= Solution
Apply the Hankel representation with $z$ replaced by $1-z$:
$$
\zeta(1-z)=\frac{\Gamma(z)}{2\pi i}
\int_H\frac{t^{-z}}{e^{-t}-1}\,dt.
$$
Enlarge $H$ to a modified Hankel contour that encloses the simple poles $t=2\pi i n$, $n\in\mathbb Z\setminus\{0\}$. In a strip where the deformation and residue sum converge, the outer pieces vanish and the orientation gives the Hankel integral as minus $2\pi i$ times the sum of the enclosed residues. Part (b) gives
$$
\begin{aligned}
\sum_{n\ne0}\operatorname{Res}_{t=2\pi i n}
\frac{t^{-z}}{e^{-t}-1}
&=-\sum_{n=1}^{\infty}\left[(2\pi i n)^{-z}+(-2\pi i n)^{-z}\right]\\
&=-2(2\pi)^{-z}\cos\!\left(\frac{\pi z}{2}\right)\zeta(z).
\end{aligned}
$$
Substitution therefore yields
$$
\boxed{\zeta(1-z)=2^{1-z}\pi^{-z}
\cos\!\left(\frac{\pi z}{2}\right)\Gamma(z)\zeta(z).}
$$
Both sides are meromorphic functions of $z$, so the <identity theorem> extends the equality from the initial strip to every point where either side is defined.
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