= Solution
Along $z=e^{i\theta}$ with $0<\theta<2\pi$, the chosen branch has $L(z)-i\pi=i(\theta-\pi)$. Use part (iii) and the circular estimate in part (a):
$$
\left|\int_0^1f(x)\,dx\right|\leq\frac1{2\pi}\int_0^{2\pi}|f(e^{i\theta})|\,|\theta-\pi|\,d\theta\leq\boxed{\frac12\int_0^{2\pi}|f(e^{i\theta})|\,d\theta.}
$$
The second step uses $|\theta-\pi|\leq\pi$. The single branch-cut point does not affect the integral. Keeping that weight gives the sharper intermediate bound.
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