# On Powers and Roots of Split Octonions

Authors: Mücahit Akbıyık
Journal: Journal of Mathematics
 (2023)

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## Abstract

In this article, we obtain the polar forms for two types of split octonions. We calculate De Moivre’s formulas for all polar forms of split octonions. Thus, we give thenth− powers and roots of split octonions and the matrix representation of split octonions. In addition, we present an illustrative example with Matlab codes.


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## Body

## 1. Introduction

De Moivre’s formula,(1)cosα+isinαn=cosnα+isinnα,is used to calculate the nth− powers and nth− roots of a complex number in polar form. Therefore, it has been one of the important subjects studied after some fundamental operations in mathematics. Many studies have been carried out on the adaptation of the formula to the lots of number systems, which are the expansion of the complex number system (references [1–7] and therein).Split octonions have been worked on physics and mathematics [8–20]. Split octonions represent Minkowski space with (4 + 4)-signature. However, split octonions have also been studied with different orders of base elements in the articles in different fields. So split octonions have been studied as split octonions, countercomplex octonions, hyperbolic octonions, or split type octonions. But, it should be emphasized that there are direct relationships between all octonions in the literature. The geometrical fundamental properties for split octonions are located in the article [8–12].In this study, we first give polar forms for two types of split octonions whose polar forms have not been given before. After that, we obtain De Moivre’s formulas of the split octonions and the8×8 matrices related with the split octonions in all polar forms. Finally, we present the nth− roots of the split octonions and the matrix representation of split octonions in all polar forms.


## 2. Preliminaries

A split octonions [8, 9] is introduced as(2)s=w+V,where the number w is called scalar part of the split octonion and(3)V=λ1J1+λ2J2+λ3J3+x1j1+x2j2+x3j3+ctI,is called vector parts of split octonion.In (2), c, t, and xm denote the speed of the light, time, and space coordinates; w and λm are considered as the phase (classical action) and the wavelengths associated with octonionic signals in geometric application. The set of all split octonions is denoted by S. The fundamental algebraic properties for split octonions are located in Table 1. The conjugate of a split octonion, denoted by sc, is introduced by the following equation:(4)sc=w−λ1J1−λ2J2−λ3J3−x1j1−x2j2−x3j3−ctI.Table 1
Split octonion multiplications.
×1J1J2J3j1j2j3I11J1J2J3j1j2j3IJ1J11j3−j2I−J3J2j1J2J2−j31j1J3I−J1j2J3J3j2−j11−J2J1Ij3j1j1−I−J3J2−1j3−j2J1j2j2J3−I−J1−j3−1j1J2j3j3−J2J1−Ij2−j1−1J3II−j1−j2−j3−J1−J2−J31The inner product over split octonions is defined by the following equation [12]:(5)Is=ssc=scs=w2−V2,where V2=λ2−x2+c2t2 (where λ2=λ12+λ22+λ33 and x2=x12+x22+x32). So split octonions are divided into three classes having positive, negative, or zero value with respect to the inner product. In this case, similar to the literature of the split quaternions, if Is<0, Is>0, and Is=0, s can be also called a spacelike split octonion, a timelike split octonion, and a lightlike split octonion, respectively. Also, the split octonions can be divided into three classes with respect to the vector part: if V2<0, V2>0, and V2=0, then it is called a split octonion with spacelike, timelike, and lightlike vector parts, respectively. The norm of the split octonion, denoted by Ns or for brevity N, is defined by N=ssc [12].The polar representation of a split octonion depends on its inner product and its vector part. So, it can be summarized in 5-case as follows:(i)
Every split octonion with the inner product having a negative value is(6)s=NsinhΥ+UcoshΥ,where(7)sinhΥ=wN,coshΥ=VN,U=∑m=13λmJm+xmjm+ctIV,andU is a unit timelike 7− vector [12].(ii)
Every split octonion with the inner product having a positive value and timelike vector part is(8)s=NcoshΥ+UsinhΥ,where(9)coshΥ=wN,sinhΥ=VN,U=∑m=13λmJm+xmjm+ctIV,andU again is a unit timelike 7− vector [12].(iii)
Every split octonion with the inner product having a positive value and spacelike vector part is(10)s=NcosΥ+UsinΥ,where(11)cosΥ=wN,sinΥ=VN,U=∑m=13λmJm+xmjm+ctIV,andU now is a unit spacelike 7− vector [12].Now, we will define the polar forms for two types of split octonions which is not given before as follows.(iv)
Every split octonion with the inner product having zero value,w2=V2, is(12)s=w1+U,where(13)U=∑m=13λmJm+xmjm+ctIw,andU now is a unit 7− vector and U2=+1 . Indeed, the following decomposition for the split octonion with the inner product having zero value, w2=V2, can be written as follows:(14)s=w+V,=w1+Vw,and sincew2=V2,(15)U2=Vw2=+1,is satisfied.(v)
Every split octonion with the lightlike vector part,V2=0, is written as(16)s=Nsgnw+U,where(17)U=∑m=13λmJm+xmjm+ctIN,andU2=0. Because the split octonion s with the lightlike vector part, V2=0, can be determined as(18)s=w+λ1J1+λ2J2+λ3J3+x1j1+x2j2+x3j3+ctI,=w2sgnw+∑m=13λmJm+xmjm+ctIw2,and sinceV2=0,(19)U2=Vw22=0,is provided.


