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