# Thread: show that the conjugate of w is also a root for p(z)

1. ## show that the conjugate of w is also a root for p(z)

Show that if w is a root of the polynomial p(z), that is p(w)=0, where p(z) has real coefficients, then the conjugate of w is also a root of p(z).

I know that for polynomial of order 2, we can just use the quadratic formula, the two roots are conjugate of each other, but how do I prove it for general polynomial of order n?

the hint i was given is the conjugate of a is itself when a is real, and the conjugate of the product of 2 complex number is the product of each of the conjugate.

2. ## Re: show that the conjugate of w is also a root for p(z)

Originally Posted by wopashui
Show that if w is a root of the polynomial p(z), that is p(w)=0, where p(z) has real coefficients, then the conjugate of w is also a root of p(z).

I know that for polynomial of order 2, we can just use the quadratic formula, the two roots are conjugate of each other, but how do I prove it for general polynomial of order n?

the hint i was given is the conjugate of a is itself when a is real, and the conjugate of the product of 2 complex number is the product of each of the conjugate.
The hint tells you every thing.
$\displaystyle P(z)=\sum\limits_{k = 0}^N {\alpha _k z^k }$ where each $\displaystyle \alpha_k\in\mathbb{R}$

then $\displaystyle \overline{P(z)}=\overline{\sum\limits_{k = 0}^N {\alpha _k z^k }}=\sum\limits_{k = 0}^N {\alpha _k \overline{z}^k }$

3. ## Re: show that the conjugate of w is also a root for p(z)

Originally Posted by Plato
The hint tells you every thing.
$\displaystyle P(z)=\sum\limits_{k = 0}^N {\alpha _k z^k }$ where each $\displaystyle \alpha_k\in\mathbb{R}$

then $\displaystyle \overline{P(z)}=\overline{\sum\limits_{k = 0}^N {\alpha _k z^k }}=\sum\limits_{k = 0}^N {\alpha _k \overline{z}^k }$
hmm ic, how does the second part of the hint apply here btw? (the conjugate of the product of 2 complex number is the product of each of the conjugate. )

4. ## Re: show that the conjugate of w is also a root for p(z)

because from that, it follows that the conjugate of z^k is the k-th power of the conjugate of z.

on the complex numbers, conjugation is a field automorphism, so the sum of the conjugates is the conjugate of the sum, and the conjugate of the product is the product of the conjugates (and thus the conjugate of a power is the same power of the conjugate).

since polynomials involve only summing and products and powers (of z), for a real polynomial p, $\displaystyle p(\overline{w}) = \overline{p(w)}$.

since p(w) = 0, the conjugate of p(w) is 0 (since 0 is real), and so $\displaystyle p(\overline{w}) = 0$.

for this to actually work, the coefficients of p have to be real, so that the coefficients of $\displaystyle p(z)$ and $\displaystyle \overline{p(z)}$ are the same.