## 3. De Moivre’s Formulas of Split Octonions

In this section, we express De Moivre’s formulas when split octonions are given in polar forms.Theorem 1.
Given the polar forms=sinhΥ+UcoshΥ for a unit split octonion with the inner product having a negative value. Then, we have the following equation:(20)sn=sinhnΥ+UcoshnΥ.Ifn is an odd integer, then the equation (20) is provided.(21)sn=coshnΥ+UsinhnΥ.Ifn is an even integer, then the equation (21) is provided.Proof.
Forn=1, it is trivial and for n=2 and n=3, with the hyperbolic angle sum formulas and considering that U is timelike, we have the following equation:(22)s2=cosh2Υ+Usinh2Υ,s3=sinh3Υ+Ucosh3Υ.
Now, let us assume that the formula is valid forn=2l,l∈ℤ+, that is,(23)sn=coshnΥ+UsinhnΥ.
Then,(24)s2ls=cosh2lΥ+Usinh2lΥsinhΥ+UcoshΥ,or by using the hyperbolic angle sum formulas and U2=1, we obtain the following equation:(25)s2l+1=sinh2l+1Y+Ucosh2l+1Y.
Similarly, suppose that the formula is provided forn=2l+1,l∈ℤ+, that is,(26)sn=sinhnΥ+UcoshnΥ.
Then,(27)s2l+1s=sinh2l+1Υ+Ucosh2l+1ΥsinhΥ+UcoshΥ,or by using the hyperbolic angle sum formulas and U2=1, we have the following equation:(28)s2l+2=cosh2l+2Υ+Usinh2l+2Υ.
Also, we can calculate for negative integer power by a similar method. Forn=−1, by the properties of the multiplicative inverse of the split octonion s, we have s−1=−sinhΥ+UcoshΥ and s−2=cosh−2Υ+Ucosh−2Υ. If the calculation process is repeated as mentioned above, the formula can be obtained similarly.Theorem 2.
Lets=coshΥ+UsinhΥ be a unit split octonion with the inner product having positive value and time-like vector part. Then, we have the following equation:(29)sn=coshnΥ+UsinhnΥ,for all n∈ℤ.Proof.
Like the proof of Theorem1, by using the induction and the hyperbolic angle sum formulas, the proof can be seen clearly.Theorem 3.
Given the polar forms=cosΥ+UsinΥ for a unit split octonion with the inner product having a positive value and spacelike vector part. Then, we have the following equation:(30)sn=cosnΥ+UsinnΥ,for all n∈ℤ.Proof.
Like the steps of the proof of Theorem1, by applying the induction, the poof can be obtained.Theorem 4.
Let us consider the polar forms=w1+U for a split octonion with the inner product having zero. Then, we have the following equation:(31)sn=wn2n−11+U,for all n∈ℤ.Proof.
Let us apply the induction method. Suppose thats is a split octonion with the inner product having zero and sn=wn2n−11+U holds. Then, U2=1 and(32)sn+1=wn2n−11+Uw1+U,=wn+12n−11+U2,=wn+12n1+U.Theorem 5.
Assume thats=sgnw+U is a unit split octonion with the light-like vector part. Then, we have the following equation:(33)sn=sgnwn+nwn−1U,for all n∈ℤ.Proof.
By applying the induction method and consideringU2=0, the proof can be seen directly.


## 4. Roots of Split Octonions

In this section, the roots of a split octonion or in other words the solutions of the equationxn=s will be examined.Theorem 6.
Suppose thats=sinhΥ+UcoshΥ is a unit split octonion with the inner product having a negative value. Let us consider the equation xn=s.(i) Ifn is an even number, then there is no root(ii) Ifn is an odd number, then there exist only 1− root as follows:(34)x0=sinhϑ+Ucoshϑ,whereϑ=Y/n.Proof.
Lets=sinhΥ+UcoshΥ be a unit split octonion with the inner product having a negative value and n is an odd number. Then, U2=1 and using Theorem 1, we can write(35)x0n=sinhnϑ+Ucoshnϑ=s.
The other case is clear from Theorem1.Theorem 7.
Lets=coshΥ+UsinhΥ be a unit split octonion with the inner product having a positive value and timelike vector part. Let us consider the equation of xn=s.(i)
Ifn is an even number, then the 4-distinct roots are given in the following forms:(36)x0=sinhϑ+Ucoshϑ,x1=−sinhϑ+Ucoshϑ,x2=coshϑ+Usinhϑ,x3=−coshϑ+Usinhϑ,whereϑ=Υ/n.(ii)
Ifn is an odd number, then the only 1− root is(37)x0=sinhϑ+Ucoshϑ,whereϑ=Υ/n.Proof.
Lets=sinhΥ+UcoshΥ be a unit split octonion with the inner product having a positive value and timelike vector part:(i)
Assume thatn is an even number. And, let(38)x0=sinhϑ+Ucoshϑ,x1=−sinhϑ+Ucoshϑ,be roots of the equationxn=s, where ϑ=Y/n. Then, using Theorem 1, we can obtain the following equation:(39)x0n=sinhnϑ+Ucoshnϑ=s,x1n=+sinhnϑ+Ucoshnϑ=s.Thus,x0 and x1 hold the equation xn=s. The other cases can be obtained by using Theorems 1 and 2.Theorem 8.
Lets=coshΥ+UsinhΥ be a unit split octonion with the inner product having a positive value and spacelike vector part. Then, there exist n− roots(40)xp=cosϑ+Usinϑ,where ϑ=Υ+2pπ/n,p=0,1,…,n−1.Proof.
One can directly verify this result with a similar way of the proof of Theorem6.Theorem 9.
Lets=w1+U be a light-like split octonion with the inner product having zero. Then,(i)
Ifn is an even number, then 2− roots of the equation xn=s are in the following forms:(41)x0=Ω1+U,x1=−Ω1+U,whereΩ=2ωn/2.(ii)
Ifn is an odd number, then only 1− root of the equation xn=s is(42)x0=Ω1+U,whereΩ=2ωn/2.Proof.
The proof can be easily obtained by using the same way of Theorem6.Theorem 10.
Lets=sgnw+U be a unit split octonion with the light-like vector part.(i)
Ifn is an even number, there are 2− roots of the equation xn=s for sgnw>0. The roots are in the following forms:(43)x0=wn1+Un,x1=−wn1+Un.But there exist no roots of the equationxn=s for sgnw<0.(ii)
Ifn is an odd number, the only 1− root of the equation of xn=s is in the form as follows:(44)x0=wn1+Un.Proof.
One can prove the formula by using the same way of Theorem6.


