Solving an equation in quaternions

Question

How to solve equation $x^4=1$ in quaternions? I know how to solve that equation in complex numbers but I have no idea how to do it in quaternions. I have also a question about general way to solve equations in quaternions. And is there similar theorem to fundamental algebra theorem which connects degree of polynomial equation with number of quaternion roots of that equation?

EDIT: My task was to find roots $x^2+1=0$ EDIT2: How to prove that $x^2+1=0$ has infinite number of roots?
I represented x as $a+bi+cj+dk$ and get equations: $a^2−b^2−c^2−d^2=−1$ and $ab+cd=0$, $ac−bd=0$, $ad+cb=0$. Is it worth to solve that or is there a smarter way?

Answer 1

Let me consolidate my two comments into an answer that may be a bit clearer to the OP.

You want to find all the roots of the polynomial $x^4 - 1$ in the quaternions $\Bbb{H}$. Note that $x^4 - 1 = (x^2 - 1)(x^2 + 1)$, just as in $\Bbb{R}$ or $\Bbb{C}$ and note that $\Bbb{H}$ has no non-trivial zero divisors: if $X, Y \in \Bbb{H}$, then $XY = 0$ implies $X = 0$ or $Y = 0$. So $X^4 -1 = 0$ iff either $X^2 - 1 = 0$ or $X^2 + 1 = 0$.

For a general element $X = a + bi + cj+ dk$ of $\Bbb{H}$, where $a, b, c, d \in \Bbb{R}$, we have the identity: $$ X^2 = (a + bi + cj+ dk)^2 = a^2 - b^2 - c^2 - d^2 + 2a(bi + cj + dk) $$ which you can verify by multiplying out using the formulas $ij = k$, $jk = i$, $ji = -k$ etc.

Thus, comparing cooefficients, $X^ 2 - 1 = 0$ iff: $$ a^2 - b^2 - c^2 - d^2 = 1 \\ ab = ac = ad = 0 $$ From the first of those equations, we must have $a \neq 0$, but then from the other equations we must have $b = c = d = 0$, implying that $a^2= 1$. So the only solutions to $x^2 - 1 = 0$ are the quaternions $X = \pm 1$.

Likewise, if $X^2 + 1 = 0$, we must have: $$ a^2 - b^2 - c^2 - d^2 = -1 \\ ab = ac = ad = 0 $$ And then the first equation tells us that at least one of $b$, $c$ or $d$ is non-zero, from which one of the the other equations will tell us that $a = 0$. So the solutions of $x^2 + 1 = 0$, are the quaternions $X = bi + cj + dk$ where $b^2 + c^2 + d^2 = 1$.

Putting that all together gives us all the solutions of $x^4 + 1 = 0$: they are the quaternions $X$ such that $X = \pm 1$ or $X = bi + cj + dk$ where $b^2 + c^2 + d^2 = 1$.

Answer 2

There is a continuum of ways to imbed the (algebraically closed) field $\mathbb C$ into the Hamiltonian quaternions. Namely, letting $j(a+bi)=a+bi+0j+0k$ be the "obvious" imbedding, for every non-zero quaternion $\beta$, $\beta\cdot j(a+bi)\cdot \beta ^{-1}$ is another imbedding of a copy of $\mathbb C$.

Answer 3

Since all the coefficients are real, the roots will have the form $a+bu$ where $u$ is any pure imaginary/vector quaternion whose squared components sum to unity, and $a,b$ are the values for which $a+bi$ is a root in $\mathbb{C}$. Note that real roots with $b=0$ remain discrete.

Thus for instance if $(1+i)/\sqrt2$ is a root in $\mathbb{C}$, they you have quaternion roots of the form $(1+u)/\sqrt2$.

Answer 4

$$x^4=1$$

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First, note that the quaternions are equivalent to the following algebra: $$i^2=j^2=-1,\ ij=-ji.$$ So you effectively have two complex units which anticommute. Then your ordinary quaternions are defined in relation to these as $$\mathbf i\overset{\text{def}}{\equiv}i,\mathbf j\overset{\text{def}}{\equiv}j,\mathbf k\overset{\text{def}}{\equiv}ij.$$ From here you can expand $x$ in its real and "imaginary" parts, $$x=r+\mathbf I$$ which commute, and from their vector multiplication defined by $$q_1q_2=(r_1+\mathbf I_1)(r_2+\mathbf I_2)=\underbrace{(r_1r_2-\mathbf I_1\cdot \mathbf I_2)}_{\in \mathbb R}+ \underbrace{(r_1\mathbf I_2+r_2\mathbf I_1+\mathbf I_1\times \mathbf I_2)}_{\in \mathbb H\setminus \mathbb R}$$ we have the square as $$x^2=r^2-|\mathbf I|^2+2r\mathbf I$$ and the square of this as $$(x^2)^2\equiv x^4=(r^2-|\mathbf I|^2)^2-4r^2|\mathbf I|^2+4r(r^2-|\mathbf I|^2)\mathbf I$$ we have the system $$\begin{align}(r^2-|\mathbf I|^2)^2-4r^2|\mathbf I|^2 &= 1\\ 4r(r^2-|\mathbf I|^2)\mathbf I &= 0\end{align}$$ which leads to a few cases.

Case $r=0$, we have $|\mathbf I|^4=1$ which leads to, if the imaginary part is $\mathbf I=ai+bj+cij,$ $$(a^2+b^2+c^2)^2=1\implies a^2+b^2+c^2=1,$$ a sphere in the imaginary part. The possibility of the RHS being $-1$ is impossible in the quaternions on the assumption of real $a,b,c.$

Case $r\neq 0,\mathbf I\neq 0$. Briefly, in the case $\mathbf I=0,r^4=1\implies x=\pm 1$. Assuming this is not the case, we have $$r^2-|\mathbf I|^2=0$$ by the second equation and so the first becomes $$4r^2|\mathbf I|^2=-1$$ or $$4r^4=-1$$ which is a contradiction since we assumed $r\in\mathbb R.$ We therefore conclude that this case is impossible and the only two solutions are given by $$\boxed{\boxed{x\in\left\{ai+bj+cij\ |\ a^2+b^2+c^2=1\right\}\cup \{1,-1\}}}.$$

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$$x^2+1=0$$

Let $x\in\mathbb H$, where $x^2+1=0.$ Decompose $x$ into its real and quaternion part, $$\begin{align}x&=r+a\mathbf i+b\mathbf j+c\mathbf k\\ &= r+\mathbf I,\text{ where }\mathbf I\equiv a\mathbf i+b\mathbf j+c\mathbf k\end{align}.$$ By quaternion vector multiplication, $$x^2=r^2-|\mathbf I|^2+2r\mathbf I,$$ and by the assumption $x^2+1=0,$ we have $$r^2-|\mathbf I|^2+2r\mathbf I=-1$$ or that $$\begin{align}r^2-|\mathbf I|^2&=-1\\ 2r\mathbf I&=0\end{align}.$$ Assume $r\neq 0,$ then $\mathbf I=0$ by the second equation, and $r^2=-1.$ But $r\in\mathbb R$, a contradiction, so it is necessarily the case that $r=0.$ Thus the second equation is redundant and the first becomes $$|\mathbf I|^2=1$$ or that $$a^2+b^2+c^2=1,$$ which is a condition on the point $x=\mathbf I=a\mathbf i+b\mathbf j+c\mathbf k$, or that the solutions to $x^2+1=0$ are given by $$x\in\left\{a\mathbf i+b\mathbf j+c\mathbf k\ |\ a^2+b^2+c^2=1\right\},$$ which is an infinite set, QED.