The correct answer is: $$v\left( t \right) = \int\limits_0^t {i\left( \tau \right)} z\left( {t – \tau } \right)d\tau $$
The Laplace transform of a function $f(t)$ is denoted by $F(s)$ and is defined as:
$$F(s) = \int_0^\infty f(t) e^{-st} dt$$
The inverse Laplace transform of a function $F(s)$ is denoted by $f(t)$ and is defined as:
$$f(t) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} F(s) e^{st} ds$$
where $c$ is a real number such that all the poles of $F(s)$ lie to the left of $c$.
In the given question, we are given that $V(s) = Z(s) I(s)$. We can write this as:
$$V(s) = \int_0^\infty z(t) e^{-st} dt \int_0^\infty i(t) e^{-st} dt$$
Using the convolution theorem, we can write this as:
$$V(s) = \int_0^\infty \left( \int_0^t z(\tau) d\tau \right) i(t-\tau) e^{-st} dt$$
Taking the inverse Laplace transform of both sides, we get:
$$v(t) = \int_0^t z(\tau) i(t-\tau) d\tau$$
This is the convolution of the functions $z(t)$ and $i(t)$.