Q: Complex roots occur in conjugate pairs?

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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.]

There are 3 cube roots and these are:the real root -1.2599and the complex roots 0.6300 - 1.0911i and its conjugate, 0.6300 + 1.0911i.

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.

The equation is:x3 - 4x2i - 9xi2+ 36i3 = 0Note: The ' i ' must be an ordinary constant, and can't be sqrt(-1).If it were sqrt(-1), then -4i would also be a root of the equation.Imaginary or complex roots always occur in conjugate pairs.If -4i is also a root of the equation and the questioner just forgot to include it,then ' i ' can be sqrt(-1), and the equation can bex4 + 25x2 + 144 = 0

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.

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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.

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.]

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.

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.

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

There are 3 cube roots and these are:the real root -1.2599and the complex roots 0.6300 - 1.0911i and its conjugate, 0.6300 + 1.0911i.

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.

A cubic function is a polynomial function of degree 3. So the graph of a cube function may have a maximum of 3 roots. i.e., it may intersect the x-axis at a maximum of 3 points. Since complex roots always occur in pairs, a cubic function always has either 1 or 3 real zeros.