I am studying CUTLAND "Computability-An introduction to recursive function theory" book.
In this book two sets are introduced: $$ Z = \{x| \phi_x(x)=0\} $$ $$ R=\{x|\phi_x=0\} $$
the set $ \ Z $ is a creative set and the set $ \ R $ is a productive set. I cannot understand what is difference between them. Why $ \ Z $ is r.e or creative and $ \ R $ is not r.e
Thanks for your attention.
The difference is that in order for $x$ to be in $Z$, $\phi_x$ has only to take the value $0$ at one particular point, while for $x$ to be in $R$, $\phi_x$ has to be identically 0.
Note that $Z$ is r.e. by dovetailing: interleave all computations $\phi_x(x)$, and if you see one of them converge to $0$, this means that $x \in Z$; moreover, for any $x \in Z$, eventually we will see $\phi_x(x)\downarrow = 0$.
We can also show that $R$ is not r.e., by giving a 1-reduction of the complement of the halting problem, $\overline{K}$ (which is not r.e.) to $R$: for given $e$, let $f(e)$ be such that $\phi_{f(e)}(n) = 0$ if $\phi_{e,n}(e)\uparrow$; that is, if $\phi_e(e)$ has not yet converged after $n$ steps, and $\phi_{f(e)}(n) = 1$ if $\phi_{e,n}(e)\downarrow$. $f$ is computable and one to one, and $f(e) \in R$ if and only if $e \in \overline{K}$.
Additionally, one can show that $K$ is reducible to $Z$: let $f(e)$ be such that $\phi_{f(e)}(n) = 0$ if $\phi_e(n)\downarrow$ and $\phi_{f(e)}(n)\uparrow$ otherwise. Then $f$ is one to one and $f(e) \in Z$ if and only if $e \in K$.
This is actually enough to deduce that $Z$ is creative and $R$ is productive; see for instance the equivalence cited in the Wikipedia article, though this can also be proved directly without too much trouble by exhibiting productive functions for $\overline{Z}$ and $R$.