[proofplan] The naive series $\sum_{j=1}^{\infty} P_j(1/(z - z_j))$ has the correct principal parts but need not converge. We fix this by subtracting from each $P_j$ a "correcting polynomial" $Q_j$ — a partial Taylor expansion of $P_j$ around $0$ — that makes the difference $P_j - Q_j$ small for $|z|$ bounded. Specifically, for each $j$ with $z_j \neq 0$, we expand $P_j(1/(z - z_j))$ in a Taylor series around $z = 0$ and truncate to degree $n_j$, choosing $n_j$ large enough to make $|P_j(1/(z - z_j)) - Q_j(z)| < 2^{-j}$ on $\overline{B}(0, j)$. The corrected series $\sum_j (P_j - Q_j)$ then converges uniformly on compact subsets of $\mathbb{C} \setminus \{z_j\}$, giving a meromorphic function with the desired principal parts. Uniqueness up to an entire function follows because the difference of two solutions is holomorphic everywhere. [/proofplan] [step:Separate the possible pole at $z = 0$ and set up the correcting polynomials] If $z_j = 0$ for some $j$ (at most one, since $|z_j| \to \infty$ implies the $z_j$ are eventually non-zero), relabel so that $z_0 = 0$ is this point, and set $P_0 := P_0(1/z)$. Since $|z_j| \to \infty$, for each $R > 0$ only finitely many $z_j$ satisfy $|z_j| \le R$. For each $j \ge 1$ with $z_j \neq 0$, the function $z \mapsto P_j(1/(z - z_j))$ is holomorphic on $\mathbb{C} \setminus \{z_j\}$, and in particular on $B(0, |z_j|)$. By [Taylor's Theorem for Holomorphic Functions](/theorems/348), $P_j(1/(z - z_j))$ has a convergent Taylor expansion around $z = 0$ valid on $B(0, |z_j|)$. Define $Q_j: \mathbb{C} \to \mathbb{C}$ to be the partial sum of this Taylor series up to degree $n_j$, where $n_j \in \mathbb{N}$ is to be chosen. [guided] The construction begins with the observation that the naive sum $\sum_{j=1}^{\infty} P_j(1/(z - z_j))$ — which formally has all the correct principal parts — may diverge because the "tails" of the $P_j$ can contribute unbounded terms far from the pole $z_j$. The fix is to subtract from each $P_j$ a polynomial $Q_j(z)$ that does not change the principal part at $z_j$ (since $Q_j$ is entire, it has no poles) but cancels enough of $P_j$'s growth on compact sets to force convergence. Where does $Q_j$ come from? Since $P_j(1/(z - z_j))$ is holomorphic in a neighbourhood of $0$ (specifically on $B(0, |z_j|)$ for $z_j \neq 0$), it has a Taylor expansion around $z = 0$ by [Taylor's Theorem for Holomorphic Functions](/theorems/348). The polynomial $Q_j$ is a truncation of this Taylor series to degree $n_j$. If $z_j = 0$ for some $j$, the principal part $P_j(1/z)$ already has its pole at the origin, and no correction is needed — we simply include it directly. Since $|z_j| \to \infty$, there is at most one such $j$, and we may relabel it as $j = 0$. [/guided] [/step] [step:Choose the truncation degrees $n_j$ to ensure rapid decay of $P_j - Q_j$ on compact sets] Fix $j \ge 1$ with $z_j \neq 0$. The Taylor series of $P_j(1/(z - z_j))$ converges to $P_j(1/(z - z_j))$ uniformly on $\overline{B}(0, \rho)$ for any $0 < \rho < |z_j|$. In particular, since $|z_j| \to \infty$, for all sufficiently large $j$ we have $|z_j| > j$, so the Taylor series converges uniformly on $\overline{B}(0, j)$. Choose $n_j \in \mathbb{N}$ large enough that the $n_j$-th partial sum $Q_j(z)$ satisfies \begin{align*} \sup_{|z| \le j} \left| P_j\!