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Math Help - 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 c\in\mathcal{R} then \overline c  = c.

    So if P(x)= \sum\limits_{k = 0}^n {a_k x^k } such that P(a)=0 then show that  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
    <br />
P(t)= \sum\limits_{k = 0}^n {a_k {\overline t }^k }<br />

    why it has to be equal

    <br />
\overline{P(t)}=\overline{ \sum\limits_{k = 0}^n {a_k { t }^k }}<br />
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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
    <br />
P(t)= \sum\limits_{k = 0}^n {a_k {\overline t }^k }<br />

    why it has to be equal

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

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

    If particular, if x= a+ bi and y= c+ di, then x+ y= (a+ c)+ (b+ d)i so \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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