# Rationnal functions

1. Feb 26, 2006

### quasar987

Consider $p_n(z)$ and $q_d(z)$ two polynomials over $\mathbb{C}[/tex], which can be factorized like so: $$p_n(z) = a_n (z-z_1)^{n_1}...(z-z_{k})^{n_k}$$ $$q_d(z) = b_d(z-\zeta_1)^{d_1}...(z-\zeta_{m})^{d_m}$$ ([itex]\sum^k n_i =n \ \ \ \sum^m d_i =d$)

and the rationnal function $R: \mathbb{C}\cup \{\infty\} \rightarrow \mathbb{C}\cup \{\infty\}$ defined by

$$R(z) = \frac{p_n(z)}{q_d(z)}$$ if $$z \neq \zeta_i, \infty$$

$$R(\zeta_i) = \infty$$

$$R(\infty) = \left\{ \begin{array}{rcl} \infty & \mbox{if} & n>d \\ \frac{a_n}{b_n} & \mbox{if} & n=d \\ 0 & \mbox{if} & n<d \end{array}\right$$

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.

Last edited: Feb 26, 2006
2. Feb 26, 2006

### shmoe

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

3. Feb 26, 2006

### Hurkyl

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

Last edited: Feb 26, 2006
4. Feb 26, 2006

### quasar987

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

5. Feb 26, 2006

### Hurkyl

Staff Emeritus
You have to count multiplicity at infinity too.

6. Feb 26, 2006

### quasar987

What does that mean?

7. Feb 26, 2006

### Hurkyl

Staff Emeritus
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.

8. Feb 26, 2006

### 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?!

9. Feb 26, 2006

### Hurkyl

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

10. Feb 26, 2006

### 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.

11. Feb 26, 2006

### Hurkyl

Staff Emeritus
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.

12. Feb 26, 2006

### quasar987

Is that a formal definition?

13. Feb 26, 2006

### Hurkyl

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

14. Feb 26, 2006

### 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.

15. Feb 27, 2006

### 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.