- #1
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Consider [itex]p_n(z)[/itex] and [itex]q_d(z)[/itex] two polynomials over [itex]\mathbb{C}[/tex], which can be factorized like so:
[tex]p_n(z) = a_n (z-z_1)^{n_1}...(z-z_{k})^{n_k}[/tex]
[tex]q_d(z) = b_d(z-\zeta_1)^{d_1}...(z-\zeta_{m})^{d_m}[/tex]
([itex]\sum^k n_i =n \ \ \ \sum^m d_i =d[/itex])
and the rationnal function [itex]R: \mathbb{C}\cup \{\infty\} \rightarrow \mathbb{C}\cup \{\infty\}[/itex] defined by
[tex]R(z) = \frac{p_n(z)}{q_d(z)}[/tex] if [tex]z \neq \zeta_i, \infty[/tex]
[tex] R(\zeta_i) = \infty[/tex]
[tex]R(\infty) = \left\{ \begin{array}{rcl}
\infty & \mbox{if}
& n>d \\ \frac{a_n}{b_n} & \mbox{if} & n=d \\
0 & \mbox{if} & n<d
\end{array}\right[/tex]
I fail to see why R(z) has exactly [itex]max\{n,d\}[/itex] roots and poles. It seems to me the number of roots is equal to k or k+1 in the case of n<d and the number of poles is m or m+1 in the case of n>d.
[tex]p_n(z) = a_n (z-z_1)^{n_1}...(z-z_{k})^{n_k}[/tex]
[tex]q_d(z) = b_d(z-\zeta_1)^{d_1}...(z-\zeta_{m})^{d_m}[/tex]
([itex]\sum^k n_i =n \ \ \ \sum^m d_i =d[/itex])
and the rationnal function [itex]R: \mathbb{C}\cup \{\infty\} \rightarrow \mathbb{C}\cup \{\infty\}[/itex] defined by
[tex]R(z) = \frac{p_n(z)}{q_d(z)}[/tex] if [tex]z \neq \zeta_i, \infty[/tex]
[tex] R(\zeta_i) = \infty[/tex]
[tex]R(\infty) = \left\{ \begin{array}{rcl}
\infty & \mbox{if}
& n>d \\ \frac{a_n}{b_n} & \mbox{if} & n=d \\
0 & \mbox{if} & n<d
\end{array}\right[/tex]
I fail to see why R(z) has exactly [itex]max\{n,d\}[/itex] roots and poles. It seems to me the number of roots is equal to k or k+1 in the case of n<d and the number of poles is m or m+1 in the case of n>d.
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