By Convolution Theorem:

Convolution of two signals in time domain is equivalent to multiplication in frequency domain.

$$x_1(n) * x_2(n) \leftarrow FT \rightarrow X_1(k) . X_2(k)$$

where, $X_1(k)$ & $X_2(k)$ are DFTs of $x_1(n)$ &. $x_2(n)$ respectively.

$X_1(k)$ ={14, 3-5j, 0, 3+5j}

$X_2(k)$ = {10, 2+2j, -2, 2-2j}

$x_1(n) * x_2(n) ≡ X_1(k) . X_2(k)$ = {14, 3-5j, 0, 3+5j}.{10, 2+2j, -2, 2-2j}

$y(n) = x_1(n) * x_2(n)$ = {140, 16-4j, 0, 16+4j}

Result: y(n) ={140, 16-4j, 0, 16+4j}