Find the algebric form of the zeros(roots) of the following polynomial: $\left(\:z^2+iz+2\right)\left(z^3-8i\right)$

Question

Good morning to everyone. I don't know how to find the zeros(roots) of the following polynomial function: $$\left(\:z^2+iz+2\right)\left(z^3-8i\right)$$. What I've tried: The zero(root) of the second term has to be : $\frac{8}{i}$. But I don't know how to compute the root of the first term. Can someone explain me how to compute that root. Thanks for any possible answers.

Answer 1

Hint: You have a product of two polynomials. Crop the problem into smaller problems in finding the zeros of each polynomial.

For $z^2+iz+2=0$: Complete the square and solve for z or simply use the quadratic formula: $z_{1/2}=\frac{-i\pm\sqrt{i^2-4\cdot \cdot 2}}{2\cdot 1}=\frac{-i\pm\sqrt{-9}}{2\cdot 1}=\frac{-i\pm 3i}{2\cdot 1}$

For $z^3-8i=0$: Rewrite $z_k^3=8i=2^3e^{\pi/2+2\pi k}$ and take the third root. $z_k=2e^{\pi/6+2\pi/3k}$, in which $k=0,1,2$.

Answer 2

Assuming the whole thing is equated to zero.

$$z^2+iz+2=0$$

$$z^2+iz =-2 $$

$$z^2 + iz +\frac{i^2}{4} =-2+\frac{i^2}{4} $$

$$ (z+\frac{i}{2})^2=\frac{-9}{4}$$

$$ z= \frac{-i}{2} \pm \frac{3}{2} i $$

Answer 3

The first term is a quadratic in $z$, so you can apply the quadratic formula. The second term is a cubic so you can apply Cardano's formulae, which I don't recommend, or you could note that $8i=(-2i)^3$ to factor out a root and get another quadratic.

Also note that $\tfrac{8}{i}$ is not a root of the second term; $$\left(\frac8i\right)^3-8i=\frac{512}{i^3}-8i=-512i-8i=-520i.$$