Question

Let \( z = \frac{1 - i\sqrt{3}}{2}, \, i = \sqrt{-1} \). Then the value of

\[ 21 + \left( z + \frac{1}{z} \right)^3 + \left( z^2 + \frac{1}{z^2} \right)^3 + \left( z^3 + \frac{1}{z^3} \right)^3 + \dots + \left( z^{21} + \frac{1}{z^{21}} \right)^3 \] is _________.

Hint:

The sum of trigonometric functions over a full period is always zero. This property simplifies many complex number series.

Answer 1

Correct Answer: 13

Step 1: Understanding the Concept:
We use the exponential form of complex numbers. $z = e^{-i\pi/3}$. Then $z^k + \frac{1}{z^k} = 2 \cos\left(\frac{k\pi}{3}\right)$. We evaluate the sum using trigonometric identities and periodic properties.
Step 2: Detailed Explanation:
$z^k + z^{-k} = 2 \cos(k\pi/3)$. We need $S = \sum_{k=1}^{21} (2 \cos(k\pi/3))^3 = 8 \sum_{k=1}^{21} \cos^3(k\pi/3)$.
Using $\cos^3 \theta = \frac{1}{4} (3 \cos \theta + \cos 3\theta)$:
$S = 8 \cdot \frac{1}{4} \sum_{k=1}^{21} (3 \cos(k\pi/3) + \cos(k\pi)) = 2 \left[ 3 \sum_{k=1}^{21} \cos(k\pi/3) + \sum_{k=1}^{21} (-1)^k \right]$.
- $\sum_{k=1}^{21} \cos(k\pi/3)$: The period is $6$. Sum over one period is $0$. $21 = 3 \times 6 + 3$.
The sum is $\sum_{k=1}^3 \cos(k\pi/3) = \cos(60^\circ) + \cos(120^\circ) + \cos(180^\circ) = \frac{1}{2} - \frac{1}{2} - 1 = -1$.
- $\sum_{k=1}^{21} (-1)^k = -1$ (since 21 is odd).
$S = 2[3(-1) + (-1)] = 2[-4] = -8$.
Final value $= 21 + S = 21 - 8 = 13$.
Step 3: Final Answer:
The final value is 13.