Finding all roots of $z^4-4z^3+9z^2-4z+8$

Question

I need to know all the roots of $z^4-4z^3+9z^2-4z+8$. I know only one root: z=i. Is there an easy way to find the 3 roots that are unknown?

thanks.

Answer 1

Hint

When you have complex roots (of real polynomials as DonAntonio pointed out in his comment), they come by conjugate pairs : so, if $z=i$ is a root, $z=-i$ is another one. So, doing the long division, you are left with a quadratic in $z$.

Answer 2

Using Complex conjugate root theorem, the other root is $-i$

So, the rest two roots will be available from $$\frac{z^4-4z^3+9z^2-4z+8}{(z-i)(z+i)}=0$$

Answer 3

Hint: There is also the conjugate root $-i$ and you can divide by $x^2+1$ and get the factorization $(x^2+1)(x^2-4x+8).$

Answer 4

We know two roots, $i$ and $-i$. Their sum is $0$ and their product is $1$.

The sum of all the roots of the quartic is the negative of the coefficient of $x^3$, that is $4$. The product of all the roots is $8$.

Thus the sum of the two missing roots is $4$, and their product is $8$. So the missing roots are the roots of $z^2-4z+8=0$.