I need the answers typed or in LATEX pls.Math 185 Problem Set 5. Due Sunday 10/5 at 3pm In the first...

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I need the answers typed or in LATEX pls.Math 185 Problem Set 5. Due Sunday 10/5 at 3pm In the first 5 problems of this homework set, we will study rational functions. We will use Liouville’s theorem and some polynomial algebra to develop the partial fraction decomposition theorem for rational functions and deduce some facts about complex integrals. I will begin by giving you some definicions and facts about rational functions that you can use. Definition. We will be working with complex coefficients. A rational function is a function α(z) = f(z) g(z) with f(z) = fdz d + fd−1z d−1 + · · · + f0 and g(z) = geze + ge−1ze−1 + · · · + g0(z) polynomials with complex coefficients, and g 6= 0. We say that two rational functions are equal if their values are equal wherever both are defined. So the function z1 is equal to the function z2 z , even though the latter is (technically) only defined on C ∗. Facts about polynomials and rational functions (you may assume these without proof; we will prove some of them in class.) • Every polynomial f(z) can be uniquely expressed as a product of linear factors, as follows: f(z) = fd · k∏ j=1 (z − zj)nj , where fd is the leading term, the distinct complex numbers zj are called the roots and nj is the multiplicity of the root zj . The degree d of the polynomial f (i.e., the largest d such that fd 6= 0) is equal to the sum of all the multiplicities. (This sum is also called the “number of roots counted with multiplicity”.) • For any pair of polynomials f, g with f, g 6= 0, there is a unique “division with remainder” expression f(z) g(z) = q(z) + r(z) g(z) with q(z), the polynomial quotient, is a polynomial and r(z), the residue g mod f, is a polynomial of degree less than the degree of f . The quotient r(z)g(z) is called the fractional part of the rational function f(z)g(z) . Example: z2 + 1 z + 1 = z − 1 + 2 z + 1 ; here z − 1 is the polynomial quotient and 2 is the remainder z2 + 1 mod z + 1. • An expression f(z)g(z) for a rational function is said to be in lowest terms if g(z) is monic (i.e., g(z) = 1 · ze + ge−1ze−1 + · · · + g0z0) and f, g have no roots in common. For α(z) = f(z)g(z) a rational function in lowest terms, roots of g are called the poles of the function α. The quotient formula for derivatives implies that a complex derivative exists everywhere except where the denominator is zero, i.e., α(z) is holomorphic everywhere except its poles. Problem 1. (a) Show that every rational function f(z)g(z) can be expressed as a fraction of two polynomials f ′(z) g′(z) for f ′, g′ in lowest terms (the prime here does not denote derivative). Hint: decompose f, g into linear factors. 1 (b) If fg is a rational function in lowest terms and γ is a simple closed curve which does not go through any roots of g, use either Cauchy’s theorem (for example Cauchy II) or the fundamental theorem of line integrals to show that ∫ γ f(z) g(z) dz = ∫ γ r(z) g(z) dz for r(z) = g(z) mod f(z) the remainder. (c) Assume deg(f) = d,deg(g) = e and the degrees satisfy the inequality e ≥ d. Let α(z) = f(z)g(z) . Define α(∞) = fe ge . Note in particular that if the inequality is strict e > d then α(∞) = 0 since fe = 0 (the eth term is past the highest term of f(z)). Show that lim |z|→∞ α(z) = α(∞) Here the limit statement means that you must prove that for any real � > 0 there exists real N such that |z| > N implies |α(z)− α(∞)|  deg f(z) (i.e., it is its own fractional part). Let z1, . . . , zk be the roots of g with degrees n1, . . . , nk. Let α1, . . . , αk be the singular parts of α at z1, . . . , zk. Prove that the function F (z) := α(z)− k∑ j=1 αj is an entire function with lim|z|→∞ F (z) = 0. Deduce using Liouville’s theorem that α(z) = k∑ j=1 αj(z). (d) Now let α(z) = f(z)g(z) be any rational function in lowest terms (no degree inequalities on deg f, deg g). Let z1, . . . , zk be roots of g with multiplicities n1, . . . , nk as before. Show that there is an expression α(z) = q(z) + ∑ rj(z) (z − zj)nj , for q(z) a polynomial function and rj(z) polynomial functions of degree deg rj

Rajeswari · Accepted Answer

71753 assignment
Q.no.1
a) Let a(z) =  be a rational function where f and g are polynomials in z.
If f(z) and g(z) both or any one cannot be factorised then this is the lowest form.  Otherwise if both f(z) and g(z) can be factorised and there are no common factors then this is the simplest form.
Lastly, if both are factorized and there is a common factor say z-k atleast one then we can cancel both since g(z) not equal to 0 by definition.  So z-k will not be 0 and we cancel it and get another polynomial for f and g with 1 degree less. Again repeat this process till we get no common factor.
The rational here is suppose (z-a) (z-b) (z-c) are common factors for both numerator and denominator, we first cancel z-a, and then check whether any other factor common is there, thus divide by z-b, and so on.  We stop only when the numerator and denominator does not have any common factor.
Hence it follows any rational function with polynomials in numerator and denominator can be written in the simplest form.
To explain in simpler terms we check for common factor one by one and cancel them till we do not have any common factor.
b) Given that f/g is a rational function in its lowest terms and C is a closed curve within which g not equals 0.So integral of f/g is well defined in the closed curve C
f/g can be written as by long division as a polynomial+{remainder/g(z)}
	
The first term polynomial when integrated over a closed curve the value would be 0.
Hence we get 
c) here denominator degree is more than that of numerator. Hence a(z) = f(z)/g(z) is a proper fraction.(no quotient is available)
As z tends to infinity we may have f/g if degrees are equal the ratio of leading terms or i.e. ratio of leading terms where d=e
i.e for Z>N for any large N, we can find epsilon such that
In the other case, if dd, then numerator will have some z terms in the numerator and hence when we take limit this would tend to infinity.
Q.no.2
Here we are going to use the Chinese remainder theorem as
Now consider the rational function 
==This would be equal to another rational function in which there would be no pole at zj since (z-zj) would get cancelled in simplified form.
b) When limit |z| tends to infinity is 0, i.e. whenever z becomes very large or very small then the function tends to 0.
This means denominator has a more degree than numerator since then only as per solution in question 1, f_e/g_e, where f_e =0 and hence 0.
This implies that degree of denominator is more whenever limit is 0.  In these cases, alpha(z) would tend to 0.
In other places, d>e.  
Then we have , where p is the polynomial of degree d-e.
When z tends to – or + infty ,this limit would not be 0 because of polynomial p present.
Since alpha(z) is holomorphic and defined except at the poles this polynomial is also holomorphic.

Math 185 Problem Set 5. Due Sunday 10/5 at 3pm In the first 5 problems of this homework set, we will study rational functions. We will use Liouville’s theorem and some polynomial algebra to develop...

Answer To: Math 185 Problem Set 5. Due Sunday 10/5 at 3pm In the first 5 problems of this homework set, we will...

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