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Prove. If a polynomial equation G(x) = 0 has real coefficients and if the imaginary number a + bi is a root of G(x) = 0, then the conjugate imaginary a - bi number is also a root.

Proof. We will show that G(x) is exactly divisible by the product

D(x) [x - (a + bi)][x - (a - bi)] x2 - 2ax + a2 + b2 .

This will constitute a proof that a bi are both roots of G(x) = 0.

Since D(x) is a quadratic and has real coefficients, we can divide G(x) by D(x) until we obtain a quotient Q(x) and a linear remainder, cx + d, where c and d are real. Thus we get

1)        G(x) = D(x)∙Q(x) + cx + d

or

2)        G(x) = [x - (a + bi)][x - (a - bi)]∙Q(x) + cx + d

Let us now substitute the value x = a + bi into 2). Since a + bi is a root of G(x), the left member of 2) becomes 0 and the [x - (a + bi)] factor becomes [a + bi - (a + bi)] = 0 and 2) becomes

3)        0 = 0∙[x - (a - bi)]∙Q(a + bi ) + ca + icb + d

or

4)        ac + d + bci = 0

Separating reals and imaginaries, we have

5)        ac + d = 0

6)        bc = 0

Since b≠0, we deduce from 6) that c = 0. Substituting c = 0 into 5) gives d = 0.

Thus we have shown that c and d in 1) above are both 0 and consequently G(x) = D(x)∙Q(x). Thus D(x) is a factor of G(x).

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