Let $j : \mathbb C^* \to \mathbb C$ be the inclusion. Write $U = \mathbb C^*$ and $X = \mathbb C$. I want to calculate the cohomology groups $H^*(X,j_* L)$ where $L$ is a local system on $U$.
It is clear that $H^0 (X,j_* L)$ = $H^0(U,L)$ by definition. However I'm less sure about $H^1(X,j_*L)$.
Using the sequence $0 \to j_! L \to j_* L \to (j_*L)_0 \to 0 $ we see that $H^0(X,j_!L) = 0$ and deduce $H^1(X,j_!L) \cong H^1(X,j_*L)$. However it doesn't seem to help much. Also I am not sure about a good cover to compute the cohomology. I feel the answer should be $H^1(U,L)$ but I can't see it unfortunately.
Edit : if I'm not mistaken, looking at the Leray spectral sequence shows that (as vector spaces) $H^1(U,L) \cong H^1(X,j_*L) \oplus H^0(X,R^1 j_* L)$. However an easy calculation gives $H^0(X,R^1 j_*L) \cong H^1(U,L)$ hence we must have $H^1(X,j_*L) = 0$ which really surprises me. I would love to see if someone agrees with my argument, and also have a more elementary argument to obtain the same result.