Let \(r_1,\dots,r_n\) be distinct complex numbers. Show that every rational function of the form \[ f(z)=\frac{b_0+b_1z+\cdots+b_{n-1}z^{n-1}} {(z-r_1)(z-r_2)\cdots(z-r_n)} \] can be written \[ f(z)=\frac{A_1}{z-r_1}+\frac{A_2}{z-r_2}+\cdots+\frac{A_n}{z-r_n} \] for suitable constants \(A_1,\dots,A_n\).
Solution
Set \[ P(z)\coloneqq b_0+b_1z+\cdots+b_{n-1}z^{n-1}, \qquad Q(z)\coloneqq\prod_{j=1}^n(z-r_j). \]
1. For each \(i\), define\[ A_i \coloneqq \frac{P(r_i)} {\displaystyle\prod_{\substack{1\le j\le n\\j\ne i}}(r_i-r_j)}. \]These constants are well-defined.
The numbers \(r_1,\ldots,r_n\) are distinct, so every factor \[ r_i-r_j \] with \(j\ne i\) is nonzero. Hence the denominator in the definition of \(A_i\) is nonzero.
2. The polynomial identity\[ P(z) = \sum_{i=1}^n A_i \prod_{\substack{1\le j\le n\\j\ne i}}(z-r_j) \]holds for every \(z\in\mathbb C\).
Let \[ H(z) \coloneqq P(z) - \sum_{i=1}^n A_i \prod_{j\ne i}(z-r_j). \] Each summand has degree at most \(n-1\), so \[ \deg H\le n-1 \] unless \(H=0\) already.
Fix \(k\in\{1,\ldots,n\}\). Evaluating at \(z=r_k\), every summand in the sum vanishes except the one with \(i=k\). Thus \[ \begin{aligned} H(r_k) &= P(r_k) - A_k\prod_{j\ne k}(r_k-r_j)\\ &= P(r_k)-P(r_k) =0, \end{aligned} \] by the definition in step 1.
Hence \(H\) has the \(n\) distinct roots \[ r_1,\ldots,r_n. \] A nonzero polynomial of degree at most \(n-1\) cannot have \(n\) distinct roots, so \(H\equiv0\).
3. Dividing the identity in step 2 by \(Q(z)\) gives\[ \boxed{ f(z) = \frac{A_1}{z-r_1} +\cdots+ \frac{A_n}{z-r_n} }. \]
For \(z\notin\{r_1,\ldots,r_n\}\), \[ \frac1{Q(z)} \prod_{j\ne i}(z-r_j) = \frac1{z-r_i}. \] Dividing step 2 by \(Q(z)\) therefore yields the displayed identity on the common domain of the rational functions.
4. Q.E.D.
Step 3 gives the required partial-fraction decomposition with the explicit constants from step 1.