# 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

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 $-1-\sqrt{3}$.
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 $1 \le k < n$.