Key Concepts and Formulas

• Euler's Formula: $e^{i\theta} = \cos\theta + i\sin\theta$
• Sum of Roots of Unity: For $n \in \mathbb{Z}^+$, $\sum_{k=0}^{n-1} e^{i\frac{2k\pi}{n}} = 0$
• Properties of Complex Numbers: $i^2 = -1$, $i = \frac{1}{-i}$, $-i = \frac{1}{i}$

Step-by-Step Solution

Step 1: Rewrite the General Term

The general term is given by $\sin\left(\frac{2k\pi}{11}\right) + i\cos\left(\frac{2k\pi}{11}\right)$. We want to express this in terms of complex exponentials using Euler's formula. Factoring out $i$, we have: $\sin\left(\frac{2k\pi}{11}\right) + i\cos\left(\frac{2k\pi}{11}\right) = i\left(\cos\left(\frac{2k\pi}{11}\right) - i\sin\left(\frac{2k\pi}{11}\right)\right)$ Using the properties $\cos(\theta) = \cos(-\theta)$ and $\sin(-\theta) = -\sin(\theta)$, we get: $i\left(\cos\left(\frac{2k\pi}{11}\right) - i\sin\left(\frac{2k\pi}{11}\right)\right) = i\left(\cos\left(-\frac{2k\pi}{11}\right) + i\sin\left(-\frac{2k\pi}{11}\right)\right)$ Applying Euler's formula, we obtain: $i\left(\cos\left(-\frac{2k\pi}{11}\right) + i\sin\left(-\frac{2k\pi}{11}\right)\right) = i e^{-i\frac{2k\pi}{11}}$ This allows us to rewrite the summation in terms of complex exponentials.

Step 2: Express the Summation

Substitute the result from Step 1 into the original summation: $\sum_{k=1}^{10} \left(\sin\left(\frac{2k\pi}{11}\right) + i\cos\left(\frac{2k\pi}{11}\right)\right) = \sum_{k=1}^{10} i e^{-i\frac{2k\pi}{11}}$ Since $i$ is a constant, we can factor it out: $\sum_{k=1}^{10} i e^{-i\frac{2k\pi}{11}} = i \sum_{k=1}^{10} e^{-i\frac{2k\pi}{11}}$ Let $z = e^{-i\frac{2\pi}{11}}$. Then the summation becomes: $i \sum_{k=1}^{10} z^k = i(z^1 + z^2 + \dots + z^{10})$ This is a geometric series.

Step 3: Apply the Sum of Roots of Unity

The terms $z^k = e^{-i\frac{2k\pi}{11}}$ for $k = 0, 1, \dots, 10$ are the complex conjugates of the 11th roots of unity. We know that the sum of the $n$th roots of unity is zero. Therefore, $\sum_{k=0}^{10} e^{-i\frac{2k\pi}{11}} = 0$ Our sum starts from $k=1$, so we can write: $\sum_{k=1}^{10} e^{-i\frac{2k\pi}{11}} = \left(\sum_{k=0}^{10} e^{-i\frac{2k\pi}{11}}\right) - e^{-i\frac{2(0)\pi}{11}} = 0 - e^0 = 0 - 1 = -1$ We used the property that any number raised to the power of 0 is 1.

Step 4: Calculate the Final Result

Substitute the result from Step 3 back into the expression: $i \sum_{k=1}^{10} e^{-i\frac{2k\pi}{11}} = i(-1) = -i$

Common Mistakes & Tips:

• Sign Errors with Euler's Formula: Be extremely careful with the signs when applying Euler's formula, particularly when manipulating expressions like $\sin\theta + i\cos\theta$. Remember that $\sin\theta + i\cos\theta = i(\cos\theta - i\sin\theta) = ie^{-i\theta}$.
• Summation Index: Always pay attention to the starting index of the summation. The sum of roots of unity property applies when the summation starts from 0. Adjust accordingly if it starts from a different index.
• Recognizing Roots of Unity: Familiarize yourself with the form $e^{i\frac{2k\pi}{n}}$. Recognizing this pattern allows you to directly apply the sum of roots of unity property.

Summary:

By rewriting the general term using Euler's formula and leveraging the property that the sum of the roots of unity is zero, we simplified the summation and calculated its value. The sum evaluates to $-i$.

The final answer is \boxed{-i}, which corresponds to option (D).