Cube roots of complex numbers

Question

I need help with finding the cube roots of the complex number 27... I know that the obvious answer is three, but what is the less simple method to solving this?

Answer 1

From Euler formula you have: $$ 1=\cos(2k\pi)+i\sin(2k\pi)=e^{i2k\pi} $$

Write: $27=27\times 1=27e^{i2k\pi}$ and find:

$$ \sqrt[3]{27e^{i2k\pi}}=3(e^{i2k\pi})^{\frac{1}{3}}= 3e^{\frac{i2k\pi}{3}} $$ and for $k=1,2,3$ you find the three complex cubic roots of $27$.

Answer 2

Well you can solve the equation $x^3-27=0$

$$x^3-27=(x-3)(x^2+3x+9).$$

So, of course we get $3$ as a solution from the first factor. To get the remaining solutions use the quadratic formula.

Answer 3

Tim's answer works, but another way to do it is using De Moivre's theorem. Using this theorem we find that the three complex roots of $27=27\left(\cos 0+i\sin 0\right)$ are given by: \begin{equation} \begin{aligned} z_k&=\sqrt[3]{27}\left(\cos\frac{0+2\pi k}{3}+i\sin\frac{0+2\pi k}{3}\right)\\ &=3\left(\cos\frac{2\pi k}{3}+i\sin\frac{2\pi k}{3}\right) \end{aligned} \end{equation} where $k=0,1,2$. This therefore gives us that the three cube roots of $27$ are: \begin{equation} \begin{aligned} z_0=3\left(\cos\frac{2\pi\times 0}{3}+i\sin\frac{2\pi\times 0}{3}\right)&=3\\ z_1=3\left(\cos\frac{2\pi\times 1}{3}+i\sin\frac{2\pi\times 1}{3}\right)&=\frac{3\left(i\sqrt{3}-1\right)}{2}\\ z_2=3\left(\cos\frac{2\pi\times 2}{3}+i\sin\frac{2\pi\times 2}{3}\right)&=-\frac{3\left(i\sqrt{3}+1\right)}{2} \end{aligned} \end{equation}