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Thread: roots prove question..

  1. #1
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    roots prove question..

    there is a polynomial p(z)
    in which every coefficient are real.
    there is a complex root called "a"

    prove that the complement of "a"
    is also a root
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    there is a polynomial p(z)
    in which every coefficient are real. There is a complex root called "a"
    Prove that the complement of "a" is also a root
    You mean conjugate not complement. You need to learn correct vocabulary.

    The proof is easy. Notice that if $\displaystyle c\in\mathcal{R}$ then $\displaystyle \overline c = c$.

    So if $\displaystyle P(x)= \sum\limits_{k = 0}^n {a_k x^k } $ such that $\displaystyle P(a)=0$ then show that $\displaystyle P\left( {\overline a } \right) = \overline {P(a)} = 0$
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  3. #3
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    yes sorry conjugate.
    if t is the root
    $\displaystyle
    P(t)= \sum\limits_{k = 0}^n {a_k {\overline t }^k }
    $

    why it has to be equal

    $\displaystyle
    \overline{P(t)}=\overline{ \sum\limits_{k = 0}^n {a_k { t }^k }}
    $
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  4. #4
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    Quote Originally Posted by transgalactic View Post
    yes sorry conjugate.
    if t is the root
    $\displaystyle
    P(t)= \sum\limits_{k = 0}^n {a_k {\overline t }^k }
    $

    why it has to be equal

    $\displaystyle
    \overline{P(t)}=\overline{ \sum\limits_{k = 0}^n {a_k { t }^k }}
    $
    By the definition of "complement" of a complex number.

    If x= a+ bi, then $\displaystyle \overline{x}= a- bi$.

    If particular, if x= a+ bi and y= c+ di, then x+ y= (a+ c)+ (b+ d)i so $\displaystyle \overline{x+ y}= (a+ c)- (b+d)i= (a- bi)+ (c- di)= \overline{x}+ \overline{y}$ and xy= (ac- bd)+ (ad+ bc)i so [tex]\overline(xy)= (ac-bd)- (ad+ bc)i= (a- bi)(c- di)= \overline{xy}[/itex].
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