# Prescribing zeros and poles of a rational function on $\bar {\mathbb C}$ at once

I have to show that for any points $P_1$, $\ldots$, $P_n$ and $Q_1$, $\ldots$, $Q_n$ ($P_i \neq Q_j$ for all $i$, $j$) on $\bar{\mathbb C}$ there exists a rational function $f$ with poles at $P_j$, $j = 1$, $\ldots$, $n$ (with order of pole $P$ = number of points $P_j = P$) and with zeros at $Q_j$, $j = 1$, $\ldots$, $n$ (with order of zero $Q$ = number of points $Q_j = Q$). For me it is clear how to construct a function with given poles or with given zeros separately (using Mittag-Leffler's or Weierstrass' theorem). It's also clear how to build a rational function with given zeros and poles on $\mathbb C$ without a zero or a pole at $\infty$. We can also build a function with given poles using Mittag-Leffler, find it zeros, divide and multiply by appropriate linear functions to get a desired function. But this procedure is not explicit. But is it possible to do it explicitly?

• If none of the $P_i$ or $Q_j$ is $\infty$, a product representation is straightforward. Is your problem what to do if some $P_i$ or $Q_j$ is/are $\infty$? – Daniel Fischer Oct 22 '14 at 19:34
• @DanielFischer have you read the question? I wrote that this case is clear – Appliqué Oct 22 '14 at 20:01
• Ah, sorry, that sentence must have hidden. – Daniel Fischer Oct 22 '14 at 20:02
• The explicit construction is $$f(z) = \frac{\displaystyle\prod_{\substack{i=1\\Q_i\neq\infty}}^n (z-Q_i)}{\displaystyle\prod_{\substack{i=1\\P_i\neq\infty}}^n (z-P_i)}.$$ – Daniel Fischer Oct 22 '14 at 20:05
• @DanielFischer, thank you very much! – Appliqué Oct 22 '14 at 20:31

If you already know how to build your given function when none of the specified points are at $\infty$, then just rotate the Riemann sphere such that a point distinct from the $P$s and $Q$s end up at $\infty$, construct your function with the rotated points, and then compose it with the opposite rotation.