Let $f(z) = \frac{1}{1+z^n}$ for $n \ge 3$. How can we compute poles and residue in first point over real axis?
Firstly, we have to solve $1+z^n=0$. I think, that solution of this are $$z_k = e^{2 \pi i k /n} \cdot i^{2/n}$$ for $k=0,...,n-1$.
Then, first $z_k$ over real axis will for $k=0$. Thus we would have residue in $z_0 = i^{2/n}$.
So $$Res(i^{2/n}, \frac{1}{1+z^n}) = \lim_{z \to i^{2/n}} \frac{z-i^{2/n}}{1+z^n}$$ After using L'Hôpital's rule we have $$Res(i^{2/n}, \frac{1}{1+z^n}) = \frac{1}{n \cdot \left( (i)^{2/n} \right)^{n-1}}$$
What can we do with this next? Can we simplify it?
It's probably easier to simplify it in more generality. Let $\zeta$ be a zero of $h(z) = 1+z^n$. Since all zeros of $h$ are simple, we have
$$\operatorname{Res} \left(\zeta; \frac{1}{1+z^n}\right) = \frac{1}{h'(\zeta)} = \frac{1}{n\zeta^{n-1}} = \frac{\zeta}{n\zeta^n} = -\frac{\zeta}{n}.$$