\left(\frac{1}{z - z_j}\right) - Q_j(z) \right| < 2^{-j}. \end{align*} This is possible because the Taylor partial sums converge uniformly to $P_j(1/(z - z_j))$ on $\overline{B}(0, j)$. For the finitely many $j$ with $|z_j| \le j$, we simply set $Q_j := 0$ (these terms will be handled separately as a finite sum). [guided] Why can we achieve the bound $2^{-j}$? The function $z \mapsto P_j(1/(z - z_j))$ is holomorphic on $B(0, |z_j|)$, so by [Taylor's Theorem for Holomorphic Functions](/theorems/348), its Taylor series around $z = 0$ converges on the entire disc $B(0, |z_j|)$. By the definition of convergence of a power series, the partial sums converge uniformly on any closed sub-disc $\overline{B}(0, \rho)$ with $\rho < |z_j|$. Since $|z_j| \to \infty$, for all sufficiently large $j$ we have $|z_j| > j + 1 > j$, so the closed disc $\overline{B}(0, j)$ lies strictly inside the disc of convergence. The uniform convergence on $\overline{B}(0, j)$ means that for large enough $n_j$, the remainder $|P_j(1/(z - z_j)) - Q_j(z)|$ is bounded by any prescribed $\varepsilon > 0$ on $\overline{B}(0, j)$. Taking $\varepsilon = 2^{-j}$ gives the desired estimate. The choice of $2^{-j}$ is not special — any summable sequence would work. The point is to get $\sum_j \sup_{|z| \le j} |P_j - Q_j| < \infty$ so that the [Weierstrass M-Test](/theorems/261) applies. For the finitely many $j$ with $|z_j| \le j$, no correction is needed: these contribute a finite sum, which is automatically meromorphic with the correct poles. [/guided] [/step] [step:Prove that the corrected series converges uniformly on compact subsets of $\mathbb{C} \setminus \{z_j\}$] Define the candidate meromorphic function by \begin{align*} f(z) := P_0\!\left(\frac{1}{z}\right) + \sum_{j \in F} P_j\!\left(\frac{1}{z - z_j}\right) + \sum_{j \notin F,\, j \ge 1} \left[ P_j\!\left(\frac{1}{z - z_j}\right) - Q_j(z) \right], \end{align*} where $F$ is the finite set of indices $j \ge 1$ with $|z_j| \le j$ (for which we set $Q_j = 0$), and $P_0(1/z)$ is included only if some $z_j = 0$. Let $K \subset \mathbb{C} \setminus \{z_j : j \in \mathbb{N}\}$ be compact. Choose $N \in \mathbb{N}$ such that $K \subset B(0, N)$ and $|z_j| > N$ for all $j > N$ (possible since $|z_j| \to \infty$). For $j > N$ with $j \notin F$, we have $|z_j| > j \ge N$, so $z_j \notin \overline{B}(0, N) \supset K$, and the bound from the previous step gives \begin{align*} \sup_{z \in K} \left| P_j\!\left(\frac{1}{z - z_j}\right) - Q_j(z) \right| \le \sup_{|z| \le j} \left| P_j\!\left(\frac{1}{z - z_j}\right) - Q_j(z) \right| < 2^{-j}, \end{align*} where the first inequality uses $K \subset B(0, N) \subset B(0, j)$ for $j > N$. Since $\sum_{j > N} 2^{-j} < \infty$, the [Weierstrass M-Test](/theorems/261) implies that the tail $\sum_{j > N,\, j \notin F} (P_j(1/(z - z_j)) - Q_j(z))$ converges uniformly on $K$. The remaining terms (the finite set of indices $j \le N$ or $j \in F$) form a finite sum of functions each holomorphic on $K$ (since $z_j \notin K$). Therefore the full series converges uniformly on $K$. [guided] We must show that $f$ is well-defined and converges uniformly on compact subsets of the punctured plane $\mathbb{C} \setminus \{z_j\}$. The first sum ($P_0(1/z)$) and the sum over $F$ are finite and contribute no convergence issues. For the infinite series, let $K \subset \mathbb{C} \setminus \{z_j\}$ be compact. Since $K$ is bounded and closed, $K \subset B(0, N)$ for some $N$. Since $|z_j| \to \infty$, only finitely many $z_j$ satisfy $|z_j| \le N$, so we can choose $N$ large enough that $|z_j| > N$ for all $j > N$. For such $j$ with $j \notin F$: - The inequality $j > N$ and $K \subset B(0, N) \subset B(0, j)$ gives $\sup_{z \in K} |P_j(1/(z - z_j)) - Q_j(z)| \le \sup_{|z| \le j} |P_j(1/(z - z_j)) - Q_j(z)| < 2^{-j}$. We apply the [Weierstrass M-Test](/theorems/261) with $M_j := 2^{-j}$. The condition $\sum_{j > N} M_j = \sum_{j > N} 2^{-j} < \infty$ is satisfied, so the series $\sum_{j > N,\, j \notin F} (P_j(1/(z - z_j)) - Q_j(z))$ converges uniformly on $K$. The finitely many remaining terms (indices $j \le N$ or $j \in F$) each have poles only at the corresponding $z_j$, which lie outside $K$ (by