= Solution
Let $M(x)=\sum_{n\le x}\mu(n)$ be the <Mertens function>. <Partial summation> gives
$$
\sum_{n\le X}\frac{\mu(n)}{n^s}=M(X)X^{-s}+s\int_1^X M(x)x^{-s-1}\,dx.
$$
For any $\Re s>0.9$, choose a sufficiently small $\epsilon>0$ with $0.9+\epsilon<\Re s$. The assumed bound makes the boundary term tend to zero and makes the integral absolutely convergent. Moreover, on each <compact set> of this half-plane, one can choose the same $\epsilon$, so the integral has <locally uniform convergence>. Hence
$$
F(s)=s\int_1^\infty M(x)x^{-s-1}\,dx
$$
is a <holomorphic function> on $\Re s>0.9$, agreeing with the reciprocal <Dirichlet series> from part (b) on $\Re s>1$.
The <identity theorem> applied to the <holomorphic functions> $(s-1)\zeta(s)F(s)$ and $s-1$ extends their equality throughout $\Re s>0.9$; the assumed simple <pole> makes the first function holomorphic at $1$. Thus $\zeta(s)F(s)=1$ away from $1$, proving \b[a zero-free half-plane]:
$$
\boxed{\zeta(s)\ne0\quad\text{if }\Re s>0.9,\ s\ne1.}
$$
To obtain the strongest symmetric conclusion, use the following basic facts explicitly: the <functional equation of the Riemann zeta function> gives the <entire function> $\xi(s)=\tfrac12s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s)$ with $\xi(s)=\xi(1-s)$, and the <trivial zeros of the Riemann zeta function> are the negative even integers. Every <Nontrivial zero of the Riemann zeta function> is therefore accompanied by its reflection $1-s$. A <Nontrivial zero of the Riemann zeta function> with real part below $0.1$ would reflect to one above $0.9$, which is impossible. Consequently \b[all nontrivial zeros lie in the closed strip]
$$
\boxed{0.1\le\Re\rho\le0.9.}
$$
The $\epsilon$ in the hypothesis prevents this argument from excluding either boundary line; the <trivial zeros of the Riemann zeta function> remain present.
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