= Solution
Fix $\operatorname{Im}\lambda>-c$. Truncate $C_1$ at $\pm R-ic$ and close it with the upper semicircle centred at $-ic$, oriented counterclockwise. For large $R$ the only pole of the integrand inside is $z=\lambda$. Uniformly on the arc, $|z-\lambda|$ is comparable to $R$ and $f(z)=o(R^{-1})$, so the arc integral is $o(R^{-1})$ and tends to zero. The line integral is absolutely convergent at infinity for the same decay reason. The <Cauchy integral formula> gives
$$
\boxed{F(\lambda)=f(\lambda),\qquad\operatorname{Im}\lambda>-c}.
$$
The right-hand side is <holomorphic> on this open half-plane by the hypothesis, proving the required <holomorphic function>. The contour never passes through $\lambda$ because its imaginary part is strictly above the line.
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