## anonymous 5 years ago \sum_{k=1}^{10} [\sin (2k \pi/11) +i\cos (2k \pi/11) ]

1. anonymous

\sum_{k=1}^{10} [\sin (2k \pi/11) +i\cos (2k \pi/11) ]

2. anonymous

$\sum_{k=1}^{10} [\sin (2k \pi/11) +i\cos (2k \pi/11) ]$

3. anonymous

$\sum_{k=1}^{10} [\sin (2k \pi/11) +i\cos (2k \pi/11) ]$

4. anonymous

I guess it doesn't work when you are entering your question. So do you know what $\sum_{k=1}^{10} [\sin (2k \pi/11)$ is?

5. anonymous

yeah that should be zero right?

6. anonymous

Yeah. So what is $\sum_{k=1}^{10} [\cos (2k \pi/11)$

7. anonymous

10?

8. anonymous

It is cos(2pi/11) + cos(4pi/11) + ... + cos(20pi/11) which I don't like at all. Are you sure you have the right question?

9. anonymous

yeah.. isn't cos(2kpi) =1?? where k is any constant

10. anonymous

Apparently it is -10 I plugged it into MAthematica to be honest. So 0 is the real part and -10 is the imaginary part. I'm not aware of any easy way of solving the summation of cos(2kpi/11)

11. anonymous

Uh quick question. Is the /11 outside or inside the cosine?

12. anonymous

you know what my book says the correct answer is -i. i dont know how the hell that shows up!

13. anonymous

inside. $\sum_{k=1}^{10} [\sin (2k \pi/11) +i\cos (2k \pi/11) ]$

14. anonymous

Weird, mathematica is telling me otherwise. Sorry.