What is the DFT of a binomial filter?
I don't really understand the thing about cosines (as in: how is that helpful?) – a DFT is really just a mapping from a complex vector with $N$ elements to a complex vector with $N$ elements; and your calculation seems to be wrong, and I'm not sure where, but doing two of the three elements of the DFT manually might actually be enough to clear things up.

\begin{align}
X_0 &amp;= \sum\limits_{n=0}^{N-1} x_n e^{-i2\pi\frac nN\cdot 0}\\
&amp;= \sum\limits_{n=0}^{N-1} x_n e^0\\
&amp;= \frac 14 (1+2+1)\\
&amp;=1\\[2em]
X_1 &amp;=\sum\limits_{n=0}^{N-1} x_n e^{-i2\pi\frac nN\cdot 1}\\
&amp;=\sum\limits_{n=0}^{N-1} x_n e^{-i2\pi\frac n3}\\
&amp;=\frac14 \left( 1 \cdot e^{-i2\pi\frac 03} + 2 e^{-i2\pi\frac 13} + 1 e^{-i2\pi\frac 23}\right)\\
&amp;= \frac 14\left[1+ 2\cos\left(2\pi\frac13\right)+ i2\sin\left(2\pi\frac13\right)
+ \cos\left(2\pi\frac23\right)+ i\sin\left(2\pi\frac23\right)\right]\\
&amp;\text{bit of basic trigonometry}\\
&amp;=\frac14\left[
1+ 2\cos\left(2\pi\frac13\right)+ i2\sin\left(2\pi\frac13\right)
- \cos\left(2\pi\frac13\right) + i\sin\left(2\pi\frac13\right)
\right]\\
&amp;=\frac14\left[
1+ 1\cos\left(2\pi\frac13\right)+ i3\sin\left(2\pi\frac13\right)
\right]\\
&amp;=\frac14\left[
1-\frac12+ i3\frac{\sqrt3}2
\right]\\
&amp;=\frac18+i\frac{3\sqrt3}8
\end{align}

Note how there's no taking of real parts or anything: the DFT of a sequence is complex in general, and you can't just drop the imaginary part. So, not quite sure why we're talking in control theory terms here, but no, you cannot just drop it. And: the discrete Fourier is not the same as the Laplace transform.