## 5. De Moivre’s Formulas of the Matrices of Split Octonions

Lets and x be any two split octonions in S. Given the following two linear transformations in S,(45)sL,sR:S⟶S,sLx=sx,sRx=xs.Then, using the transformationsL and basis vectors 1,Jm,jm,I, for m=1,2,3, the left matrix representation of the transformation sL is determined by(46)sL=wλ1λ2λ3−x1−x2−x3ctλ1wx3−x2−ctλ3−λ2x1λ2−x3wx1−λ3−ctλ1x2λ3x2−x1wλ2−λ1−ctx3x1−ct−λ3λ2w−x3x2λ1x2λ3−ct−λ1x3w−x1λ2x3−λ2λ1−ct−x2x1wλ3ct−x1−x2−x3λ1λ2λ3w.Similarly, using the transformationsR and basis vectors 1,Jm,jm,I, for m=1,2,3, the right matrix representation of the transformation sR is given by(47)sR=wλ1λ2λ3−x1−x2−x3ctλ1w−x3x2ct−λ3λ2−x1λ2x3w−x1λ3ct−λ1−x2λ3−x2x1w−λ2λ1ct−x3x1ctλ3−λ2wx3−x2−λ1x2−λ3ctλ1−x3wx1−λ2x3λ2−λ1ctx2−x1w−λ3ctx1x2x3−λ1−λ2−λ3w.In addition to this, we can define an isomorphismϕ:S⟶ML, where(48)ML=wλ1λ2λ3−x1−x2−x3ctλ1wx3−x2−ctλ3−λ2x1λ2−x3wx1−λ3−ctλ1x2λ3x2−x1wλ2−λ1−ctx3x1−ct−λ3λ2w−x3x2λ1x2λ3−ct−λ1x3w−x1λ2x3−λ2λ1−ct−x2x1wλ3ct−x1−x2−x3λ1λ2λ3w|w,λm,xm∈ℝ,m=1,2,3.Similarly, we can define a bijective and surjective transformationψ:S⟶MR, where(49)MR=wλ1λ2λ3−x1−x2−x3ctλ1w−x3x2ct−λ3λ2−x1λ2x3w−x1λ3ct−λ1−x2λ3−x2x1w−λ2λ1ct−x3x1ctλ3−λ2wx3−x2−λ1x2−λ3ctλ1−x3wx1−λ2x3λ2−λ1ctx2−x1w−λ3ctx1x2x3−λ1−λ2−λ3w|w,λm,xm∈ℝ,m=1,2,3.But note that,ψs1+s2=ψs1+ψs2 and ψs1s2=ψs2ψs1, for ∀s1,s2∈S are hold.Lemma 1.
Lets be a split octonion. Then,(50)detsL=detsR=N4,where N is the norm of s.Proof.
One can easily prove with some algebraic operations.
From now on, we consider only the right matrix representationsR of any split octonion s through all our calculations. But all results also can be easily obtained for the left matrix representations of any split octonion by the same methods:(i)
Ifs=sinhΥ+UcoshΥ is a unit split octonion with the inner product having a negative value and a unit timelike 7− vector U , the right representation of the split octonion sR can be written as follows:(51)sinhΥλ1VcoshΥλ2VcoshΥλ3VcoshΥ−x1VcoshΥ−x2VcoshΥ−x3VcoshΥctVcoshΥλ1VcoshΥsinhΥ−x3VcoshΥx2VcoshΥctVcoshΥ−λ3VcoshΥλ2VcoshΥ−x1VcoshΥλ2VcoshΥx3VcoshΥsinhΥ−x1VcoshΥλ3VcoshΥctVcoshΥ−λ1VcoshΥ−x2VcoshΥλ3VcoshΥ−x2VcoshΥx1VcoshΥsinhΥ−λ2VcoshΥλ1VcoshΥctVcoshΥ−x3VcoshΥx1VcoshΥctVcoshΥλ3VcoshΥ−λ2VcoshΥsinhΥx3VcoshΥ−x2VcoshΥ−λ1VcoshΥx2VcoshΥ−λ3VcoshΥctVcoshΥλ1VcoshΥ−x3VcoshΥsinhΥx1VcoshΥ−λ2VcoshΥx3VcoshΥλ2VcoshΥ−λ1VcoshΥctVcoshΥx2VcoshΥ−x1VcoshΥsinhΥ−λ3VcoshΥctVcoshΥx1VcoshΥx2VcoshΥx3VcoshΥ−λ1VcoshΥ−λ2VcoshΥ−λ3VcoshΥsinhΥ.(ii)
