Convolution in frequency domain
If you have two DFTs $A[k]$ and $B[k]$ (note the correct representation of a sinusoid at DFT bin number $1$)

A = [0,-j,0,0,j]; B = [1,1,1,1,1];

with the corresponding time-domain sequences $a[n]$ and $b[n]$


a = ifft(A);    % [0, 0.38042, 0.23511, -0.23511, -0.38042];
b = ifft(B);    % [1,0,0,0,0];


then the multiplication of the time-domain sequences $c[n]=a[n]b[n]$ corresponds to the cyclic (or circular) convolution of the DFTs $A[k]$ and $B[k]$:

$$\text{DFT}\{c[n]\}=C[k]=\frac{1}{N}\sum_{n=0}^{N-1}A[n]B[k-n]_{\text{mod} N},\quad k=0,1,\ldots,N-1\tag{1}$$

where the indices are taken modulo $N$ ($N=5$ in this case), such that they remain in the range $[0,N-1]$:

$$\begin{align}C[0]&amp;=\left(A[0]B[0]+A[1]B[N-1]+\ldots+A[N-1]B[1]\right)\cdot\frac{1}{N}\\
C[1]&amp;=\left(A[0]B[1]+A[1]B[0]+\ldots+A[N-1]B[2]\right)\cdot\frac{1}{N}\\&amp;\vdots\\
C[N-1]&amp;=\left(A[0]B[N-1]+A[1]B[N-2]+\ldots+A[N-1]B[0]\right)\cdot\frac{1}{N}
\end{align}$$

It's easy to see that the cyclic convolution of $A[k]$ and $B[k]$ as given above results in the zero vector, because due to $B[k]=1$ for all values of $k$, each element of the result is the sum of all elements of $A[k]$, which is always zero.