Proving that $\sum^n_{k=1} e^{ik\theta}=\sum^n_{i=1}\cos k\theta +i\sum^n_{k=1}\sin k\theta$.

Question

Prove:

$$\sum^n_{k=1} e^{ik\theta}=\sum^n_{i=1}\cos k\theta +i\sum^n_{k=1}\sin k\theta$$ Thanks a lot!!

I tried:

With Euler's identity I can get $\sin x= \dfrac{e^{ix} - e^{-ix}}{2i}$ and the cosine too. But i'm lost trying to development it.

Answer 1

Start manipulating on the LHS until you can show that it's equal to the RHS. Note that:

$$e^{ik\theta} = \cos(k\theta) + i \sin(k\theta)$$

Therefore:

$$\sum_{k=1}^n e^{ik\theta} = \sum_{k = 1}^n \Bigg(\cos(k\theta) + i\sin(k\theta)\Bigg)$$

And from here we can break the sum into two, which is just the commutative property of addition:

$$= \sum_{k = 1}^n \Bigg(\cos(k\theta)\Bigg) + \sum_{k=1}^n \Bigg(i\sin(k\theta)\Bigg)$$

And now you can factor $i$ out of the entire latter sum to finish up.

Answer 2

Given that

$$ e^{ik\theta} = \cos(k\theta) + \textbf{i} \sin(k\theta) $$

Then we get

$$ \sum_{k=1}^n e^{ik\theta} = \sum_{k=1}^n \Big( \cos(k\theta) + \textbf{i} \sin(k\theta) \Big) $$

which gives

$$ \sum_{k=1}^n e^{ik\theta} = \sum_{k=1}^n \cos(k\theta) + \textbf{i} \sum_{k=1}^n \sin(k\theta) $$