The value of $\sum_{k=1}^{10} [\sin\frac{2\pi k}{11} + i\cos\frac{2\pi k}{11}]$

Question

Multiplying and dividing by $-i^3$

$$\frac {1}{-i^3} \sum \left[e^{i \frac{2k\pi}{11}}\right]$$

It can be seen as $w+w^2+w^3....w^{10}$, where $w$ is a complex root of the complex equation. Also $$1+w+w^2+w^3....w^{10}=0$$

Since sum is till 10, the final value would be -1

$$\frac{-1}{-i^3}=\frac{1}{-i}=\frac{i}{-i^2}=i$$

But the answer given is $-i$. What is going wrong?

Answer 1

Your error seems to be that you are assuming

$$\sin\left(2\pi k\over11\right)+i\cos\left(2\pi k\over 11\right)={-1\over i^3}\left(\cos\left(2\pi k\over11\right)+i\sin\left(2\pi k\over11\right) \right)$$

You can see that this is wrong if you multiply both sides by $i^3$ and note that the real part on the left becomes $i^4\cos(2\pi k/11)=\cos(2\pi k/11)$, which is not equal to the real part on the right, $-\cos(2\pi k/11)$.

Rather than multiplying and dividing by a power of $i$, it's easier to simply factor the $i$ out front:

$$\begin{align} \sum_{k=1}^{10}\left(\sin\left(2\pi k\over11\right)+i\cos\left(2\pi k\over11\right)\right) &=i\sum_{k=1}^{10}\left(\cos\left(2\pi k\over11\right)-i\sin\left(2\pi k\over11\right)\right)\\ &=i\sum_{k=1}^{10}e^{-2\pi ik/11}\\ &=i\left(\sum_{k=0}^{10}e^{-2\pi ik/11}-1 \right)\\ &=i(0-1)\\ &=-i \end{align}$$

Answer 2

Use Intuition behind euler's formula

$$\sum_{k=1}^{10}e^{i (\pi/2-2\pi k/11)}=-e^{i\pi/2}+e^{i\pi/2}\cdot\dfrac{e^{-2\pi i}-1}{\cdots}=?$$

Answer 3

Let $a$ be $\exp(2\pi i/11)$. Then we have: $$ \begin{aligned} 0&=\frac {1-a^{11}}{1-a}=1+a+a^2+\dots+a^{10}\ ,\text{ so }\\ -1&=a+a^2+\dots+a^{10}\ . \end{aligned} $$ Now multiply with $i$.