Complex Fourier series has a singular term

Ragnord
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Homework Statement



Find the complex Fourier series for f(t)=t(1-t), 0<t<1

Homework Equations



\sum_{n=-\infty}^{\infty}c_{n}e^{2in\pi t}

where c_{n}=\int_{0}^{1}f(t)e^{-2in\pi t}dt


The Attempt at a Solution



I've worked out that c_{n}=-1/(2n^2 \pi^2). The problem is that for n=0, it is singular. Is there some way around this or does it mean that the complex Fourier series doesn't exist?
I tried using maple to graph the series with the n=0 term omitted and it comes out to the right shape, but is shifted vertically down some, leading me to believe that the singular term should be replaced by a constant or something.
 
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c_0=\int_0^1 f(t)\,dt
 
Well that was easy, just like a real Fourier series. Thanks.
I'm interested in knowing why that's the case though, I haven't seen anything about doing anything special for c_{0} in anything I've seen about complex Fourier series.
 
Just plug in n=0 before integrating instead of after!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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