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$n$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.

Kenta S
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user1091990
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  • $r$ is a square root, and the cube root of a square root is a sixth root. – greelious Feb 04 '19 at 02:21
  • Don't express irrational trig values as decimal approximations. NOBODY cares about what the approximate values are. – fleablood Feb 04 '19 at 02:27
  • $r = \sqrt 8$. So you need to find $\sqrt[3] {r} = \sqrt[3]{\sqrt 8}$. And $\sqrt[3]{\sqrt{8}} = \sqrt[6]{8}$. – fleablood Feb 04 '19 at 02:33
  • $n = 3$ 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 04 '19 at 02:36

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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 $\sqrt[n]{\sqrt[m]{a}} = \sqrt[nm]{a}$.

Alessio Del Vigna
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