# Complex values of the cube root

I just learned that the cube root has 2 complex roots. For example, the cube root of 8 has : 2 , -1 plus or minus square root of 3 *i

I was wondering, how do you find those conjugate complex values ??? Is the method the same for all odd roots? Thankyou!

• I recommend Googling "roots of unity" – GFauxPas Apr 14 '15 at 0:33
• In general for X^n=A>0→X=√(n&A)√(n&1)=√(n&A)e^(2kiπ/n)=√(n&A)(cos 2kiπ/n+isin 2kiπ/n); k=0,1,2,…,(n-1) – Piquito Apr 14 '15 at 1:30

## 3 Answers

Whilst its certainly true that you see more structure if you pursue a "modulus and argument" approach to this sort of question, a more primitive methodology is also possible.

Let $(a+ bi)^3 = 8$ where $a, b\in \mathbb{R}$. Now, $$(a+ bi)^3 = a^3 + 3a^2bi + 3ab^2 i^2 + b^3 i^3 = (a^3 - 3ab^2) + (3a^2 b -b^3)i = 8 + 0i.$$ Equating imaginary parts, we have $b(3a^2 - b^2) = 0$. So we have two cases: either $b=0$, the root is real, and $(a+bi)^3 = a^3 = 8$, so $a = 2$; or $3a^2 = b^2$, $b = \pm \sqrt{3} a$, $a^3 -9a^3 = -8a^3 = 8$ and $a = -1$. So the roots are $2, -1 +\sqrt{3}$ and $-1-\sqrt{3}$.

Now, like I said, there are more structurally informative techniques, but sometimes just banging on with the techniques you already have can get you the answers you seek. I'd recommend doing some more calculations yourself before you dive into the whole "modulus and argument" deal. You'll get a stronger feel for the domain, and be more confident in finding things out for yourself.

Using Euler's formula, any complex number can be written in the form $$x+iy = r\cos{\theta}+ir\sin{\theta} = re^{i\theta}.$$ Any complex number of the form $$z_k = a^{1/n} e^{2\pi i k/n},$$ where $k$ is an integer, is an $n$th root of a real number $a$, because $$(a^{1/n}e^{2\pi i k/n})^n = ae^{2\pi i k} = a.$$ Then Euler's formula tells you that the real and imaginary parts of $z_k$ are $$a^{1/n} \cos{(2\pi k/n)}, \quad a^{1/n} \sin{(2\pi k/n)},$$ respectively.

Complex roots come in pairs, so if $z$ is a root then $\bar{z}$ is another root (where $\bar{z}$ is the complex conjugate of $z$). In general, you can obtain all the roots of a rational number by multiplying by powers of the primitive roots of unity $\zeta^i$, keeping in mind that $\zeta^n = 1$ where $n$ is the root you are considering. For example $x^3 - 2 = 0$ has three solutions, $x = 2^{\frac{1}{3}}$ (the real root), $\zeta 2^{\frac{1}{3}}$ and $\zeta^2 2^{\frac{1}{3}}$. I'm not sure how much you know about complex numbers but the primitive $n$-th root of unity can be expressed in terms of the exponential as $e^{\frac{2ki\pi}{n}}$, where $1 \le k < n$.