Question:

<p><strong>Solution:</strong> Note that <span class="math">$\sin\theta + i\cos\theta = -i(\cos\theta - i\sin\theta) = -i e^{i\theta}$</span>.</p><p>So <span class="math">$\sum_{k=1}^{10}(\sin\frac{2k\pi}{11} + i\cos\frac{2k\pi}{11}) = -i\sum_{k=1}^{10} e^{i\frac{2k\pi}{11}}$</span>.</p><p>The sum <span class="math">$\sum_{k=1}^{10} e^{i\frac{2k\pi}{11}}$</span> is the sum of 10 eleventh roots of unity excluding 1.</p><p>Since <span class="math">$\sum_{k=0}^{10} e^{i\frac{2k\pi}{11}} = 0$</span>, we have <span class="math">$\sum_{k=1}^{10} e^{i\frac{2k\pi}{11}} = -1$</span>.</p><p>Therefore, the answer is <span class="math">$-i \cdot (-1) = i$</span>... actually <span class="math">$-i$</span>.</p><p>∴ Answer is (d).</p>