# Complex Fourier series of sin (t)

1. Nov 11, 2013

### unscientific

In finding Cn, I arrived at a different answer. I got an extra factor of (1/i) instead, which came when you do the integral of each exponential with respect to t; so you get a factor of 1/i(1-n) and 1/i(1+n) respectively..

Did they intentionally leave that out?

Last edited: Nov 11, 2013
2. Nov 11, 2013

### mathman

Last edited: Nov 11, 2013
3. Nov 11, 2013

### mathman

Addition thought. f(t) = sint is its own Fourier series. exp(it) = cost + isint. Therefore c1 needs a coefficient 1/i. I can't see what your disagreement is about. Perhaps you can give the details of your derivation.

4. Nov 11, 2013

### unscientific

Sorry, I spotted my mistake!