# What are the 4th degree roots of $1$?

So my question is the fourth root of $1 = i$?

Where $i^4 = 1$? or if it is still $1$ nonetheless?

• First, please use informative titles which relate to the question bodies. Second, this belongs on math.stackexchange as it is unrelated to physics. Third, googling "fourth root of one" gives many satisfactory answers, for example this one. Commented Dec 4, 2015 at 2:52
• oops.... didnt knew that there was a math stackoverflow. thanks! or what do you mean by your first condition.
– Gareth Compton
Commented Dec 4, 2015 at 2:54
• there are four square roots of 1: 1, -1, i, -i Commented Dec 4, 2015 at 3:00
• I'm voting to close this question as off-topic because it's a pure math question. However, I'm not voting for migration because it's not up to quality standards and I don't think we should be migrating questions until fix them up. Commented Dec 4, 2015 at 3:11

There are four fourth roots of $1$ in $\mathbb C$. Those are $\pm1$ and $\pm i$. Two of them are "primitive" fourth roots of $1$, meaning you don't get $1$ if you raise them to any lower power than $4$. You can raise $-1$ to a lower power than $4$, namely $2$, and get $1$, and you can raise $1$ to a lower power than $4$, namely $1$, and get $1$.

However, by convention, the notation $\sqrt[4]{1}$ means the positive fourth root of $1$.

Remember $i$ is an imaginary number defined as $$\color{red}{i=\sqrt{-1}}\ \ \$$$$\text{or } i^2=-1$$ The forth roots of $1$ are given as $$1^{1/4}=(\cos0+i\sin 0)^{1/4}$$ $$=\left(\cos(2k\pi)+i\sin(2k\pi)\right)^{1/4}$$ $$=\left(\cos\left(\frac{k\pi}{2}\right)+i\sin\left(\frac{k\pi}{2}\right)\right)$$ where, $k=0, 1, 2, 3$ hence, all forth roots of $1$ are given as $$1, \ \ i, \ \ -1, \ \ -i$$

hence, $i$ is one of the fourth roots of $1$ (unity)

and, one should have $$i^4=(i^2)(i^2)=(-1)(-1)=1$$

It's not necessarily correct to talk about the fourth root of a number, in the same way that it's not necessarily correct to talk about the second (square) root of a number. In the latter case, if $x^2 = y$, then $(-x)^2 = y$, so both $x$ and $-x$ are second roots of $x$. We'll often talk about "the" square root of $x$ as the output $\sqrt{x}$ of the "square root function" by picking a "handy" second root of $x$ for every $x$ we define it on. For example, I could've defined \begin{align*} g(x) & = \begin{cases} \sqrt{x} & x \geq 0, x \in \mathbb{Q} \\ - \sqrt{x} & x \geq 0, x \not \in \mathbb{Q}, \end{cases} \end{align*} and $g$ would still spit out a square root of every input. In the case of $\mathbb{C}$, every non-zero complex number has $n$ unique $n^{\textrm{th}}$ roots. So though $i$ is a fourth root of $1$, it is not correct to call it the fourth root of $1$, as $- i, \pm 1$ are also fourth roots of $1$.

In order to solve $\sqrt[n]{1}$:

• Draw the unit circle
• Draw the first solution, which is obviously $1+0i=\cos(0)+\sin(0)i$
• Repeat $n-1$ times: find the next solution by rotating the previous solution $\frac{2\pi}{n}$ radians

For example, $\sqrt[5]{1}$:

• $\cos(0)+\sin(0)i$
• $\cos(\frac{2\pi}{5})+\sin(\frac{2\pi}{5})i$
• $\cos(\frac{4\pi}{5})+\sin(\frac{4\pi}{5})i$
• $\cos(\frac{6\pi}{5})+\sin(\frac{6\pi}{5})i$
• $\cos(\frac{8\pi}{5})+\sin(\frac{8\pi}{5})i$