$\sum _{k=1}^n\:\left(\cos\left(\frac{2\cdot k\cdot \pi }{n}\right)-2\:+\:i\cdot \sin\left(\frac{2\cdot k\cdot \pi }{n}\right)\right)$

Question

$\sum _{k=1}^n\:\left(\cos\left(\frac{2\cdot k\cdot \pi }{n}\right)-2\:+\:i\cdot \sin\left(\frac{2\cdot k\cdot \pi }{n}\right)\right)$

Normally the general factor is $a(n)=\cos\left(\frac{2k\pi }{n}\right)-2+i\cdot \sin\left(\frac{2k\pi }{n}\right)$

All I was able to do to the sum above was rewrite it as:

$\sum _{k=1}^n\:\left(\cos\left(\frac{2k\pi }{n}\right)\right)\:+\:\sum _{k=1}^n\left(i\cdot \sin\left(\frac{2k\pi }{n}\right)\right)\: - 2n$

Expecting the sums to cancel each other out and a result of something around -2n but not sure how to do that.

Answer 1

$$S = \sum_{k =1}^n\left[\cos\left(\frac{2k\pi}{n}\right) + i \sin\left(\frac{2k\pi}{n}\right)\right]$$

Consider vectors in complex plane:

• Analytical $$\begin{align*}S &= \sum_{k =1}^n\left[\cos\left(\frac{2k\pi}{n}\right) + i \sin\left(\frac{2k\pi}{n}\right)\right] = \sum_{k =1}^ne^{i\frac {2k\pi}{n}}\\ & =\frac {e^{i(2\pi + \frac {2\pi}{n})}-e^{i(\frac {2\pi}{n})}}{e^\theta-1} =0 \text{ :Numerator = 0}\\ \end{align*}$$

$$\implies \sum _{k=1}^n\:\left(\cos\left(\frac{2\cdot \:k\cdot \pi }{n}\right)-2\:+\:i\cdot \:\sin\left(\frac{2\cdot \:k\cdot \pi }{n}\right)\right) = -2n$$