Find roots of a complex quadratic equation having one purely imaginary root

Question

Consider : $$P(z)= z^4 - 2z^3 + 6z^2 - 8z + 8$$ As the title says, find the roots of this complex quadratic equation having one purely imaginary root.

I need help with this problem, i am new with the ''complex world''. This is what i thought: Given a complex number: $z= a + bi$ where $a,b\in\mathbb C$, we know $P(z)$ has a pure imaginary root, then: $$P(bi)= (bi)^4 - 2(bi)^3 + 6(bi)^2 - 8(bi) + 8= 0$$ But i am stuck in here, i dont know how to proceed or if reasoning was correct. Any help would be helpful.

Answer 1

Hint:

$i^2=-1$

This gives $$b^4+2b^3 i-6b^2-8bi+8=0 $$ Now use the fact that a complex number is zero if and only if its real and imaginary part are zero.

This will give you a system of two equations. Find the common solution to them.

Answer 2

Since the coefficients of $P$ are purely real numbers, if $bi$ is a root of $P(z)$ then so is $\bar{bi}=-bi$. Hence, $(z-bi)(z+bi)=z^2+b^2\mid P(z)$. So $$P(z)=(z^2+b^2)(z^2+az+c)=z^4+az^3+(c+b^2)z^2+ab^2z+b^2c$$for some $a,c\in\Bbb C$. Compare coefficients with what we know $P(z)$ to be.

Answer 3

HINT...If $z=ai$ is a purely imaginary root, then so is $z=-ai$ since all the coefficients of the polynomial are real.

Therefore $(z^2+a^2)$ is a factor.

Therefore we can factorise the polynomial into the form $$(z^2+a^2)(z^2+bz+c)$$

it is then a simple matter to compare coefficients and find $a,b,c$

Answer 4

Hint:

If $ki$ is a solution also $-ki$ is a solution and the polynomial can be factorized as: $$(z^2+k^2)(az^2+bz+c)$$

can you find $a,b,c,k$?