(Complex) circular approximation to $\gamma$ using the zeta-function around $s=1$ Just playing around considering some older hint from some answer here in MSE (a couple of years ago) I came to the following observation.
Assume a small $\epsilon >0 $ then the arithmetic mean $$ {\zeta(1+\epsilon) + \zeta(1-\epsilon) \over 2} \approx \gamma = 0.577... \tag 1$$ 
and there is even the well known equality in the limit for $\epsilon \to zero$ which I've seen some times stated here and elsewhere for to explain the (Ramanujan's) evaluation/replacement of $\zeta(1)\underset{\mathfrak R}= \gamma$.
Well, for $\epsilon \to 0$ this approximates nicely, but of course, if $\epsilon$ increases, say $\epsilon = \frac12$ or even greater $\epsilon = 6 $ then this approximation worsens rapidly (and of course follows the oscillatory divergent values for $\zeta$ at the negative integers).                   
Now I tried what would happen, if I do not take $\pm \epsilon$ which means $\epsilon \cdot (1,-1)$ but higher complex unit-roots $\omega_k$ say, for order $d$, the vector $\epsilon \cdot [ 1, \omega, \omega^2, ... \omega^{d-1}]$ and compute the arithmetic mean and compare to the Eulerconstant $\gamma$.     
Interestingly this approximates $\gamma$ again, and taking $d=5$, $d=50$, or even $d=150$ we can even have $\epsilon =1 $ to have approximation to more than 100 digits (if $d=150$ for instance) such that
$$\lim_{d \to \infty} \frac 1d \sum_{k=0}^{d-1} \zeta(1+ r \exp(2 i \pi \frac kd)) = \gamma \tag 2$$ 
where $r$ may be non-infinitesimal and for instance $r=1$ or $r=13$.              
This is much surprising for me - and I'd like to know

a) is this true?
  b) how can this be generalized/extended to $d \to \infty$ : does this mean to introduce an integral instead of the sum/mean? (I've no idea yet how this should actually be written - this is, perhaps, "contour-integration" ?)

 A: Presumably there should be a $\frac{1}{d}$ in front of your sum (so that it is an average, as in your first example). Then, viewing your integral as a Riemann sum, the limit as $d \to \infty$ will be
$$\lim_{d \to \infty} \frac{1}{d}\sum_{k=0}^{d-1}\zeta(1+re^{2\pi i k/d}) \to \int_{0}^{1}\zeta(1+re^{2\pi i x})d x \\ = \frac{1}{2\pi}\int_{0}^{2\pi}\zeta(1+re^{i \theta})d \theta$$
By making the substitution $z=re^{i \theta}$, $0 \le \theta \le 2\pi$ this integral may be naturally viewed as the contour integral
$$\frac{1}{2\pi i}\int_{\gamma}\frac{\zeta(1+z)}{z}dz$$
where $\gamma$ is the circle of radius $r$ centred at the origin. By the residue theorem, this is equal to
$$\sum_{i} Res(f,z_{i})$$
where the sum is over all poles of $f(z)=\frac{\zeta(1+z)}{z}$ inside $\gamma$. For any value of $r$, there is exactly one pole in this region, and it has order $2$. It follows that the residue of the pole (and hence the contour integral) is the principal part of $\zeta(1)$, which is the constant $\gamma$.  
I get the impression you have not seen contour integration before, so this argument might not be as helpful as it could be - I'd be happy to explain, but if I give a pre-emptive explanation of all of it then this answer would be far too long. All the theorems used would be found in any book/course on complex analysis.
A: You should read a course on complex analysis.
$F(s)=(s-1)\zeta(s)$ is entire (analytic on $\mathbb{C}$), in particular it is holomorphic at $s=1$ with $F(1) = 1, F'(1) = \gamma$. 


*

*Thus for any complex number $a \ne 0$
$$\lim_{h \to 0} \frac{F(1+ah)-F(1)}{ah} = F'(1) = \gamma$$

*Also note we have the Taylor series $F(s) = \sum_{n=0}^\infty \frac{F^{(n)}(1)}{n!} (s-1)^n$
so that
$$\sum_{k=0}^{d-1} \zeta(1+ a \exp(2 i \pi \frac kd))=  \sum_{k=0}^{d-1} \sum_{n=0}^\infty \frac{F^{(n)}(1)}{n!} a^{n-1} \exp( i (n-1) 2\pi \frac kd) \\=  \sum_{n=0}^\infty \frac{F^{(n)}(1)}{n!} a^{n-1} \underbrace{(\sum_{k=0}^{d-1} \exp( i (n-1) 2\pi \frac kd)}_{=0 \text{ if } \frac{n-1}{d} \not \in \mathbb{Z}} = d \sum_{n=0}^\infty \frac{F^{(nd+1)}}{(nd+1)!} a^{nd}$$

*Dividing by $d$ and letting $d \to \infty$ we obtain the Fourier series theorem
$$\frac{1}{2\pi}  \int_0^{2\pi} \frac{F(1+a e^{it})}{a e^{it}} dt = \frac{1}{2\pi }  \int_0^{2\pi} (\sum_{n=0}^\infty \frac{F^{(n)}(1)}{n!} a^{n-1} e^{int}) e^{-it} dt \\ =\sum_{n=0}^\infty \frac{F^{(n)}(1)}{n!}\frac{1}{2\pi}   a^{n-1}\underbrace{\int_0^{2\pi}  e^{i(n-1) t}  dt}_{= 0 \text{ if } n - 1 \ne 0}= F'(1)$$
