Fourier transformations of discrete time signals
When finding the DTFT of short discrete sequences like this, you can directly compute the sum. 

For example, say you had x[n] = [1 0 2] for n=0,1,2. Notice that I defined the values for n. If you had different n's, then the answer would be different. 

You can compute the sum as
\begin{align}
X(\omega) &amp;= \sum_n x[n] e^{-j\omega n} \\
&amp; = 1e^{-j\omega 0} + 0e^{-j\omega 1} + 2e^{-j\omega 2} \\
&amp; = 1 + 2e^{-j\omega 2}
\end{align}

Another note, you can sometimes simplify the answer by combining two exponentials. Be on the lookout for exponentials with the same coefficient. An example would be (using $e^{-j0}$ = 1)

\begin{align}
X(\omega) &amp;= 2 + 2e^{-j\omega 4} \\
&amp;= 2e^{-j\omega 0} + 2e^{-j\omega 4} \\
&amp;= 2e^{-j\omega 2} (e^{j\omega2 } + e^{-j\omega 2}) \\
&amp;= 2e^{-j\omega 2} (2\cos(\omega2)) \\
&amp;= 4e^{-j\omega 2} \cos(\omega2)
\end{align}
Hope this helps.