Finding the complex roots of $t^3+2$ Finding the complex roots of an equation: $t^3+2$.
Well typically you would set $t^3+2$ to zero and solve for $t$. Where in this case you would get $t=\sqrt[3]{2}$. However I want to find solutions that are in $\mathbb{Q}$ or $\mathbb{C}$. What is the trick to get those solutions?
 A: Actually, you would get $t=\sqrt[3]{-2}$, not $\sqrt[3]{2}$.
However, in general, you can always find $n$ solutions to the equation $z^n = w$. if you are given $w$.
In particular, the solutions are found by first writing $w$ in a polar format, i.e.
$$w=r\cdot (\cos \phi + i\sin\phi)$$
where $r$ is a positive real number and $\phi$ is a real number between $0$ and $2\pi$.
Then, the solutions of $z^n = w$ are:
$$z = \sqrt[n]{r}\left(\cos \left(\frac{\phi}{n} + \frac{2\pi k}{n}\right) + i\sin\left(\frac{\phi}{n} + \frac{2\pi k}{n}\right)\right)$$
where $k=0,1,2,3,\dots, n-1$
A: Hint:
1)factorize
$$
t^3+2=(t+\sqrt[3]{2})[t^2-\sqrt[3]{2}t+(\sqrt[3]{2})^2]
$$
2) solve  $$(t+\sqrt[3]{2})=0 \qquad \lor \qquad t^2-\sqrt[3]{2}t+(\sqrt[3]{2})^2=0$$
obviously you have not solutions in $\mathbb{Q}$ but you can find solutions in $\mathbb{R}$ and $\mathbb{C}$.
A: Nth Roots lie evenly on a circle in the complex plane.
So the solutions are the negative third root of 2, and the two complex numbers that will 120 degrees from that.  I.e.  (1/2 +- sqrt (3)/2)*cube root (2)
