I have a stochastic variable $Z$ and know that that $Z(t)/Z(0)$ is a martingale. Now, my book claims that this means that it has mean 1. I assume this is because of the martingale-property:
$$ E[Z(t)/Z(0)] = Z(0)/Z(0) = 1 $$
However, the above I guess only makes sense if we take the expectation value up to and include time $0$. But I would think that when we refer to the mean of a variable, then we are taking the expectation up to infinity, and not 0?
As @charlus said, for every $t \in \mathbb R^+$, $Z(t)$ is a random variable. So when we take the expectation $\mathbb E[Z(t)]$, we integrate with respect to $\omega$ and not $t$. In other words, $$\mathbb E[Z(t)] := \int_{\omega \in \Omega} Z(t,\omega) d\mathbb P(\omega) \quad \forall t\in\mathbb R^+ $$ So there isn't any notion of "taking expectation up to infinity", we are simply computing the expectation of every random variable $Z(t)$ in the family $\{Z(t),t\in\mathbb R^+\}$.
As for the result you want to prove, it is indeed a consequence of the martingale property :
I assume that $Z(0)$ is a constant almost surely, and that you are working with the canonical filtration $(\mathcal F_t)_{t\ge0}$ for which the process $Z(t)/Z(0)$ is a martingale.
The martingale property tells you that for every $s\le t$, $\mathbb E[Z(t)/Z(0)\ |\ \mathcal F_s] = Z(s)/Z(0)$
Now noticing that $\mathcal F_0 = \{\varnothing, \Omega\}$, what your book says is that
$$\begin{align}\mathbb E[Z(t)/Z(0)] &= \mathbb E [Z(t)/Z(0)\ |\ \Omega] \\ &=\mathbb E [Z(t)/Z(0) \ |\ \mathcal F_0] \\ &= Z(0) / Z(0) \ \text{ by the martingale property} \\ &= 1 \end{align} $$
Edit : As suggested by @Brian Moehring I'll do the proof without the assumption of almost surely constant $Z(0)$, which is unnecessary. So now we just assume that $Z(0) \ne 0$ almost surely : $$\begin{align}\mathbb E[Z(t)/Z(0)] &= \mathbb E[\mathbb E [Z(t)/Z(0)\ |\ \mathcal F_0]] \text{ (law of total expectation)} \\ &=\mathbb E [Z(0)/Z(0)] \text{ by the martingale property} \\ &= 1 \end{align} $$