Analytic expression for non-trivial commutators Motivated by a previous question, consider bosonic creation/anihilation operators $a, a^+$ such that $[a, a^+]=1$, and $N = a^+a$.
Is there an analytic expression for the following commutators: 
$[e^{za}, e^{wN}]$  and $[e^{za^+}, e^{wN}]$
where $z$ and $w$ are complex (belong to $\mathbb{C}$). 
 A: There are no unique representations, so we will assume that OP is mostly interested in normal-ordered expressions between the three operators $a^{\dagger}$, $a$ and $N:=a^{\dagger}a$. Here $[a,a^{\dagger}]=1$. The underlying identities are
$$ \tag{1} [a,N] ~=~a \qquad\text{and}\qquad [N,a^{\dagger}] ~=~a^{\dagger}, $$ 
which lead to 
$$ \tag{2}  f(a)e^{zN}~=~ e^{zN} f(e^{z} a)\qquad\text{and}\qquad 
 e^{zN}f(a^{\dagger})~=~ f(e^{z} a^{\dagger})e^{zN}, $$
respectively. Here $z\in \mathbb{C}$ is a complex number, and $f:\mathbb{C}\to \mathbb{C}$ is a sufficiently well-behaved function, e.g. an exponential function. So the sought-for commutators read in normal-order form
$$ 
\tag{3}  [f(a),e^{zN}]~=~ e^{zN} \{f(e^{z} a)-f(a)\},
$$
and 
$$ 
\tag{4}  [e^{zN},f(a^{\dagger})]~=~  \{f(e^{z} a^{\dagger})-f(a^{\dagger})\}e^{zN},
$$
respectively. More generally, one has 
$$ 
\tag{5} e^{zN}f(a^{\dagger},a)~=~ f(e^{z} a^{\dagger},e^{-z} a)e^{zN}, $$
with corresponding commutator
$$ 
\tag{6}  [e^{zN},f(a^{\dagger},a)]~=~  \{f(e^{z} a^{\dagger},e^{-z} a)-f(a^{\dagger},a)\}e^{zN}.
$$
