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Q: Complex roots occur in conjugate pairs?

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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.

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.

It is a polynomial of odd power - probably a cubic. It has only one real root and its other two roots are complex conjugates. It could be a polynomial of order 5, with two points of inflexion, or two pairs of complex conjugate roots. Or of order 7, etc.

In the complex field, the two numbers are the same. If you restrict yourself to real solutions, the relationship is as follows: A polynomial of degree p has p-2k real solutions where k is an integer such that p-2k is non-negative. [There will be 2k pairs of complex conjugate roots.]

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.

No. Complex zeros always come in conjugate pairs. So if a+bi is one zero, then a-bi is also a zero.The fundamental theorem of algebra says"Every polynomial equation of degree n with complex coefficients has n roots in the complex numbers."If you want to know how many complex root a given polynomial has, you might consider finding out how many real roots it has. This can be done with Descartes Rules of signsThe maximum number of positive real roots can be found by counting the number of sign changes in f(x). The actual number of positive real roots may be the maximum, or the maximum decreased by a multiple of two.The maximum number of negative real roots can be found by counting the number of sign changes in f(-x). The actual number of negative real roots may be the maximum, or the maximum decreased by a multiple of two.Complex roots always come in pairs. That's why the number of positive or number of negative roots must decrease by two. Using the two rules for positive and negative signs along with the fact that complex roots come in pairs, you can determine the number of complex roots.

There cannot be such a polynomial. If a polynomial has rational coefficients, then any complex roots must come in conjugate pairs. In this case the conjugate for 2-3i is not a root. Consequently, either (a) the function is not a polynomial, or (b) it does not have rational coefficients, or (c) 2 - 3i is not a root (nor any other complex number), or (d) there are other roots that have not been mentioned. In the last case, the polynomial could have any number of additional (unlisted) roots and is therefore indeterminate.

In the case of real roots, you could, but the second part of the ordered pair (the ordinate) will always be zero, so there is not much point.In the case of complex roots (or real roots in the complex field), you could list them as ordered pairs: with (a, b) representing a + bi where i is the imaginary square root of -1..

real roots= Overdamped equal roots= critically damped complex roots /imaginary roots = Underdamped

Sort of... but not entirely. Assuming the polynomial's coefficients are real, the polynomial either has as many real roots as its degree, or an even number less. Thus, a polynomial of degree 4 can have 4, 2, or 0 real roots; while a polynomial of degree 5 has either 5, 3, or 1 real roots. So, polynomial of odd degree (with real coefficients) will always have at least one real root. For a polynomial of even degree, this is not guaranteed. (In case you are interested about the reason for the rule stated above: this is related to the fact that any complex roots in such a polynomial occur in conjugate pairs; for example: if 5 + 2i is a root, then 5 - 2i is also a root.)

It can be used as a convenient shortcut to calculate the absolute value of the square of a complex number. Just multiply the number by its complex conjugate.I believe it has other uses as well.

It's the same as the negative of the cube root of 2.

It has two complex roots.

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) = axÂ³ + bxÂ² + 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}.

The complex roots of an equation are the complex numbers that are solutions to the equation.

I posted an answer about cube roots of complex numbers. The same info can be applied to square roots. (see related links)

yes it does occur in the roots as it takes nutrients from the soil.

A rational equation can be multiplied by the least common multiple of its denominators to make it into a polynomial equation. The degree of this polynomial is the highest power (of the variable) that appears in it. It can be proven that a polynomial of degree n must have n roots in the complex domain. However, there may be fewer roots in the real domain. This is because if the coefficients are real then there may be pairs of complex roots [conjugates] which will not count as real roots. Also, there may be identical roots of multiple order. For example, x4 - 1 = 0 has 4 complex roots. These are 1, -1, i and -i where i is the imaginary root of -1. There are only 2 real roots -1 and +1. x4 = 0 has 4 multiple roots, each of which is 0. Thus x = 0 is a root of multiplicity 4.

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.

5n2 + 2n + 6 = 0 is a quadratic equation in the variable n.The equation does not have any real roots. The roots are the complex conjugate pair -0.2 Â± 1.077i where i is the imaginary square root of -1.

No real roots but the roots are a pair of complex conjugates.

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.

Short answer: No.Assuming that the original quadratic is completely real, complex roots always come in conjugate pairs - meaning that if you multiplied both of the complex roots together, you would get a real number. For example, if one root was 2 + 3i, then you know that another root will be 2 - 3i, because those two multiplied together give you -5 (thanks to (x2 - y2) = (x+y)(x-y). You see how math all fits together? It's great!)Therefore, a real quadratic can only have 2 real roots or 2 complex roots. If you have one of each, either something has gone horribly wrong or your teacher is a sadist.Also, bear in mind that I've only done A level (American translation: late high school/early college) math, so this might be wrong if you're past that level.

31 pairs. And they arise from the combination of the ventral and dorsal roots.

No.The roots are the complex conjugate pair 5 Â± 2.4495iwhere i is the imaginary square root of -1.

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