1. ## Complex Number Questions

Hi, I have been unable to do these questions (any help would be great):
1. Simplify sin(x)+sin(2x)+...+sin(nx)

2. Let z^13=w and w^11=z. Prove that Im(z)=sin{(m*pi)/n} if both z and w are complex numbers, where m and n are positive integers with no common factors.

Thanks

2. ## Re: Complex Number Questions

$\text{The trick for question 1 is to use trig identities to make it into a telescoping sum so that the terms cancel out:}$
$\text{Note that: } \cos((k-1)x) - \cos((k+1)x) = \cos(kx-x) - \cos(kx+x)$
$= 2\sin(kx)\sin(x)$
$\text{Therefore, } \sin(kx) = \dfrac{\cos((k-1)x) - \cos((k+1)x)}{2\sin(x)}$
$\text{Now, by substituting this into the sum a lot of terms will cancel out. However, it can be awkawrd to write the series so summation notation is helpful here: }$
$\sum_{k=1}^{n} \sin(kx)$
$= \sum_{k=1}^{n} \dfrac{\cos((k-1)x) - \cos((k+1)x)}{2\sin(x)}$
$= \dfrac{1}{2\sin(x)} \left( \sum_{k=1}^{n} \cos((k-1)x) - \cos((k+1)x) \right)$
$= \dfrac{1}{2\sin(x)} \left( \sum_{k=1}^{n} \cos((k-1)x) - \sum_{k=1}^{n} \cos((k+1)x) \right)$
$= \dfrac{1}{2\sin(x)} \left( \sum_{k=1}^{n} \cos((k-1)x) - \sum_{k=3}^{n+2} \cos((k-1)x) \right) \text{ Now we have the same function in the sum so we can cancel}$
$= \dfrac{1}{2\sin(x)} \left( \cos(0x) + \cos(1x) + \sum_{k=3}^{n} \cos((k-1)x) - [\sum_{k=3}^{n} \cos((k-1)x)] - \cos(nx) - \cos((n+1)x) \right)$
$= \dfrac{1 + \cos(x) - \cos(nx) - \cos((x+1)x)}{2\sin(x)}$

$\text{For the second problem we have: } (z^{11} )^{13} = z$
$z^{143} = z$
$(r\cdot cis\theta)^{143} = r\cdot cis\theta$
$r^{143} \cdot cis 143\theta = r \cdot cis\theta \text{ Therefore } r = 1$
$143\theta = \theta + 2n\pi$
$142\theta = 2n\pi$
$71\theta = n\pi$
$\theta = \frac{n\pi}{71}$
$Im(z) = \sin(\frac{n\pi}{71}) \text{Note that 71 is prime so no common factors}$

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