definition of $K \subset \mathbb{C} \setminus \{z_j\}$), so they are holomorphic on an open set containing $K$. Adding finitely many such functions to a uniformly convergent series does not affect uniform convergence on $K$. [/guided] [/step] [step:Verify that $f$ is meromorphic with the prescribed principal parts] Since the series converges uniformly on compact subsets of $\mathbb{C} \setminus \{z_j\}$, and each summand $P_j(1/(z - z_j)) - Q_j(z)$ is holomorphic on $\mathbb{C} \setminus \{z_j\}$, the limit function is holomorphic on $\mathbb{C} \setminus \{z_j : j \in \mathbb{N}\}$ by [Preservation of Holomorphicity under Locally Uniform Convergence](/theorems/365). At each $z_j$, isolate the term with the pole: write \begin{align*} f(z) = P_j\!\left(\frac{1}{z - z_j}\right) + h_j(z), \end{align*} where $h_j(z)$ is the sum of all other terms (including the $-Q_j(z)$ correction for the $j$-th term). Each other summand $P_i(1/(z - z_i)) - Q_i(z)$ for $i \neq j$ is holomorphic at $z_j$ (since $z_i \neq z_j$), and the partial sum over $i \neq j$ converges uniformly on compact subsets of a neighbourhood of $z_j$. The function $-Q_j(z)$ is a polynomial, hence entire. Therefore $h_j$ is holomorphic at $z_j$. Since $P_j(1/(z - z_j))$ is precisely the prescribed principal part (a polynomial in $(z - z_j)^{-1}$ with no constant term), the Laurent expansion of $f$ at $z_j$ has principal part exactly $P_j(1/(z - z_j))$. [guided] We verify two things: (1) $f$ is holomorphic away from the $z_j$, and (2) the principal part of $f$ at each $z_j$ is exactly $P_j(1/(z - z_j))$. **(1) Holomorphicity.** On any compact subset $K$ of $\mathbb{C} \setminus \{z_j\}$, the series converges uniformly. Each partial sum is holomorphic on $K$ (each summand is holomorphic away from its own pole, and none of the $z_j$ lie in $K$). By [Preservation of Holomorphicity under Locally Uniform Convergence](/theorems/365), the uniform limit of holomorphic functions is holomorphic. Therefore $f$ is holomorphic on $\mathbb{C} \setminus \{z_j\}$. **(2) Correct principal parts.** Fix $z_j$. We decompose $f$ as \begin{align*} f(z) = P_j\!\left(\frac{1}{z - z_j}\right) + h_j(z), \end{align*} where $h_j$ collects all other terms. Why is $h_j$ holomorphic at $z_j$? Each summand $P_i(1/(z - z_i)) - Q_i(z)$ for $i \neq j$ is holomorphic at $z_j$ because $z_i \neq z_j$ (the poles are distinct since $|z_j| \to \infty$ forces the sequence to have distinct terms eventually, and any repeated values can be combined). The correction $-Q_j(z)$ is a polynomial and thus entire. The convergence of these remaining terms is uniform on compact subsets of a neighbourhood of $z_j$ (by the same Weierstrass M-Test argument, since $z_j$ is not a pole of any of the other terms). By [Preservation of Holomorphicity under Locally Uniform Convergence](/theorems/365), $h_j$ is holomorphic at $z_j$. Since $P_j(1/(z - z_j))$ is a polynomial in $(z - z_j)^{-1}$ with no constant term, and $h_j$ is holomorphic at $z_j$, the Laurent expansion of $f$ at $z_j$ has principal part exactly $P_j(1/(z - z_j))$. [/guided] [/step] [step:Prove uniqueness up to an entire function] Let $f$ and $\tilde{f}$ be two meromorphic functions on $\mathbb{C}$ with poles exactly at the $z_j$ and with the same principal parts $P_j(1/(z - z_j))$ at each $z_j$. Define $g := f - \tilde{f}$. At each $z_j$, the principal parts cancel: \begin{align*} g(z) = \left[P_j\!\left(\frac{1}{z - z_j}\right) + h_j(z)\right] - \left[P_j\!\left(\frac{1}{z - z_j}\right) + \tilde{h}_j(z)\right] = h_j(z) - \tilde{h}_j(z), \end{align*} where $h_j$ and $\tilde{h}_j$ are holomorphic at $z_j$. Therefore $g$ extends holomorphically across each $z_j$, so $g$ is holomorphic on all of $\mathbb{C}$, i.e., $g$ is entire. [/step]