De Moivre's Theorem to calculate the fourth roots of 8

Question

I was given this question:

Use de Moivre's theorem to derive a formula for the $4^{th}$ roots of 8.

As far as my understanding of this theorem goes, it is only applicaple to complex numbers. How am I supposed to use it for 8?

My initial thought was use 8 to make $z = 8 + i0$. Determining the polar form:

$$z = 8(\cos \pi + i \sin \pi )$$

from the polar form a I then get that

$$z^4 = 8^4 (\cos (4\pi) + i \sin(4\pi))$$

This is as far as I can go. I only recently started working with de Moivre's theorem so I'm not sure about my calculations and would appreciate any clarification on how to go about answering this question.

Answer 1

Even though you were told to use the estimable Mr. de Moivre, it’s easiest without:

You know that if $\rho$ is one fourth root of a number, the others are $-\rho$ and $\pm i\rho$. Since one fourth root of $8$ is the real number $2^{3/4}$, you have them all.

Answer 2

Note that $z = 8+0i$ is equivalent to $z = 8(\cos(2k\pi) + i\sin(2k\pi))$

And then for the fourth roots:

$$z^{\frac{1}{4}}=8^\frac{1}{4}\left(\cos\left(\frac{k\pi}{2}\right) + i\sin\left(\frac{k\pi}{2}\right)\right)$$

For $k=0$: $z^{\frac{1}{4}} = 8^\frac{1}{4}(\cos(0) + i\sin(0)) = 8^\frac{1}{4}$

For $k=1$: $z^{\frac{1}{4}} = 8^\frac{1}{4}(\cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})) = 8^\frac{1}{4}i$

For $k=2$: $z^{\frac{1}{4}} = 8^\frac{1}{4}(\cos(\pi) + i\sin(\pi)) = -8^\frac{1}{4}$

For $k=3$: $z^{\frac{1}{4}} = 8^\frac{1}{4}(\cos(\frac{3\pi}{2}) + i\sin(\frac{3\pi}{2})) = -8^\frac{1}{4}i$

Answer 3

Your trying to find $z$ such that,

$$z^4=8$$

We know that we can write $z=r(\cos \theta+i\sin \theta)$. Then we may write,

$$z^4=r^4(\cos 4 \theta+i \sin 4\theta)=8+0i$$

In order for this to be true we definitely need $\sin 4\theta=0$ (otherwise $z^4$ would have nonzero imaginary part or we would have $z^4=0$) which then tells us $\cos 4\theta=1$ (can't be negative $1$ because $r^4$ is always positive) and $r^4=8$. Then you can go from there.

Answer 4

Indeed $8$ is a complex number.

de Moivre's rule says that $(z^w)^n=z^{wn}$, $\forall n\in\mathbb{N}$

A complex number has $4$ fourth roots, in particular:

$$\sqrt[4]{8}=(8e^{jk2\pi})^{\frac{1}{4}}\stackrel{de\ M.}=\sqrt[4+]{8}e^{jk\pi/2}, \forall k\in\mathbb{Z}$$ where $\sqrt[4+]{x}$ is the positive $4$-th root of a positive real $x\in\mathbb{R}_+$

And so:

$$\sqrt[4]{8}=\sqrt[+]{2\sqrt[+]{2}}e^{jk\pi/2},\ k = 0,\ 1,\ 2,\ 3$$

where $\sqrt[+]{x}$ is the positive square root of a positive real $x\in\mathbb{R}_+$