Ifs=coshΥ+UsinhΥ is a split octonion with the inner product having a positive value and a unit timelike vector part U, then the right representation of the split octonion sR can be written as follows:(52)coshΥλ1VsinhΥλ2VsinhΥλ3VsinhΥ−x1VsinhΥ−x2VsinhΥ−x3VsinhΥctVsinhΥλ1VsinhΥcoshΥ−x3VsinhΥx2VsinhΥctVsinhΥ−λ3VsinhΥλ2VsinhΥ−x1VsinhΥλ2VsinhΥx3VsinhΥcoshΥ−x1VsinhΥλ3VsinhΥctVsinhΥ−λ1VsinhΥ−x2VsinhΥλ3VsinhΥ−x2VsinhΥx1VsinhΥcoshΥ−λ2VsinhΥλ1VsinhΥctVsinhΥ−x3VsinhΥx1VsinhΥctVsinhΥλ3VsinhΥ−λ2VsinhΥcoshΥx3VsinhΥ−x2VsinhΥ−λ1VsinhΥx2VsinhΥ−λ3VsinhΥctVsinhΥλ1VsinhΥ−x3VsinhΥcoshΥx1VsinhΥ−λ2VsinhΥx3VsinhΥλ2VsinhΥ−λ1VsinhΥctVsinhΥx2VsinhΥ−x1VsinhΥcoshΥ−λ3VsinhΥctVsinhΥx1VsinhΥx2VsinhΥx3VsinhΥ−λ1VsinhΥ−λ2VsinhΥ−λ3VsinhΥcoshΥ.(iii)
Ifs=cosΥ+UsinΥ is a split octonion with the inner product having a positive value and spacelike vector part, then the right representation of the split octonion sR can be obtained as follows:(53)cosΥλ1VsinΥλ2VsinΥλ3VsinΥ−x1VsinΥ−x2VsinΥ−x3VsinΥctVsinΥλ1VsinΥcosΥ−x3VsinΥx2VsinΥctVsinΥ−λ3VsinΥλ2VsinΥ−x1VsinΥλ2VsinΥx3VsinΥcosΥ−x1VsinΥλ3VsinΥctVsinΥ−λ1VsinΥ−x2VsinΥλ3VsinΥ−x2VsinΥx1VsinΥcosΥ−λ2VsinΥλ1VsinΥctVsinΥ−x3VsinΥx1VsinΥctVsinΥλ3VsinΥ−λ2VsinΥcosΥx3VsinΥ−x2VsinΥ−λ1VsinΥx2VsinΥ−λ3VsinΥctVsinΥλ1VsinΥ−x3VsinΥcosΥx1VsinΥ−λ2VsinΥx3VsinΥλ2VsinΥ−λ1VsinΥctVsinΥx2VsinΥ−x1VsinΥcosΥ−λ3VsinΥctVsinΥx1VsinΥx2VsinΥx3VsinΥ−λ1VsinΥ−λ2VsinΥ−λ3VsinΥcosΥ.(iv)
Ifs=w1+U is a split octonion with the inner product having zero. Then, the right representation of the split octonion sR can be given as follows:(54)sR=wλ1λ2λ3−x1−x2−x3ctλ1w−x3x2ct−λ3λ2−x1λ2x3w−x1λ3ct−λ1−x2λ3−x2x1w−λ2λ1ct−x3x1ctλ3−λ2wx3−x2−λ1x2−λ3ctλ1−x3wx1−λ2x3λ2−λ1ctx2−x1w−λ3ctx1x2x3−λ1−λ2−λ3w.(v)
Ifs=sgnw+U is a unit split octonion with the lightlike vector part s  = 0. Then, the right representation of the split octonion sR is found as(55)sR=sgnwλ1λ2λ3−x1−x2−x3ctλ1sgnw−x3x2ct−λ3λ2−x1λ2x3sgnw−x1λ3ct−λ1−x2λ3−x2x1sgnw−λ2λ1ct−x3x1ctλ3−λ2sgnwx3−x2−λ1x2−λ3ctλ1−x3sgnwx1−λ2x3λ2−λ1ctx2−x1sgnw−λ3ctx1x2x3−λ1−λ2−λ3sgnw.Theorem 11.
Let us consider the matrixsR defined by(56)sinhΥλ1VcoshΥλ2VcoshΥλ3VcoshΥ−x1VcoshΥ−x2VcoshΥ−x3VcoshΥctVcoshΥλ1VcoshΥsinhΥ−x3VcoshΥx2VcoshΥctVcoshΥ−λ3VcoshΥλ2VcoshΥ−x1VcoshΥλ2VcoshΥx3VcoshΥsinhΥ−x1VcoshΥλ3VcoshΥctVcoshΥ−λ1VcoshΥ−x2VcoshΥλ3VcoshΥ−x2VcoshΥx1VcoshΥsinhΥ−λ2VcoshΥλ1VcoshΥctVcoshΥ−x3VcoshΥx1VcoshΥctVcoshΥλ3VcoshΥ−λ2VcoshΥsinhΥx3VcoshΥ−x2VcoshΥ−λ1VcoshΥx2VcoshΥ−λ3VcoshΥctVcoshΥλ1VcoshΥ−x3VcoshΥsinhΥx1VcoshΥ−λ2VcoshΥx3VcoshΥλ2VcoshΥ−λ1VcoshΥctVcoshΥx2VcoshΥ−x1VcoshΥsinhΥ−λ3VcoshΥctVcoshΥx1VcoshΥx2VcoshΥx3VcoshΥ−λ1VcoshΥ−λ2VcoshΥ−λ3VcoshΥsinhΥ,for a unit split octonion with the inner product having a negative value and a unit timelike 7-vector U. Then, if n∈ℤ is an odd number, the matrix sRn