A conjugate number refers to a complex number having both the imaginary and real parts of opposite signs and equal magnitude.
As much as, in these days of uncertainty, anything can be anything. As long as the constraints of a rational number are kept to, a rational number will always remain a rational number.
Yes.
Yes, but only if the rational number is 0.
Only if the rational number is 0.
It depends on what the denominator was to start with: a surd or irrational or a complex number. You need to find the conjugate and multiply the numerator by this conjugate as well as the denominator by the conjugate. Since multiplication is by [conjugate over conjugate], which equals 1, the value is not affected. If a and b are rational numbers, then conjugate of sqrt(b) = sqrt(b) conjugate of a + sqrt(b) = a - sqrt(b), and conjugate of a + ib = a - ib where i is the imaginary square root of -1.
The conjugate of an irrational number is a non-zero number such that the product of the two numbers is rational. In high school mathematics, you are usually required to know only the conjugates of surds of the form a + b*sqrt(x). The conjugate is a - b*sqrt(x) [-a + b*sqrt(x) is also a conjugate - all you need to do is to switch the sign of one of the two terms.]
Graphically, the conjugate of a complex number is its reflection on the real axis.
When a complex number is multiplied by its conjugate, the product is a real number and the imaginary number disappears.
Since the imaginary portion of a real number is zero, the complex conjugate of a real number is the same number.
It is a rational number. It can be written as a fraction.
To eliminate the radical in the denominator.
No. The product of sqrt(2) and sqrt(2) is 2, a rational number. Consider surds of the form a+sqrt(b) where a and b are rational but sqrt(b) is irrational. The surd has a conjugate pair which is a - sqrt(b). Both these are irrational, but their product is a2 - b, which is rational.
For any number (a + bi), its conjugate is (a - bi), so the real part stays the same, and the imaginary part is negated.For this one, conjugate of [-3 - 9i] is: -3 + 9i
"Conjugate" usually means that in one of two parts, the sign is changed - as in a complex conjugate. If the second part is missing, the conjugate is the same as the original number - in this case, 100.
yes
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.