Author Topic: Complex Number Questions (Read 772 times) Tweet Share

frog0101 · « **on:** October 09, 2018, 09:34:07 am »

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

RuiAce · « **Reply #1 on:** October 09, 2018, 09:52:15 am »

RuiAce · « **Reply #2 on:** October 09, 2018, 10:14:26 am »

$\text{Essentially since }z^{13} = w\text{ we have }z^{143} = w^{11}.\\ \text{Therefore }z^{143} = z \implies \boxed{z\left( z^{142}-1\right) = 0}$
$z = 0\text{ is then immediately one solution}\\ \text{although we discard it if we aren't interested in specifically real numbers.}\\ \text{The remaining complex numbers are just the 142-th roots of unity}\\ \text{which are just }\cos \frac{k\pi}{142} + i \sin \frac{k\pi}{142}\text{ where }k\text{ is an integer.}$
$\text{So clearly }\operatorname{Im}(z) = \sin \frac{k\pi}{142}.\\ \text{Depending on what }k\text{ is, this inevitably simplifies down}\\ \text{to }\sin \frac{m\pi}{n}\text{ for integers }m\text{ and }n\text{ anyway.}$
The fact that $m$ and $n$ have no common factors just means that it is a fraction simplified to its lowest terms. For example, if we had something like $ \frac{2}{142}$, we could just simplify it down to $ \frac{1}{71}$ instead.

frog0101 · « **Reply #3 on:** October 10, 2018, 08:28:26 am »

Thanks so much RuiAce!!

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