is(57)sinhnΥλ1VcoshnΥλ2VcoshnΥλ3VcoshnΥ−x1VcoshnΥ−x2VcoshnΥ−x3VcoshnΥctVcoshnΥλ1VcoshnΥsinhnΥ−x3VcoshnΥx2VcoshnΥctVcoshnΥ−λ3VcoshnΥλ2VcoshnΥ−x1VcoshnΥλ2VcoshnΥx3VcoshnΥsinhnΥ−x1VcoshnΥλ3VcoshnΥctVcoshnΥ−λ1VcoshnΥ−x2VcoshnΥλ3VcoshnΥ−x2VcoshnΥx1VcoshnΥsinhnΥ−λ2VcoshnΥλ1VcoshnΥctVcoshnΥ−x3VcoshnΥx1VcoshnΥctVcoshnΥλ3VcoshnΥ−λ2VcoshnΥsinhnΥx3VcoshnΥ−x2VcoshnΥ−λ1VcoshnΥx2VcoshnΥ−λ3VcoshnΥctVcoshnΥλ1VcoshnΥ−x3VcoshΥsinhnΥx1VcoshΥ−λ2VcoshΥx3VcoshnΥλ2VcoshnΥ−λ1VcoshnΥctVcoshnΥx2VcoshnΥ−x1VcoshnΥsinhnΥ−λ3VcoshnΥctVcoshnΥx1VcoshnΥx2VcoshnΥx3VcoshnΥ−λ1VcoshnΥ−λ2VcoshnΥ−λ3VcoshnΥsinhnΥ,and if n∈ℤ is an even number, the matrix sRn is(58)coshnΥλ1VsinhnΥλ2VsinhnΥλ3VsinhnΥ−x1VsinhnΥ−x2VsinhnΥ−x3VsinhnΥctVsinhnΥλ1VsinhnΥcoshnΥ−x3VsinhnΥx2VsinhnΥctVsinhnΥ−λ3VsinhnΥλ2VsinhnΥ−x1VsinhnΥλ2VsinhnΥx3VsinhnΥcoshnΥ−x1VsinhnΥλ3VsinhnΥctVsinhnΥ−λ1VsinhnΥ−x2VsinhnΥλ3VsinhnΥ−x2VsinhnΥx1VsinhnΥcoshnΥ−λ2VsinhnΥλ1VsinhnΥctVsinhnΥ−x3VsinhnΥx1VsinhnΥctVsinhnΥλ3VsinhnΥ−λ2VsinhnΥcoshnΥx3VsinhnΥ−x2VsinhnΥ−λ1VsinhnΥx2VsinhnΥ−λ3VsinhnΥctVsinhnΥλ1VsinhnΥ−x3VsinhnΥcoshnΥx1VsinhnΥ−λ2VsinhnΥx3VsinhnΥλ2VsinhnΥ−λ1VsinhnΥctVsinhnΥx2VsinhnΥ−x1VsinhnΥcoshnΥ−λ3VsinhnΥctVsinhnΥx1VsinhnΥx2VsinhnΥx3VsinhnΥ−λ1VsinhnΥ−λ2VsinhnΥ−λ3VsinhnΥcoshnΥ.Proof.
By applying matrix multiplication and using the hyperbolic angle sum formulas and the fact(59)λ12V2+λ22V2+λ32V2−x12V2−x22V2−x32V2+ct2V2=1,
The2nd− and 3rd− power of the given matrix is calculated as(60)cosh2Υλ1Vsinh2Υλ2Vsinh2Υλ3Vsinh2Υ−x1Vsinh2Υ−x2Vsinh2Υ−x3Vsinh2ΥctVsinh2Υλ1Vsinh2Υcosh2Υ−x3Vsinh2Υx2Vsinh2ΥctVsinh2Υ−λ3Vsinh2Υλ2Vsinh2Υ−x1Vsinh2Υλ2Vsinh2Υx3Vsinh2Υcosh2Υ−x1Vsinh2Υλ3Vsinh2ΥctVsinh2Υ−λ1Vsinh2Υ−x2Vsinh2Υλ3Vsinh2Υ−x2Vsinh2Υx1Vsinh2Υcosh2Υ−λ2Vsinh2Υλ1Vsinh2ΥctVsinh2Υ−x3Vsinh2Υx1Vsinh2ΥctVsinh2Υλ3Vsinh2Υ−λ2Vsinh2Υcosh2Υx3Vsinh2Υ−x2Vsinh2Υ−λ1Vsinh2Υx2Vsinh2Υ−λ3Vsinh2ΥctVsinh2Υλ1Vsinh2Υ−x3Vsinh2Υcosh2Υx1Vsinh2Υ−λ2Vsinh2Υx3Vsinh2Υλ2Vsinh2Υ−λ1Vsinh2ΥctVsinh2Υx2Vsinh2Υ−x1Vsinh2Υcosh2Υ−λ3Vsinh2ΥctVsinh2Υx1Vsinh2Υx2Vsinh2Υx3Vsinh2Υ−λ1Vsinh2Υ−λ2Vsinh2Υ−λ3Vsinh2Υcosh2Υ,sinh3Υλ1Vcosh3Υλ2Vcosh3Υλ3Vcosh3Υ−x1Vcosh3Υ−x2Vcosh3Υ−x3Vcosh3ΥctVcosh3Υλ1Vcosh3Υsinh3Υ−x3Vcosh3Υx2Vcosh3ΥctVcosh3Υ−λ3Vcosh3Υλ2Vcosh3Υ−x1Vcosh3Υλ2Vcosh3Υx3Vcosh3Υsinh3Υ−x1Vcosh3Υλ3Vcosh3ΥctVcosh3Υ−λ1Vcosh3Υ−x2Vcosh3Υλ3Vcosh3Υ−x2Vcosh3Υx1Vcosh3Υsinh3Υ−λ2Vcosh3Υλ1Vcosh3ΥctVcosh3Υ−x3Vcosh3Υx1Vcosh3ΥctVcosh3Υλ3Vcosh3Υ−λ2Vcosh3Υsinh3Υx3Vcosh3Υ−x2Vcosh3Υ−λ1Vcosh3Υx2Vcosh3Υ−λ3Vcosh3ΥctVcosh3Υλ1Vcosh3Υ−x3VcoshΥsinh3Υx1VcoshΥ−λ2VcoshΥx3Vcosh3Υλ2Vcosh3Υ−λ1Vcosh3ΥctVcosh3Υx2Vcosh3Υ−x1Vcosh3Υsinh3Υ−λ3Vcosh3ΥctVcosh3Υx1Vcosh3Υx2Vcosh3Υx3Vcosh3Υ−λ1Vcosh3Υ−λ2Vcosh3Υ−λ3Vcosh3Υsinh3Υ,respectively. It is assumed that the formula is valid for n=2l,l∈ℤ+. In this case, using the hyperbolic sum formulas and matrix multiplications sRnsR and sRn+1sR, the proof is completed easily for n=2l+1 and =2l+2.
