Riemann zeta, why are the residues either zero or one? One more question, probably equally simple to answer but I don't know how this is true either:
Why is the residue of Riemann zeta zero - trivial or non-trivial:
$$\text{residue}\left(\zeta (s)-\frac{\zeta '(\rho _n+s-1)}{\zeta \left(\rho _n+s-1\right)},\{s,1\}\right) = 0$$ 
equal to $0$, but the residue of some other number in the complex plane of the same expression:
$$\text{residue}\left(\zeta (s)-\frac{\zeta '(k+i n+s-1)}{\zeta (k+i n+s-1)},\{s,1\}\right)=1$$
equal to $1$?
The {$s,1$} means the Residue of the expression inside the $residue()$ parentheses at $s=1$.
Some Mathematica code:
Table[Residue[(Zeta[s] - 
    Zeta'[s - 1 + ZetaZero[n]]/Zeta[s - 1 + ZetaZero[n]]), {s, 
   1}], {n, 1, 12}]

which returns: {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
and
Monitor[Table[
  Table[Residue[(Zeta[s] - 
      Zeta'[s - 1 + n*I + k]/Zeta[s - 1 + n*I + k]), {s, 1}], {n, 1, 
    12}], {k, 1, 12}], k]

which returns: {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1},...

With and exception at $n=0$ and $k=1$ when the Residue is $2$.

After the answers below I realized I should have looked at:
Table[Residue[(Zeta'[s]/Zeta[s]), {s, ZetaZero[n]}], {n, 1, 12}]

which gives: {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}
and:
Table[Residue[(Zeta'[s - 1]/Zeta[s - 1]), {s, ZetaZero[n]}], {n, 1, 
  12}]

which gives: {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
 A: Note that the residue depends linearly on the function, so in particular we have
$$\operatorname{Res}\left(f(s) - g(s); s_0\right) = \operatorname{Res}\left(f(s);s_0\right) - \operatorname{Res}\left(g(s);s_0\right)$$
for all functions $f,g$ holomorphic in a punctured neighbourhood of $s_0$.
Now, when we have a Laurent expansion
$$f(s) = \sum_{n=-\infty}^\infty a_n (s-s_0)^n$$
of $f$ in a punctured disk around $s_0$, then $\operatorname{Res}\left(f(s);s_0\right) = a_{-1}$, the residue is the coefficient of $(s-s_0)^{-1}$ in that expansion.
For the pole of $\zeta$ in $1$, we know the start of the Laurent series (since it is a simple pole, there is only one term with a negative exponent), namely
$$\zeta(s) = \frac{1}{s-1} + \gamma + \dotsc,$$
so we have
$$\operatorname{Res}\left(\zeta(s);1\right) = 1.$$
For the other term, note that if $h$ has a pole or a removable singularity in $s_0$, then we can write $h(s) = (s-s_0)^k\cdot p(s)$, where $p$ is holomorphic in a neighbourhood of $s_0$ with $p(s_0) \neq 0$. The exponent $k$ is the order of $h$ in $s_0$. If $k > 0$, then $h$ has a zero of order $k$ in $s_0$, and if $k < 0$, $h$ has a pole of order $\lvert k\rvert$ in $s_0$. Then for the logarithmic derivative of $h$, we obtain the representation
$$\frac{h'(s)}{h(s)} = \frac{k(s-s_0)^{k-1}p(s) + (s-s_0)^kp'(s)}{(s-s_0)^kp(s)} = \frac{k}{s-s_0} + \frac{p'(s)}{p(s)},$$
where the second term is holomorphic in a neighbourhood of $s_0$, and thus
$$\operatorname{Res}\left(\frac{h'(s)}{h(s)}; s_0\right) = k.$$
In our situation, we have
$$g(s) = \frac{\zeta'(\rho_n+s-1)}{\zeta(\rho_n+s-1)},$$
so $\operatorname{Res}\left(g(s);1\right)$ is the multiplicity of the zero $\rho_n$.
The first zeros of $\zeta$ at least (maybe all, I don't know) are all simple zeros, so
$$\operatorname{Res}\left(\zeta(s) - \frac{\zeta'(\rho_n+s-1)}{\zeta(\rho_n+s-1)}; 1\right) = 1 - 1 = 0.$$
For the other values, $\zeta(k+in) \neq 0$, so there $g$ is holomorphic, hence has residue $0$, and only the residue of $\zeta$ in $1$ remains.
A: The residue of $\zeta(s)$ at $s=1$ is $1$. This is because $\lim\limits_{s\to1}(s-1)\zeta(s)=1$.
For any analytic function $f$, $\frac{f'(z)}{f(x)}$ has residue $n$ at $z=a$ where $n$ is the degree of the $0$ of $f$ at $z=a$.
What you are seeing is that $\zeta(k+in+s-1)$ does not have a $0$ at $s=0$ and $\zeta(\rho_n+s-1)$ has a $0$ of degree $1$.
