finding root of an equation with real coefficient.

Question

If the equation $x^4 + ax^3 + bx^2 + cx+ 1=0 $ (where a,b,c are real numbers) has no real roots and if at least one root is of modulus one, then a)b=c b)a=c c)a=b d)none of the above

Answer 1

Since the equation has real coefficients and no real roots, its complex roots come in conjugate pairs: $z_1, \bar z_1, z_2, \bar z_2$.

Since the independent term is $1$, we have $z_1 \bar z_1 z_2 \bar z_2=1$.

Let $z_1$ be the root with modulus $1$. Then $ z_1 \bar z_1 = 1$ and so $z_1 \bar z_1 z_2 \bar z_2=1$ implies that $z_2 \bar z_2=1$.

Therefore, if $z$ is a root, then $1/z$ is a root. This means that the coefficients are symmetric and so $a=c$.

Here we have use this fact, which is easily proved:

The equation whose roots are the inverses of the roots of $x^4 + ax^3 + bx^2 + cx+ 1=0$ is $x^4 + cx^3 + bx^2 + ax+ 1=0$.