# Rational functions

1. Aug 10, 2007

### Ratzinger

In this article, how do I rewrite (2) to get the third equation R(z)=... ?

thank you

2. Aug 10, 2007

### threetheoreom

remember the rational function will produce a quotient, which when multiplied by the divisor will yeild the original R(z). So the second expression is in the form R(z) = quotient x divisor.

More specifically (z-a_j)^uj X S_j(z) .. Where as they said S_j(z) is the rational fuction ( which was the quotient ). It is not that much about "deriving" the third form but more showing that the complex function R(z) is a product of the number of roots ( z-a_j) and the quotient S_j(z).

My 2 cents - correct if neccesary.

Last edited: Aug 10, 2007
3. Aug 11, 2007

### Ratzinger

But what do you mean by qoutient produced by a rational function? The rational is a quotient of two polynomials, so what quotient is it producing?

4. Aug 11, 2007

### CompuChip

Take
$$S_j(z) = \frac{a_0(z-\alpha_1)^{\mu_1}(z-\alpha_2)^{\mu_2}\cdots(z-\alpha_{j - 1})^{\mu_{j - 1}}(z-\alpha_{j + 1})^{\mu_{j + 1}}\cdots(z-\alpha_r)^{\mu_r}} {b_0(z-\beta_1)^{\nu_1}(z-\beta_2)^{\nu_2}\ldots(z-\beta_s)^{\nu_s}},$$

5. Aug 11, 2007

### HallsofIvy

Staff Emeritus
The equation labled (3) is not derived directly from equation (2). What they have done is write the product of all terms in the numerator of (2) as P(z) and the product of all terms in the denominator as Q(z):
R(z)= P(z)/Q(z).

Then they look at R(z)- c= P(z)/Q(z)- c. Getting the common denominator (Q(z)) you have P(z)/Q(z)- cQ(z)/Q(z)= (P(z)- cQ(z))/Q(z)