Why does a rational function have exactly max{n,d} roots and poles?

[quasar987]

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

 

[shmoe]

They are counting with multiplicity, e.g. z^2 has 2 zeros at z=0.

 

[Hurkyl]

Don't forget that you are assuming that $z_i \neq \zeta_j$. (But that has nothing to do with your confusion)

 

[quasar987]

They are counting with multiplicity, e.g. z^2 has 2 zeros at z=0.

I also investigated that possibility. But even so, counting with multiplicity, R has n or n+1 roots and d or d+1 poles.

 

[Hurkyl]

You have to count multiplicity at infinity too.

 

[quasar987]

What does that mean?

 

[Hurkyl]

Your function may have a multiple root/pole at infinity, just like it may have a multiple root/pole at any other number. You have to count the multiplicity of the root/pole at infinity, just like you have to count the multiplicity of the roots/poles at all the other numbers.

 

[quasar987]

But for the roots/poles in $\mathbb{C}$, I know what their order of multiplicity are by looking at the number $n_i/d_i$ respectively. How do I know what the multiplicity is at infinity?!

 

[Hurkyl]

I don't know how your book defines the multiplicity of a root/pole at infinity. What does its definition say?

 

[quasar987]

It is not defined. I am using the definition from a linear algebra book my Lay, which says that the order of multiplicity of an eigenvalue $a$ is the power of $(\lambda-a)$ in the caracteristic polynomial.

 

[Hurkyl]

Well, as you could guess from the answer, a function that looks asymptotically like x^k has a pole of order k at infinity, and similarly for one that looks like x^-k.

 

[quasar987]

Is that a formal definition?

 

[Hurkyl]

I don't remember what the formal definition is. I just remember that that's what you want to get out of it.

 

[quasar987]

Thanks Hurky, but it seems unlike our teacher to just throw stuff at us that we can't prove for ourself very easily. I'll ask him for more details.

 

[shmoe]

The order of the pole/zero at infinity of $$f(z)$$ is usually defined to be the order of the pole/zero of $$f\left(\frac{1}{z}\right)$$ at zero.