On the other hand, for negative integer powers, the formula can be calculated as follows. Forn=−1, the inverse of the given matrix can be found:(61)sinhΥ˜λ1VcoshΥ˜λ2VcoshΥ˜λ3VcoshΥ˜−x1VcoshΥ˜−x2VcoshΥ˜−x3VcoshΥ˜ctVcoshΥ˜λ1VcoshΥ˜sinhΥ˜−x3VcoshΥ˜x2VcoshΥ˜ctVcoshΥ˜−λ3VcoshΥ˜λ2VcoshΥ˜−x1VcoshΥ˜λ2VcoshΥ˜x3VcoshΥ˜sinhΥ˜−x1VcoshΥ˜λ3VcoshΥ˜ctVcoshΥ˜−λ1VcoshΥ˜−x2VcoshΥ˜λ3VcoshΥ˜−x2VcoshΥ˜x1VcoshΥ˜sinhΥ˜−λ2VcoshΥ˜λ1VcoshΥ˜ctVcoshΥ˜−x3VcoshΥ˜x1VcoshΥ˜ctVcoshΥ˜λ3VcoshΥ˜−λ2VcoshΥ˜sinhΥ˜x3VcoshΥ˜−x2VcoshΥ˜−λ1VcoshΥ˜x2VcoshΥ˜−λ3VcoshΥ˜ctVcoshΥ˜λ1VcoshΥ˜−x3VcoshΥ˜sinhΥ˜x1VcoshΥ˜−λ2VcoshΥ˜x3VcoshΥ˜λ2VcoshΥ˜−λ1VcoshΥ˜ctVcoshΥ˜x2VcoshΥ˜−x1VcoshΥ˜sinhΥ˜−λ3VcoshΥ˜ctVcoshΥ˜x1VcoshΥ˜x2VcoshΥ˜x3VcoshΥ˜−λ1VcoshΥ˜−λ2VcoshΥ˜−λ3VcoshΥ˜sinhΥ˜,where Υ˜=−Υ. Also, the −2th− and −3th− power of the matrix can be obtained similarly. Finally, by using similar steps above, the proof for the negative integer is completed.Theorem 12.
Assumed that the matrixsR defined by(62)coshΥλ1VsinhΥλ2VsinhΥλ3VsinhΥ−x1VsinhΥ−x2VsinhΥ−x3VsinhΥctVsinhΥλ1VsinhΥcoshΥ−x3VsinhΥx2VsinhΥctVsinhΥ−λ3VsinhΥλ2VsinhΥ−x1VsinhΥλ2VsinhΥx3VsinhΥcoshΥ−x1VsinhΥλ3VsinhΥctVsinhΥ−λ1VsinhΥ−x2VsinhΥλ3VsinhΥ−x2VsinhΥx1VsinhΥcoshΥ−λ2VsinhΥλ1VsinhΥctVsinhΥ−x3VsinhΥx1VsinhΥctVsinhΥλ3VsinhΥ−λ2VsinhΥcoshΥx3VsinhΥ−x2VsinhΥ−λ1VsinhΥx2VsinhΥ−λ3VsinhΥctVsinhΥλ1VsinhΥ−x3VsinhΥcoshΥx1VsinhΥ−λ2VsinhΥx3VsinhΥλ2VsinhΥ−λ1VsinhΥctVsinhΥx2VsinhΥ−x1VsinhΥcoshΥ−λ3VsinhΥctVsinhΥx1VsinhΥx2VsinhΥx3VsinhΥ−λ1VsinhΥ−λ2VsinhΥ−λ3VsinhΥcoshΥ,is a right matrix representation for a unit split octonion with the inner product having a positive value and timelike vector part. Then, the matrix sRn is(63)coshnΥλ1VsinhnΥλ2VsinhnΥλ3VsinhnΥ−x1VsinhnΥ−x2VsinhnΥ−x3VsinhnΥctVsinhnΥλ1VsinhnΥcoshnΥ−x3VsinhnΥx2VsinhnΥctVsinhnΥ−λ3VsinhnΥλ2VsinhnΥ−x1VsinhnΥλ2VsinhnΥx3VsinhnΥcoshnΥ−x1VsinhnΥλ3VsinhnΥctVsinhnΥ−λ1VsinhnΥ−x2VsinhnΥλ3VsinhnΥ−x2VsinhnΥx1VsinhnΥcoshnΥ−λ2VsinhnΥλ1VsinhnΥctVsinhnΥ−x3VsinhnΥx1VsinhnΥctVsinhnΥλ3VsinhnΥ−λ2VsinhnΥcoshnΥx3VsinhnΥ−x2VsinhnΥ−λ1VsinhnΥx2VsinhnΥ−λ3VsinhnΥctVsinhnΥλ1VsinhnΥ−x3VsinhnΥcoshnΥx1VsinhnΥ−λ2VsinhnΥx3VsinhnΥλ2VsinhnΥ−λ1VsinhnΥctVsinhnΥx2VsinhnΥ−x1VsinhnΥcoshnΥ−λ3VsinhnΥctVsinhnΥx1VsinhnΥx2VsinhnΥx3VsinhnΥ−λ1VsinhnΥ−λ2VsinhnΥ−λ3VsinhnΥcoshnΥ.Proof.
The formula can be shown by considering the same steps at the proof of Theorem11 and by using the hyperbolic angle sum formulas.Theorem 13.
