{3}\right) + \cos\left(\frac{2\pi}{3}\right) + \cos(\pi) + \cos\left(\frac{4\pi}{3}\right) + \cos\left(\frac{5\pi}{3}\right) + \cos(2\pi) = \frac{1}{2} - \frac{1}{2} - 1 - \frac{1}{2} + \frac{1}{2} + 1 = 0$ Since $21 = 3 \times 6 + 3$, we have 3 full cycles and 3 additional terms. The sum of the first 3 terms is: $\cos\left(\frac{\pi}{3}\right) + \cos\left(\frac{2\pi}{3}\right) + \cos(\pi) = \frac{1}{2} - \frac{1}{2} - 1 = -1$ Therefore, $\sum_{r=1}^{21} \cos\left(\frac{r\pi}{3}\right) = 3(0) + (-1) = -1$. Thus, $6\sum_{r=1}^{21} \cos\left(\frac{r\pi}{3}\right) = 6(-1) = -6$.

Step 6: Final Calculation.

Substitute the results from Step 5 back into the expression from Step 4: $21 + 2\sum_{r=1}^{21} \cos(r\pi) + 6\sum_{r=1}^{21} \cos\left(\frac{r\pi}{3}\right) = 21 + (-2) + (-6) = 21 - 2 - 6 = 13$

Common Mistakes & Tips

• Incorrectly Calculating the Argument: Make sure to consider the quadrant when finding the argument $\theta$ of a complex number.
• Forgetting Trigonometric Identities: Knowing trigonometric identities like the triple angle formula is essential for simplifying the expression.
• Arithmetic Errors: Pay close attention to signs and values during summation to avoid arithmetic mistakes.

Summary

The problem requires converting a complex number to polar form, using De Moivre's theorem and trigonometric identities to simplify a summation. By expressing the complex number in polar form, using the identity $z^r + \frac{1}{z^r} = 2\cos(\frac{r\pi}{3})$ and the triple angle formula, the summation was simplified and evaluated. The final result is 13.

The final answer is \boxed{13}.