answersLogoWhite

0

One significant feature of complex numbers is that all polynomial equations of order n, in the complex field, have n solutions. When multiple roots are

Given any set of complex numbers {a(0),  … , a(n)}, such that at least one of a(1) to a(n) is non-zero, the equation

a(n)*z^n + a(n-1)*z^(n-1) + ... + a(0) has at least one solution in the complex field.

This is the Fundamental Theorem of Algebra and establishes the set of Complex numbers as a closed field.

[a(0), ... , a(n) should be written with suffices but this browser has decided not to be cooperative!]

The above solution is the complex root of the equation.

In fact, if the equation is of order n, that is, if the coefficient a(n) is non-zero then, taking account of the multiplicity, the equation has exactly n roots (some of which may be real).

User Avatar

Wiki User

10y ago

Still curious? Ask our experts.

Chat with our AI personalities

ProfessorProfessor
I will give you the most educated answer.
Chat with Professor
FranFran
I've made my fair share of mistakes, and if I can help you avoid a few, I'd sure like to try.
Chat with Fran
RossRoss
Every question is just a happy little opportunity.
Chat with Ross

Add your answer:

Earn +20 pts
Q: What is the significance of complex roots?
Write your answer...
Submit
Still have questions?
magnify glass
imp
Continue Learning about Basic Math

What is the relationship between the degree of a polynomial and the number of roots it has?

In answering this question it is important that the roots are counted along with their multiplicity. Thus a double root is counted as two roots, and so on. The degree of a polynomial is exactly the same as the number of roots that it has in the complex field. If the polynomial has real coefficients, then a polynomial with an odd degree has an odd number of roots up to the degree, while a polynomial of even degree has an even number of roots up to the degree. The difference between the degree and the number of roots is the number of complex roots which come as complex conjugate pairs.


What are the roots of complex numbers in mathematics?

See the answer to the related question: 'How do you solve the power of an imaginary number?' (Link below)


Square root of 17 in complex number terms?

A positive real number, such as 17, has two square roots. One is the one your calculator gives you, if you use the square root function. The other is the same number, with a minus sign in front. None of these has an imaginary part. There are no additional complex roots that have a non-zero imaginary part.


How could you use Descartes' rule to predict the number of complex roots to a polynomial?

Descartes' rule of signs will not necessarily tell exact number of complex roots, but will give an idea. The Wikipedia article explains it pretty well, but here is a brief explanation:It is for single variable polynomials.Order the polynomial in descending powers [example: f(x) = ax³ + bx² + cx + d]Count number of sign changes between consecutive coefficients. This is the maximum possible real positive roots.Let function g(x) = f(-x), count number of sign changes, which is maximum number of real negative roots.Note these are maximums, not the actual numbers. Let p = positive maximum and q = negative maximum. Let m be the order (maximum power of the variable), which is also the total number of roots.So m - p - q = minimum number of complex roots. Note complex roots always occur in pairs, so number of complex roots will be {0, 2, 4, etc}.


What is a polynomial that does not factor over the real numbers referred to as?

There is no specific term for such polynomials. They may be referred to as are polynomials with only purely complex roots.