Question regarding DTFT of a complex signal
I think you almost have the solution. As you suggest in your comment, if you express your signal as :
$$
x(n) = j\cdot \delta(n) - j\cdot \delta(n-4)
$$
As DFT is linear, you will get :
$$
\begin{align}
X(k) &amp;= \sum_{n=0}^{N-1}x(n)\cdot e^{-2j\pi \frac{kn}{N}}\\
 &amp;= j\cdot \sum_{n=0}^{N-1}\delta (n)\cdot e^{-2j\pi \frac{kn}{N}} - j\cdot \sum_{n=0}^{N-1}\delta (n-4)\cdot e^{-2j\pi \frac{kn}{N}}\\
&amp;=j-j\cdot e^{-8j\pi \frac{k}{N}}\\
&amp;=\text{ }...
\end{align}
$$