Slight wording adjustments.
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@@ -153,12 +153,12 @@
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\abs{\int_{\gamma_j}\frac{f(z)}{z - z_0}dz - R_j(z)} < \frac{\eps}{n}
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\]
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so
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Thus
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\[
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\abs{f(z_0) - \sum_{j = 1}^n R_j(z)} < \eps
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\]
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for all $z_0 \in K$. By the \hyperref[pole pushing lemma]{lemma:pole-pushing}, there exists $S \in \complex(z) \cap H(\complex_\infty \setminus P)$ such that $|S(z_0) - \sum_{j = 1}^n R_j(z_0)| < \eps$ for all $z_0 \in K$.
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for all $z_0 \in K$. By the \hyperref[pole pushing lemma]{lemma:pole-pushing}, there exists $S \in \complex(z) \cap H(\complex_\infty \setminus P; \complex)$ such that $|S(z_0) - \sum_{j = 1}^n R_j(z_0)| < \eps$ for all $z_0 \in K$. Therefore $|S(z_0) - f(z_0)| < 2\eps$ for all $z_0 \in K$.
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\end{proof}
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\begin{corollary}
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