- #1
Elzair
- 11
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Homework Statement
Given [tex]f\left(x\right) \in \mathbb{R}\left[x\right][/tex], if [tex]a + bi[/tex] is a root of [tex]f\left(x\right)[/tex], then [tex]a - bi[/tex] is a root of [tex]f\left(x\right)[/tex] as well.
Homework Equations
Factor Theorem: if [tex]f\left(x\right)[/tex] is a polynomial over a ring with identity, then [tex]a[/tex] is a root of [tex]f\left(x\right)[/tex] if and only if [tex]\left(x - a\right)[/tex] is a factor of [tex]f\left(x\right)[/tex].
The Attempt at a Solution
I know I did not do this in the standard way, but I was wondering if it still worked.
If [tex]a + bi[/tex] is a root of [tex]f\left(x\right)[/tex], then [tex]f\left(a + bi\right) = 0[/tex], and so by the Factor Theorem [tex]x - \left(a + bi\right)[/tex] is a factor of [tex]f\left(x\right)[/tex].
Since [tex]\left(a + bi\right) \notin \mathbb{R}[/tex], [tex]\left(x - \left(a + bi\right)\right) \notin \mathbb{R}}\left[x\right][/tex].
Therefore, [tex]\exists \left(c + di\right)[/tex] such that [tex]\left(x - \left(a + bi\right)\right)\left(x - \left(c + di\right)\right) \in \mathbb{R}}\left[x\right][/tex].
[tex]\left(x - \left(a + bi\right)\right)\left(x - \left(c + di\right)\right)[/tex]
[tex]=x^{2} - x\left(c + di\right) - x\left(a + bi\right) + \left(a + bi\right)\left(c + di\right)[/tex]
[tex]=x^{2} - x\left(a + c\right) - x\left(b + d\right)i + \left(ac + \left(ad + bc\right)i - bd\right) [/tex]
[tex]=x^{2} - x\left(a + c\right) - x\left(b + d\right)i + \left(ac + \left(ad + bc\right)i - bd\right) [/tex]
[tex]=x^{2} - x\left(a + c\right) + ac - bd \left( ad + bc - x\left(b + d\right)\right)i[/tex]
If [tex]\left(x^{2} - x\left(a + c\right) + ac - bd + \left( ad + bc - x\left(b + d\right)\right)i\right) \in \mathbb{R}}\left[x\right][/tex], [tex]ad + bc - xb - xd = 0[/tex]
Set [tex]c = a[/tex] and [tex]d = -b[/tex].
[tex]x^{2} - x\left(2a\right) + a^{2} - b^{2} + \left( ba - ab - x\left(b - b\right)\right)i[/tex]
[tex]= x^{2} - x\left(2a\right) + a^{2} - b^{2} \in \mathbb{R}\left[x\right][/tex]
Therefore, [tex]\left(x - \left(a - bi\right)\right)[/tex] is a factor of [tex]f\left(x\right)}\left[x\right][/tex], so [tex]a - bi[/tex] is a root of [tex]f\left(x\right)[/tex] in [tex]\mathbb{R}\left[x\right][/tex].