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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.
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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).
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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.
What is the significance of the study of educational world roots to American education?
What is the significance of the study of educational world roots to American education?
How many complex roots does 6x4 plus x3 - 5 equals 0 have?
It has two complex roots.
What are complex root?
The complex roots of an equation are the complex numbers that are solutions to the equation.
How many roots in a radical problem if the index is odd?
An odd number. In the complex field, the number of roots is the same as the index. Complex (non-real) roots come in pairs (complex conjugates) so the number of real roots will also be odd.
What is the significance of cube and cube roots?
you are my friend
What do you think is true of the square roots of a complex number?
I posted an answer about cube roots of complex numbers. The same info can be applied to square roots. (see related links)
What does the discriminant tell you when solving quadratic equations for the roots?
Whether the equation has 2 distinct roots, repeated roots, or complex roots. If the determinant is smaller than 0 then it has complex roots. If the determinant is 0 then it has repeated roots. If the determinant is greater than 0 then it has two distinct roots.
If the discriminant is negative the equation has?
No real roots but the roots are a pair of complex conjugates.
How do you simplify fourth roots?
The answer will depend on the form of the fourth root. Positive real numbers will have two fourth roots which are real and two that are complex. Complex numbers will have four complex roots. However, none of these can be "simplified" in the normal sense of the term.
Can you have a quadratic function with one real root and one complex root?
No you can not. Complex roots appear as conjugates. if a root is complex so is its conjugate. so either the roots are real or are both coplex.
If the roots are real then which type of vibrations will occur in damped systems?
real roots= Overdamped equal roots= critically damped complex roots /imaginary roots = Underdamped
What the square roots of -256?
There are no real square roots of -256. But using complex numbers the square roots of -256 are 16i and -16i.