Let us consider the matrixsR defined by(64)cosΥλ1VsinΥλ2VsinΥλ3VsinΥ−x1VsinΥ−x2VsinΥ−x3VsinΥctVsinΥλ1VsinΥcosΥ−x3VsinΥx2VsinΥctVsinΥ−λ3VsinΥλ2VsinΥ−x1VsinΥλ2VsinΥx3VsinΥcosΥ−x1VsinΥλ3VsinΥctVsinΥ−λ1VsinΥ−x2VsinΥλ3VsinΥ−x2VsinΥx1VsinΥcosΥ−λ2VsinΥλ1VsinΥctVsinΥ−x3VsinΥx1VsinΥctVsinΥλ3VsinΥ−λ2VsinΥcosΥx3VsinΥ−x2VsinΥ−λ1VsinΥx2VsinΥ−λ3VsinΥctVsinΥλ1VsinΥ−x3VsinΥcosΥx1VsinΥ−λ2VsinΥx3VsinΥλ2VsinΥ−λ1VsinΥctVsinΥx2VsinΥ−x1VsinΥcosΥ−λ3VsinΥctVsinΥx1VsinΥx2VsinΥx3VsinΥ−λ1VsinΥ−λ2VsinΥ−λ3VsinΥcosΥ,for a split octonion with the inner product having a positive value and spacelike vector part. Then, the matrix sRn is(65)cosnΥλ1VsinnΥλ2VsinnΥλ3VsinnΥ−x1VsinnΥ−x2VsinnΥ−x3VsinnΥctVsinnΥλ1VsinhnΥcosnΥ−x3VsinnΥx2VsinnΥctVsinnΥ−λ3VsinnΥλ2VsinnΥ−x1VsinnΥλ2VsinnΥx3VsinnΥcosnΥ−x1VsinnΥλ3VsinnΥctVsinnΥ−λ1VsinnΥ−x2VsinnΥλ3VsinnΥ−x2VsinnΥx1VsinnΥcosnΥ−λ2VsinnΥλ1VsinnΥctVsinnΥ−x3VsinnΥx1VsinnΥctVsinnΥλ3VsinnΥ−λ2VsinnΥcosnΥx3VsinnΥ−x2VsinnΥ−λ1VsinnΥx2VsinnΥ−λ3VsinnΥctVsinnΥλ1VsinnΥ−x3VsinnΥcosnΥx1VsinnΥ−λ2VsinnΥx3VsinnΥλ2VsinnΥ−λ1VsinnΥctVsinnΥx2VsinnΥ−x1VsinnΥcosnΥ−λ3VsinnΥctVsinnΥx1VsinnΥx2VsinnΥx3VsinnΥ−λ1VsinnΥ−λ2VsinnΥ−λ3VsinnΥcosnΥ.Proof.
One can show the proof like the proof of Theorem11.Theorem 14.
Let us consider the following matrix:(66)sR=w1λ1wλ2wλ3w−x1w−x2w−x3wctwλ1w1−x3wx2wctw−λ3wλ2w−x1wλ2wx3w1−x1wλ3wctw−λ1w−x2wλ3w−x2wx1w1−λ2wλ1wctw−x3wx1wctwλ3w−λ2w1x3w−x2w−λ1wx2w−λ3wctwλ1w−x31x1w−λ2wx3wλ2w−λ1wctwx2w−x1w1−λ3wctwx1wx2wx3w−λ1w−λ2w−λ3w1,for a split octonion with the inner product having zero. In this case, the matrix sRn is(67)wn2n−11λ1wλ2wλ3w−x1w−x2w−x3wctwλ1w1−x3wx2wctw−λ3wλ2w−x1wλ2wx3w1−x1wλ3wctw−λ1w−x2wλ3w−x2wx1w1−λ2wλ1wctw−x3wx1wctwλ3w−λ2w1x3w−x2w−λ1wx2w−λ3wctwλ1w−x31x1w−λ2wx3wλ2w−λ1wctwx2w−x1w1−λ3wctwx1wx2wx3w−λ1w−λ2w−λ3w1.Proof.
Suppose thats is a split octonion with the inner product having zero and U2=1, that is,(68)λ12w2+λ22w2+λ32w2−x12w2−x22w2−x32w2+ct2w2=1.
One can verify the proof as similar to Theorem4.Theorem 15.
Let us consider the following matrix:(69)sR=sgnwλ1λ2λ3−x1−x2−x3ctλ1sgnw−x3x2ct−λ3λ2−x1λ2x3sgnw−x1λ3ct−λ1−x2λ3−x2x1sgnw−λ2λ1ct−x3x1ctλ3−λ2sgnwx3−x2−λ1x2−λ3ctλ1−x3sgnwx1−λ2x3λ2−λ1ctx2−x1sgnw−λ3ctx1x2x3−λ1−λ2−λ3sgnw,for a unit split octonion with the lightlike vector part U,U2  = 0. Then, the matrix sRn is(70)sgnwnnwn−1λ1nwn−1λ2nwn−1λ3−nwn−1x1−nwn−1x2−nwn−1x3nwn−1ctnwn−1λ1sgnwn−nwn−1x3nwn−1x2nwn−1ct−nwn−1λ3nwn−1λ2−nwn−1x1nwn−1λ2nwn−1x3sgnwn−nwn−1x1nwn−1λ3nwn−1ct−nwn−1λ1−nwn−1x2nwn−1λ3−nwn−1x2nwn−1x1sgnwn−nwn−1λ2nwn−1λ1nwn−1ct−nwn−1x3nwn−1x1nwn−1ctnwn−1λ3−nwn−1λ2sgnwnnwn−1x3−nwn−1x2−nwn−1λ1nwn−1x2−nwn−1λ3nwn−1ctnwn−1λ1−nwn−1x3sgnwnnwn−1x1−nwn−1λ2nwn−1x3nwn−1λ2−nwn−1λ1nwn−1ctnwn−1x2−nwn−1x1sgnwn−nwn−1λ3nwn−1ctnwn−1x1nwn−1x2nwn−1x3−nwn−1λ1−nwn−1λ2−nwn−1λ3sgnwn.Proof.
The formula can be found similar to Theorem5.


