# Cubic with complex roots

I have a problem figuring out how exactly I find the cube roots of a cubic with complex numbers.

I need solve the cubic equation $z^3 − 3z − 1 = 0$.

I've come so far as to calculate the two complex roots of the associated quadratic but then I'm stuck. I've got the solutions here and my lecture notes, have a look at this:

What I don't understand is how you go from $e^{i\pi/3}$ to $e^{i\pi/9}$. Because that root, as I understand it, should be the two conjugates roots added together which I believe do not add up to $e^{i\pi/9}$. What's the step going on here?

Any help is much appreciated!

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Certainly $e^{i\pi/9}$ is a cube root of $e^{i\pi/3}$. –  André Nicolas May 17 '12 at 19:09
Yeah but how do I get to $e^{i\pi/9}$ from $e^{i\pi/3}$ that's what I dont understand. –  AzaraT May 17 '12 at 19:11
$(e^{i\pi /3})^{1/3}=e^{(i\pi/3)\cdot(1/3)}=e^{i\pi/9}$. (Again, this is only one of the cube roots of $e^{i\pi/3}$.) –  anon May 17 '12 at 19:12
In general, $e^{\alpha/k}$ is a $k$-th root of $e^{\alpha}$. Just take the $k$-th power, using usual rules of exponents, to check. –  André Nicolas May 17 '12 at 19:13
ohh right, thanks! And then for the 2 others you add $2/3$ to the powers right? –  AzaraT May 17 '12 at 19:14

Certainly $e^{\pi i/9}$ is one of the cube roots of $e^{\pi i/3}$.

Note that the $3$ cube roots of $1$ are $1$, $e^{2\pi i/3}$, and $e^{4\pi i/3}$. More familiarly, they are $1$, $\frac{-1+i\sqrt{3}}{2}$, and $\frac{-1-i\sqrt{3}}{2}$.

It follows that $e^{\pi i/9}e^{2\pi i/3}$ and $e^{\pi i/9}e^{4\pi i/3}$ are also cube roots of $e^{\pi i/3}$. If we simplify a bit, we get the $e^{7\pi i/9}$ and $e^{13\pi i/9}$ circled in the post.

Remark: In general the $n$-th roots of unity are $e^{2\pi ik/n}$, where $k$ ranges from $0$ to $n-1$. For any $a\ne 0$, if we have found an $n$-th root $w$ of $a$, then the $n$-th roots of $a$ are given by $we^{2\pi ik/n}$, where $k$ ranges from $0$ to $n-1$.

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When I tried to solve the given cubic equation I found that it has three REAL roots as hereunder:

-1.532089, 1.8793852 and -0.3472962. Since you have mentioned that the equation has complex roots I tried to put the above values in the equation and found that the equation is satisfied with each value.

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OP referred to complex roots of the associated quadratic, not of the original cubic. You probably have to look at the link in the question to see what this is about. –  Gerry Myerson May 22 '12 at 4:21