= Solution
The sector contains exactly one <simple pole>, at $z_0=e^{i\pi/n}$. All roots of $1+z^n$ are simple, and its <residue> is
$$
\operatorname{Res}(f;z_0)=\frac{z_0^m}{nz_0^{n-1}}
=-\frac1n e^{i\pi(m+1)/n},
$$
since $z_0^n=-1$. For the positively oriented boundary, the <residue theorem> gives
$$
\boxed{\int_{C_R}f(z)\,dz=-\frac{2\pi i}{n}e^{i\pi(m+1)/n}}\qquad(R>1).
$$
The vertex at zero is harmless: $f$ is analytic there for nonnegative integer $m$, or a small indent may be removed because its integral is $O(\rho^{m+1})$.
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