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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- Oct 30th 2009, 02:35 PMtransgalacticroots 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 - Oct 30th 2009, 03:01 PMPlato
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 $\displaystyle P\left( {\overline a } \right) = \overline {P(a)} = 0$ - Oct 30th 2009, 06:57 PMtransgalactic
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 }}

$ - Oct 31st 2009, 01:02 AMHallsofIvy
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].