# Complex numbers and polynomials

• Mar 15th 2010, 11:58 PM
blorpinbloo
Complex numbers and polynomials
Having trouble with trying to prove this. It looks like it isn't too hard but I can't wrap my head around it.

Let f be a polynomial with real coefficients. Prove: if a complex number z satisfies f(z)=0, then the conjugate of z (call it x) also satisfies f(x)=0
• Mar 16th 2010, 12:15 AM
Prove It
• Mar 16th 2010, 12:24 AM
chisigma
Let the polynomial be of degree n so that...

$P(z)= a_{0} + a_{1}\cdot z + a_{2}\cdot z^{2} + \dots + a_{n}\cdot z^{n}$ (1)

If $z_{0} = \rho\cdot e^{i \theta}$ satisfies the condizion $P(z_{0})=0$ that means that is...

$a_{0} + a_{1}\cdot \rho\cdot e^{i \theta} + a_{2}\cdot \rho^{2}\cdot e^{2 i \theta} + \dots + a_{n}\cdot \rho^{n} \cdot e^{n i \theta} = 0$ (2)

Of course (2) is valid if both the real and imaginary parts of $P(z_{0})$ vanish and that's true if we change in (2) $\theta$ with $- \theta$, so that $\overline{ z_{0}} = \rho\cdot e^{-i \theta}$ is also a zero of $P(z)$...

Kind regards

$\chi$ $\sigma$