Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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?

Answer 5

yeah that should be zero right?

Answer 6

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

Answer 7

10?

Answer 8

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?

Answer 9

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

Answer 10

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)

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

Weird, mathematica is telling me otherwise. Sorry.