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The complex conjugate of a plus bi is?

a-bi a(bi)-1 not negative bi


Is the difference of a complex number and it's conjugate an imaginary number?

Yes. By definition, the complex conjugate of a+bi is a-bi and a+bi - (a - bi)= 2bi which is imaginary (or 0)


Is it possible for the solution set of a quadratic equation with real coefficients to consist of one real number and one nonreal number?

Complex Conjugate Root TheoremBy knowing the Complex Conjugate Root Theorem, we know that all imaginary roots in the form a + bi have a conjugate a - bi which is also a root. Thus, there are either 2 or 0 imaginary roots to the equation and it would not be possible to have only one non-real number.The DiscriminantThe discriminant of any quadratic equation in standard form is b2 - 4ac. If the discriminant is a positive number, the equation has 2 real roots. If the discriminant equals 0, there is 1 real root. If the discriminant is negative, there are 2 imaginary roots. Thus it is not possible to have one real and one non-real root to a quadratic equation with real coefficients.Proof By ContradictionAssume there exists a polynomial of degree 2 with real coefficients with one real solution x= R where R is a real number and one non-real solution x = a + bi where a and b are real numbers and i is defined as the square root of -1. This quadratic equation can now be written as:(x - R)(x - (a + bi)) = x2 - Rx - x(a + bi) + R(a + bi) = x2 - Rx - ax - (bi)x + Ra + R(bi)By combining like terms we get:x2 - (R + a + bi)x + R(a + bi)This equation does not have real coefficients which contradicts our original assumption. Thus, no such quadratic equation exists.Extending to Polynomial EquationsKnown TheoremsUsing the fundamental theorem of algebra in conjunction with the Complex Conjugate Root Theorem, we see that any odd degree polynomial must have an odd number of real roots (1, 3, 5, ...) and that an even degree polynomial must have an even number of real roots (0, 2, 4, ...).Proof by ContradictionLet's take x = -(a+bi) to be a solution where a and b are both real numbers (b is non-zero) and i is the square root of -1. Now represent the real solution with x = -R (R is a real number).Now the polynomial out in a factored form:((x+(a+bi))^n)((x+R)^m) where n > 0 and m > 0.Now multiply out all of the non-real factors collectively and all the real numbers collectively, but leave them separated.[xn + ... + ((n choose 1)(a+bi)n-1)x + (a+bi)n][xm + ... + Rm]Case 1: R is non-zeroNow multiply all the terms together:xn+m + ... + (Rm-1(a+bi)n + Rm(n choose 1)(a+bi)n-1)x + Rm(a+bi)nIf n is an even integer greater than 0 and a = 0, then (bi)n is a real number and (bi)n-1 is an imaginary number.If n is an odd integer greater than 0 and a = 0, then (bi)n-1 is a real number and (bi)n is an imaginary number.If a is non-zero, then we have (a+bi)n = (an + ... + (n choose 1)(bi)n-1(a) + (bi)n). Rm(a+bi)n is non-real.If R = 0, then we have:xn+m + ... + ((n choose 1)(a+bi)n-1)xm+1 + (a+bi)nxmand (a + bi)n will always have a non-real part.Thus, in all cases, the polynomial does not have real coefficients which contradicts our original statement. So no such polynomial exists.


Is Jordan Maron bi?

No, he is not. He's completely straight.


Why does the natural log of 0 not exist?

there is no power that you could possibly raise a base to, to equil 0. n^0=1and n^(a+bi)=a negative numberwere n is not 0were (a+bi) is a complex power