We have the functions $f(x)=1/6$ for $0 \leq x \leq 6$ and $g(x)=x^2-3ix$. I am asked to find $h=f*g$ where $*$ is the convolution operator. And then evaluate $h(4)$. The solution says that $h(4)=4-3i$.
I am truly lost about how they got that result, and am not even sure that their solution is correct, because I asked another question about convolution few days back, and it turned out that the solution our professor gave us is wrong.
I don't think I can solve that graphically if I need to deal with imaginary numbers. Any help would be appreciated.
Thanks for your help !
In order for the integrals to exists,we must have that $f=0$ outside of $[0,6]$
If so then note that $$(f\ast g)(x)=\int_0^6 \frac{1}{6}[(x-y)^2-3i(x-y)]dy$$
So $$(f\ast g)(4)=\int_0^6 \frac{1}{6}[(4-y)^2-3i(4-y)]dy$$ $$=\int_{-2}^4\frac{1}{6}[u^2-3iu]du=4-3i$$