Sum of Absolute Complex Roots

We'll show that all eight of the roots of the polynomial satisfy $|x|=1$ , so that their sum is $\boxed8$ .

Let $g(x)$ be the given polynomial. Since $x=0$ is clearly not a root, we can divide through by $20x^4$ to get $h(x)=x^4+kix^3-kix^{-3}+x^{-4}$

where $k=\frac{7}{20}$ . The roots of $h$ will be the same as those of $g$ . Now let $f(x)=h(ix)=x^4 + kx^3 + kx^{-3}+x^{-4}$

If the roots of $f$ lie on the unit circle, so do those of $h$ and $g$ . Note that $f$ has real coefficients (so its roots occur in conjugate pairs) and these coefficients are palindromic, which suggests the substitution $y=x+x^{-1}$ . Then we can rewrite $f(x)$ as $p(y)=f(x)=y^4+ky^3-4y^2-3ky+2$

It's easy to show that all four roots of $p(y)$ are real, distinct and lie in the interval $(-2,2)$ ; in fact we have

Now, since $y=x+x^{-1}$ , we can solve to get $x=\frac{y\pm\sqrt{y^2-4}}{2}$

Since for each of the roots of $p(y)=0$ we have $|y|<2$ , this is better written $x=\frac{y\pm i\sqrt{4-y^2}}{2}$

and now we see that for all $x$ , we have $|x|=\sqrt{\left(\frac{y}{2}\right)^2+\frac{4-y^2}{4}}=1$

This proves that the original polynomial has eight distinct roots all with absolute value $1$ , as claimed.