roots prove question..

**transgalactic** · October 30th 2009, 02:35 PM

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

**Plato** · October 30th 2009, 03:01 PM

Originally Posted by transgalactic

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 $P\left( {\overline a } \right) = \overline {P(a)} = 0$

**transgalactic** · October 30th 2009, 06:57 PM

yes sorry conjugate.
if t is the root
$ P(t)= \sum\limits_{k = 0}^n {a_k {\overline t }^k } $

why it has to be equal

$ \overline{P(t)}=\overline{ \sum\limits_{k = 0}^n {a_k { t }^k }} $

**HallsofIvy** · October 31st 2009, 01:02 AM

Originally Posted by transgalactic

yes sorry conjugate.
if t is the root
$ P(t)= \sum\limits_{k = 0}^n {a_k {\overline t }^k } $

why it has to be equal

$ \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 $\overline{x}= a- bi$ .

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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