Question

Please help me solve this sum on complex numbers:

Answer 1

\[C = \cos \theta + \cos 3\theta + \cos 5\theta + ... + \cos(2n-1) \theta\]
\[S = \sin \theta + \sin3\theta + \sin5\theta + ... + \sin(2n-1)\theta\]

Answer 2

I have already shown that:\[e ^{ik \theta} - 1 = 2ie^{i \theta}\sin \theta\]

Answer 3

and I've already shown that C+iS is a GP, and found the sum of the series. Now I need to show that \[\left| C+iS \right| = \frac{ \sin n \theta }{ \sin \theta }\]

Answer 4

and eventually find arg (C+iS)

Answer 5

@Country_Boy_Profile8.

Answer 6

i suck at math lol

Answer 7

but i will try my best

Answer 8

thank you :)

Answer 9

What exactly are you trying to solve?

Answer 10

my last two comments, apart from thank you :P

Answer 11

C+iS...
that can only be
\[\small\cos(\theta) + \color{blue}i\sin(\theta)+\cos(3\theta) + \color{blue}i\sin(3\theta)+\cos(5\theta) + \color{blue}i\sin(5\theta)+...+\cos[(2n-1)\theta] + \color{blue}i\sin[(2n-1)\theta]\]

Answer 12

yupyup.

Answer 13

where a = cis (theta) and r = (cis(theta))^2

Answer 14

That means
\[\Large e^{\color{blue}i\theta}+e^{3\color{blue}i\theta}+e^{5\color{blue}i\theta}+e^{7\color{blue}i\theta}+e^{9\color{blue}i\theta}...+e^{(2n-1)\color{blue}i\theta}\]

Answer 15

effectively, yes

Answer 16

A geometric progression?

Answer 17

yes, i mentioned that earlier, didn't i?

Answer 18

Well, I'd rather see where it's all coming from :D

Answer 19

okie dokie, whatever rows your boat.

Answer 20

Let's see... its sum is
\[\Large e^{\color{blue}i\theta}\left[\frac{1-e^{2n\theta}}{1-e^{2\theta}}\right]\]

Answer 21

yes.

Answer 22

my question is, how do I find the mod?? i can simply it once I get it, no probs, but how do I get it in the first place?

Answer 23

The modulus?

Answer 24

yup. well, in terms of a + ib, (a^2 + b^2)^0.5

Answer 25

Are we supposed to get something nice from this? I mean, a rather nice-looking expression?

Answer 26

yes, we get sin(n theta)/sin (theta)

Answer 27

That's supposed to be the answer?

Answer 28

yes, we just have to get it.

Answer 29

Maybe we can try...
\[\Large \left|\frac{1-e^{2ni\theta}}{1-e^{2i\theta}}\right|\]

Answer 30

\[\Large \left|\frac{1-\cos(2n\theta)-\color{blue}i\sin(2n\theta)}{1-\cos(2\theta)-\color{blue}i\sin(2\theta)}\right|\]

Answer 31

Just to let you know... I still have no idea what I'm doing :/

Answer 32

Multiply that number by its complex conjugate:
\[ \left| \frac{1-e^{2ni\theta}}{1-e^{2i\theta}}\right|^2 = \frac{1-e^{2ni\theta}}{1-e^{2i\theta}}\cdot \frac{1-e^{-2ni\theta}}{1-e^{-2i\theta}} = \frac{2 - (e^{2ni\theta} + e^{-2ni\theta})}{2 - (e^{2i\theta} + e^{-2i\theta})}\]
\[ = \frac{1-\cos(2n\theta)}{1-\cos(2\theta)} \]
but
\[ 1 - cos(2x) = 2\cdot \sin^2(x) \]
so we have
\[ \frac{\sin^2(n\theta)}{\sin^2(\theta)} \]
which gives you your result.