Kth root of $-1$

Question

I just would like to ask a simple problem. I just don't know if this was already asked and answered here since I haven't seen anything alike.

What is the $kth$ root of $-1$? I know that if $k=2$ then the answer is $i$. If $k=3$ then the answer is $-1$. Thus if $k$ is odd the $kth$ root of $-1$ is $-1$.

But what if $k$ is even?

Can anyone help me?

Answer 1

Actually, if $k=2$ there are two answers: $i$ and $-i$.

For general $k$, you can use de Moivre's formula to deduce from the fact that $-1=e^{i\pi}$ that there are $k$ answers:$$e^{\pi i/k},e^{3\pi i/k},\ldots,e^{(2k-1)\pi i/k.}$$

Answer 2

We can write $-1 = e^{\pi i +2\pi i n}$ for some $n \in \mathbb{Z}$. Taking the $k$th root of both sides, we get $(-1)^{1/k} = e^{(\pi i +2\pi i n)/k}$. Plugging in different values for $n$ will give you different $k$th roots of -1.

These can be re-written using Euler's formula, so $$(-1)^{1/k} = \text{cos}((\pi +2\pi n)/k) + i \text{sin}((\pi +2\pi n)/k)$$.

With this view, we can see that when $k$ is odd ($=2l + 1$), we can set $n = l$, and obtain $$(-1)^{1/k} = \text{cos}((\pi +2\pi l)/(2l+1)) + i \text{sin}((\pi +2\pi l)/(2l+1))$$

Note that $(\pi +2\pi l)/(2l+1) = \pi(2l + 1)/(2l+1) = \pi$, so the above expression becomes $$(-1)^{1/k} = \text{cos}(\pi) + i\text{sin}(\pi) = -1$$ Just as you said in your post.

However, using this method, you can find all the roots.

Answer 3

When talking about the $k$'th root, you imply there is only one. There is a valid use for such a defined function that gives only one result, because it allows it to be a function rather than be multivalued. (Note: such a function will fail to be analytic however)

We can choose to define a complex number, rather than being in the form $a+bi$ instead in polar form as $re^{i\theta}$ where $r$ is the "magnitude" and is a non-negative real number and $\theta$ is the "principal argument" and lies specifically in the range $0\leq \theta < 2\pi$.

(Note: in using principal arguments, there is an unambiguous way to refer to a particular complex number when unambiguity is desired. You do still have things like $e^{i\pi}=e^{3i\pi}=e^{5i\pi}$ and so on, so you can talk about a complex number having an argument outside of that range, but when talking about the principal argument, you always very specifically pick the unique choice of the value of $\theta$ within that range)

(Note: some authors might prefer instead to define principal argument as the value of $\theta$ for the number that lies in the range $-\pi<\theta\leq \pi$ instead. This difference in definition equates to a different location for the "branch cut" for our function)

We can then define the "principal $k$'th root" of a complex number to be $\sqrt[k]{r}e^{i\theta/k}$ where $\sqrt[k]{r}$ is defined as usual for non-negative real numbers. In doing so, we have the principal $k$'th root is otherwise defined as the $k$'th root which happens to have the smallest principal argument from among all $k$'th roots of our complex number.

Under such a definition, we have the $k$'th principal root of a non-negative real corresponds exactly with the usual definition of $k$'th roots for non-negative real numbers. However, under such a definition we have $\sqrt[3]{-1}=\frac{1}{2}+\frac{\sqrt{3}}{2}i$ instead of $\sqrt[3]{-1}=-1$.

Some authors then prefer to modify the definition to instead be "The principal $k$'th root of a complex number is the positive real $k$'th root of the complex number if it exists, the negative real $k$'th root of the complex number if it exists, otherwise the $k$'th root with minimum principal argument."

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It is important to note that there are a number of different choices that can be made here with definitions. Some authors refuse to work with principal roots and instead work only with the multivalued roots, and for good reason. You find that some of the nice properties we are familiar with for positive reals doesn't work when extending to complex numbers and principal roots like $\sqrt{ab}$ not necessarily equaling $\sqrt{a}\times \sqrt{b}$, which leads to incorrect calculations like $-1 = i\times i = \sqrt{-1}\times \sqrt{-1} = \sqrt{(-1)\times (-1)} = \sqrt{1}= 1$. Also, as mentioned, the choice of where to have the branch cut makes an impact as well. If you choose to work with principal roots, it requires you be aware of these changes in behavior and differences in definition.