Factoring Real and Complex polynomials.

Question

Factor:

a) $x^2 + 1 \in \mathbb{R}[x]$

b) $z^3 - i \in \mathbb{C}[x]$

Well I solved for $x^2$ and got $-i$ and $i$, but they aren't from Real. And I couldn't solve for Complex (part b).

Answer 1

I think that part $(a)$ is trying to say that there are no real roots to that equation.

For the second part, if \begin{equation} z^{3} - i = 0 \end{equation} Then, writing \begin{equation} i=\exp\left({\frac{i \pi}{2}}\right) \end{equation} means that one root is \begin{equation} z=\exp \left({\frac{i \pi}{6}}\right) \end{equation}

Can you find the other two?

EDIT: All roots to this are of the form; \begin{equation} z_{n} = \exp ({i \pi/6 + 2ni\pi/6)} \end{equation} Where I have given you the root corresponding to $n=0$. The other two are found by inserting $n=1, 2$ into the roots above. By the way, this way of finding the roots to a complex polynomial is a consequence of De Moivre's Theroem.

Answer 2

HINT: a) You cannot factorise deeper. Why? b) $z^3=i$ and you can find 3 roots of $i$.

Answer 3

For b) note that $(-i)$ is a root. Then $z^3-i = (z+i)(z^2-zi-1)$. The rest is to find the $2$ roots of $z^2-zi-1 = 0$.