Find the three cube roots of $z = -2+2i$

$$n$$th roots of a complex number:

For a positive integer $$n$$, the complex number $$z=r(\cos(\theta) + i\sin(\theta))$$ has exactly $$n$$ distinct $$n$$th roots given by

$$\sqrt[n]r\left(\cos\left(\frac{\theta + 2\pi k}n\right) + i\sin\left(\frac{\theta + 2\pi k}n\right)\right)$$ where $$k=0, 1, 2, ..., n-1$$

Find the three cube roots of $$z=-2 + 2i$$

$$r = \sqrt{(-2)^2 + 2^2} = \sqrt{8}$$

$$\tan(\theta) = \frac ba = \frac {2}{-2} = -1$$

$$\theta = 45^o$$, because $$z$$ is in Quad II, $$\theta = 135^o$$

The trigonometric form of $$z$$ is $$z = -2 + 2i = \sqrt{8}(cos135^o + isin135^o)$$

Using the formula of $$n$$th roots, the cube roots have the form

$$\sqrt[6]8\left(\cos\left(\frac{135^o + 360^ok}3\right) + i\sin\left(\frac{135^o + 360^ok}3\right)\right)$$

For $$k = 0$$ we get the root $$1 + i$$

For $$k = 1$$ we get the root $$-1.3660 + 0.3660i$$

For $$k = 2$$ we get the root $$0.3660 - 1.3660i$$

My question is, using the formula for $$nth$$ roots, why is the first part of the formula given by $$\sqrt[6]8$$ and not $$\sqrt[3]8$$

because $$n=3$$, so I'm confused as to where 6 is coming from.

• $r$ is a square root, and the cube root of a square root is a sixth root. – greelious Feb 4 at 2:21
• Don't express irrational trig values as decimal approximations. NOBODY cares about what the approximate values are. – fleablood Feb 4 at 2:27
• $r = \sqrt 8$. So you need to find $\sqrt[3] {r} = \sqrt[3]{\sqrt 8}$. And $\sqrt[3]{\sqrt{8}} = \sqrt[6]{8}$. – fleablood Feb 4 at 2:33
• $n = 3$ but $r \ne 8$. so $\sqrt[3]{r} \ne \sqrt[3] 8$. Since $r = \sqrt{8} \ne 8$ we heve $\sqrt[3]{r} = \sqrt[3]{\sqrt{8}} \ne \sqrt[3] 8$. – fleablood Feb 4 at 2:36

You have $$r=\sqrt{8}$$ and then you have to compute $$\sqrt[3]{r}=\sqrt[3]{\sqrt{8}}=\sqrt[6]{8}$$. This is the basic property that you have to use: if $$a\geq 0$$ and $$n,m\in \mathbb{N}$$, with $$n,m\geq 1$$, then $$\sqrt[n]{\sqrt[m]{a}} = \sqrt[nm]{a}$$.