prove that if r is a root of a n x n + a n-1 x n-1 +........+ a 1 x + a 0 = 0 th

Question

prove that if r is a root of an xn + an-1 xn-1 +........+ a1 x + a0 = 0

then the conjugate of r is a root of the conjugate of a.

please show a simple actual example, not a long winded proof with a bunch of variables. i want to actually see a concrete example where this is case is true.

2. using the previous question, prove that if the coeffcients of a are real and r is a root of a, then the conjufgate of r is also a root of a.

this is supposed to prove if the coefficients are real, then its imaginary roots must occur in conjugate pairs. please show a concrete example with numbers worked out.

thank you very much!

Explanation / Answer

I am giving some very simple proofs and examples:

1) Let f(x) = an xn + an-1 xn-1 +........+ a1 x + a0

Hence, f(x) = 0

Take conjugates both the sides,

(f(x))' = 0' = 0

expanding f(x) and using the distributive property of the conjugation operator over both addition and multiplication

(f(x))' = an' x'n + an-1' x'n-1 +........+ a1' x' + a0'

And hence, x' is a root with coefficients as ai'.

Take the case of x+i = 0

Here -i is the solution.

Take the conjugates of the coefficients, the equation becomes, x-i = 0

and the solution turns out to be i which is a complex conjugate of -i.

2) Put ai as a real number and then,

ai' = ai

Hence we also see that (f(x))' = f(x')

Hence f(x') = 0, which means that x' is also a root of the original equation with real coefficients.

Hence proved.

As for examples, take x2+1=0

The coefficients are real and i and -i are both roots of this equation and both are complex conjugates of each other.

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