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- Thread starter Ratzinger
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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.

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.

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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?

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CompuChip

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[tex]

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}},

[/tex]

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HallsofIvy

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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)

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