how to prove that if a quartic equation ( with real coefficients ) has 4 imaginary roots they all will be in conjugate pairs?

Question

I proved this fact for qudratic equation in the following way , let a qudratic equation have a imaginary root p+iq(q is not 0) , and let other root be (a+ib). Now here sum of roots will be a real number lets say R , => p+iq + a+ib = R , =>(p+a) + i(q+b) = R+i0 , Now this is possible only when q+b = 0 => q=-b . So the two roots are p+iq and a-iq . Now product of roots is also a real number, lets say R' , => (p+iq)(a-iq)=R' => (pa+q^2)+i(aq-pq)=0 => aq-pq=0 => aq=pq => a=p . So finally the roots are , p+iq and p-iq , hence proved . I tried to prove this fact for quartic equation in a similar way but could not reach to the conclusion . Please guide me by answering by my method or by suggesting any other simple method to prove that if a quartic equation has all imaginary roots then they will occur in conjugate pairs.

Answer 1

There's an easy, conceptual proof of this relying only on the behaviour of conjugates, which applies even more generally for polynomials of any degree.

But for your case, you need to note that for any complex number $z,$ we have that $s=z+\bar{z}$ and $p=z\bar{z}$ are always real. Also, you need to know that every quartic may be factored into two quadratics. Then if you assume that you have a real quartic with complex roots that do not occur in conjugate pairs, then the linear factors, when paired into quadratics, will end up with quadratics that are not necessarily real (since then $s, p$ are not real), which gives us a quartic that's not necessarily real, contrary to assumption.

Therefore if the coefficients must be real, then the complex roots must occur in conjugate pairs so that when the linear factors are multiplied, we have real quadratic factors, which give us a real quartic, as desired.

Answer 2

If $$P(z)=\sum_{i=0}^na_iz^i$$ is a polynomial having real coefficients with $z_0$ as a root, then taking conjugate on both sides we see that $\overline{z_0}$ is also a root.

Now, if $a+ib$ is a root of a quartic with real coefficients, then by this reasoning, $a-ib$ is also a root. Hence, given polynomial is divisible by the quadratic $(x-a-bi)(x-a+bi)=(x-a)^2+b^2$. Divide it by this quadratic, get a quadratic and apply the same procedure.

Note

The above also shows that:

If a non-real complex number $z$ is a root of a polynomial with real coefficients with multiplicity $k$, then its conjugate is also a root with same multiplicity (irrespective of the degree of the polynomial)

Answer 3

Let $P(z)$ be any polynomial with real coefficients. Suppose $\zeta$ is a complex root.

Then $P(\bar \zeta) = \overline{P(\zeta)} = \bar 0 = 0$. Hence $\bar \zeta$ is also a root.

Answer 4

Suppose $\lambda_1,..,\lambda_4$ are the 4 non real roots (possibly repeated). Since $p(z) = p(\bar{z})$, if $\lambda$ is a root then so is $\bar\lambda$.

Hence, reordering if necessary, the roots are $\lambda_1, \overline{\lambda_1}, \lambda_2, \overline{\lambda_2}$.