# Thread: complex no. (cosx+cos2x+...+cos2nx+i(sinx+...sin2nx))=0

1. ## complex no. (cosx+cos2x+...+cos2nx+i(sinx+...sin2nx))=0

i don't understand this step. sorry!

2. Let $a = \cos\theta + i\sin\theta = \text{cis} \theta$

$(\cos\theta + i\sin\theta)+(\cos\theta + i\sin\theta)^2+(\cos\theta + i\sin\theta)^3+...+(\cos\theta + i\sin\theta)^{2n}$

$a + a^2 + a^2 + ...+ a^{2n} = \sum_{k=1}^{2n} a^k$

Use the identity $\sum_{i=m}^{n}x^i = \frac{x^{n+1}-x^m}{x-1}$ to find the sum.

3. Hello, afeasfaerw23231233!

$S \;=\;(\cos\theta + i\sin\theta) + (\cos\theta + i\sin\theta)^2 + (\cos\theta + i\sin\theta)^3 + \hdots + (\cos\theta + i\sin\theta)^{2n}$

This is a geometric series . . .
. . first term: $a \:=\\cos\theta + i\sin\theta)" alt="a \:=\\cos\theta + i\sin\theta)" />
. . common ratio: $r \:=\\cos\theta + i\sin\theta)" alt="r \:=\\cos\theta + i\sin\theta)" />
. . number of terms: $2n$

Recall that the sum of a geometric series is: . $S_n \:=\:a\,\frac{1-r^n}{1-r}$

So we have: . $S \;=\;\overbrace{(\cos\theta+i\sin\theta)}^a \,\frac{\overbrace{1 -(\cos\theta + i\sin\theta)^{2n}}^{1-r^n}}{\underbrace{1 - (\cos\theta + i\sin\theta)}_{1-r}}$

Got it?

,

### cos2nx

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