Find the value of $$\sum^{10}_{k=1}\left (\sin\left (\frac{2k\pi}{11} \right )+i\cos\left (\frac{2k\pi}{11}\right ) \right)$$
My approach:
Since $\cos\theta + i\sin\theta = e^{i\theta}$, we can write the given equation as:
$$\begin{align*}
&i \left \{\sum^{10}_{k=1} \left (\cos\frac{2k\pi}{11} -i\sin\frac{2k\pi}{11} \right ) \right \}\\
= &i \left \{\sum^{10}_{k=1}\left (e^{-i\frac{2k\pi}{11}} \right ) \right \} \tag{i}
\end{align*}$$
Solving the index part only which is
$$\begin{align*}
-i\frac{2k\pi}{11} &= -i\frac{2\pi}{11}(1+2+3+\cdots+10) \quad (\text{putting the values of } k)\\
&= -i\frac{2\pi}{11}( 55) \quad \left(\text{By applying sum of first $n$ natural numbers} = \frac{n(n+1)}{2}\right )\\
&=-i10\pi
\end{align*}$$
Putting this value in $(\text{i})$ we get:
$e^{-i10\pi} = i\cos10\pi = i.$
But the answer is $-i$. Please suggest where I went wrong… Thanks..
Best Answer
I am not sure where your steps came from. You are right that you are summing
$$i \sum_{k=1}^{10} e^{-i 2 \pi k/11}$$
This is a geometric series and has value
$$i\frac{e^{-i 2 \pi/11}-e^{-i 2 \pi}}{1-e^{-i 2 \pi/11}} = i\frac{e^{-i 2 \pi/11}-1}{1-e^{-i 2 \pi/11}}= -i$$