## 6. Roots of the Matrices of Split Octonions

In this section, we investigate the solutions of the equationxn=sR. The equation has some different solutions with respect to the type of split octonion matrix representations as follows:Theorem 16.
Assume thatsR is given as a right matrix representation for a unit split octonion with an inner product having a negative value and a unit timelike 7− vector U. Then, the equation xn=sR has the following solutions:(i)
Ifn is an even number, then there is no solution(ii)
Ifn is an odd number, then there is only 1− solution x0 given by(71)sinhϑλ1Vcoshϑλ2Vcoshϑλ3Vcoshϑ−x1Vcoshϑ−x2Vcoshϑ−x3VcoshϑctVcoshϑλ1Vcoshϑsinhϑ−x3Vcoshϑx2VcoshϑctVcoshϑ−λ3Vcoshϑλ2Vcoshϑ−x1Vcoshϑλ2Vcoshϑx3Vcoshϑsinhϑ−x1Vcoshϑλ3VcoshϑctVcoshϑ−λ1Vcoshϑ−x2Vcoshϑλ3Vcoshϑ−x2Vcoshϑx1Vcoshϑsinhϑ−λ2Vcoshϑλ1VcoshϑctVcoshϑ−x3Vcoshϑx1VcoshϑctVcoshϑλ3Vcoshϑ−λ2Vcoshϑsinhϑx3Vcoshϑ−x2Vcoshϑ−λ1Vcoshϑx2Vcoshϑ−λ3VcoshϑctVcoshϑλ1Vcoshϑ−x3Vcoshϑsinhϑx1Vcoshϑ−λ2Vcoshϑx3Vcoshϑλ2Vcoshϑ−λ1VcoshϑctVcoshϑx2Vcoshϑ−x1Vcoshϑsinhϑ−λ3VcoshϑctVcoshϑx1Vcoshϑx2Vcoshϑx3Vcoshϑ−λ1Vcoshϑ−λ2Vcoshϑ−λ3Vcoshϑsinhϑ,whereϑ=Υ/n.Proof.
Using Theorem11, we can write(72)sinhnϑλ1Vcoshnϑλ2Vcoshnϑλ3Vcoshnϑ−x1Vcoshnϑ−x2Vcoshnϑ−x3VcoshnϑctVcoshnϑλ1Vcoshnϑsinhnϑ−x3Vcoshnϑx2VcoshnϑctVcoshnϑ−λ3Vcoshnϑλ2Vcoshnϑ−x1Vcoshnϑλ2Vcoshnϑx3Vcoshnϑsinhnϑ−x1Vcoshnϑλ3VcoshnϑctVcoshnϑ−λ1Vcoshnϑ−x2Vcoshnϑλ3Vcoshnϑ−x2Vcoshnϑx1Vcoshϑsinhnϑ−λ2Vcoshnϑλ1VcoshnϑctVcoshnϑ−x3Vcoshnϑx1VcoshnϑctVcoshnϑλ3Vcoshnϑ−λ2Vcoshnϑsinhnϑx3Vcoshnϑ−x2Vcoshnϑ−λ1Vcoshnϑx2Vcoshnϑ−λ3VcoshnϑctVcoshnϑλ1Vcoshnϑ−x3Vcoshϑsinhnϑx1Vcoshnϑ−λ2Vcoshnϑx3Vcoshnϑλ2Vcoshnϑ−λ1VcoshnϑctVcoshnϑx2Vcoshnϑ−x1Vcoshnϑsinhnϑ−λ3VcoshnϑctVcoshnϑx1Vcoshnϑx2Vcoshnϑx3Vcoshnϑ−λ1Vcoshnϑ−λ2Vcoshnϑ−λ3Vcoshnϑsinhnϑ,
for thenth− power of x0. Hence, x0n=sR is obtained. The other case is clear from Theorem 11.Theorem 17.
Assume thatsR is given as a right matrix representation for a unit split octonion with an inner product having a positive value and a unit timelike 7− vector U. Then, the equation xn=sR has the following solutions:(i)
Ifn is an even number, then the 4− distinct solutions are(73)x0=sinhϑλ1Vcoshϑλ2Vcoshϑλ3Vcoshϑ−x1Vcoshϑ−x2Vcoshϑ−x3VcoshϑctVcoshϑλ1Vcoshϑsinhϑ−x3Vcoshϑx2VcoshϑctVcoshϑ−λ3Vcoshϑλ2Vcoshϑ−x1Vcoshϑλ2Vcoshϑx3Vcoshϑsinhϑ−x1Vcoshϑλ3VcoshϑctVcoshϑ−λ1Vcoshϑ−x2Vcoshϑλ3Vcoshϑ−x2Vcoshϑx1Vcoshϑsinhϑ−λ2Vcoshϑλ1VcoshϑctVcoshϑ−x3Vcoshϑx1VcoshϑctVcoshϑλ3Vcoshϑ−λ2Vcoshϑsinhϑx3Vcoshϑ−x2Vcoshϑ−λ1Vcoshϑx2Vcoshϑ−λ3VcoshϑctVcoshϑλ1Vcoshϑ−x3Vcoshϑsinhϑx1Vcoshϑ−λ2Vcoshϑx3Vcoshϑλ2Vcoshϑ−λ1VcoshϑctVcoshϑx2Vcoshϑ−x1Vcoshϑsinhϑ−λ3VcoshϑctVcoshϑx1Vcoshϑx2Vcoshϑx3Vcoshϑ−λ1Vcoshϑ−λ2Vcoshϑ−λ3Vcoshϑsinhϑ,x1=−sinhϑλ1Vcoshϑλ2Vcoshϑλ3Vcoshϑ−x1Vcoshϑ−x2Vcoshϑ−x3VcoshϑctVcoshϑλ1Vcoshϑsinhϑ−x3Vcoshϑx2VcoshϑctVcoshϑ−λ3Vcoshϑλ2Vcoshϑ−x1Vcoshϑλ2Vcoshϑx3Vcoshϑsinhϑ−x1Vcoshϑλ3VcoshϑctVcoshϑ−λ1Vcoshϑ−x2Vcoshϑλ3Vcoshϑ−x2Vcoshϑx1Vcoshϑsinhϑ−λ2Vcoshϑλ1VcoshϑctVcoshϑ−x3Vcoshϑx1VcoshϑctVcoshϑλ3Vcoshϑ−λ2Vcoshϑsinhϑx3Vcoshϑ−x2Vcoshϑ−λ1Vcoshϑx2Vcoshϑ−λ3VcoshϑctVcoshϑλ1Vcoshϑ−x3Vcoshϑsinhϑx1Vcoshϑ−λ2Vcoshϑx3Vcoshϑλ2Vcoshϑ−λ1VcoshϑctVcoshϑx2Vcoshϑ−x1Vcoshϑsinhϑ−λ3VcoshϑctVcoshϑx1Vcoshϑx2Vcoshϑx3Vcoshϑ−λ1Vcoshϑ−λ2Vcoshϑ−λ3Vcoshϑsinhϑ,x2=coshϑλ1Vsinhϑλ2Vsinhϑλ3Vsinhϑ−x1Vsinhϑ−x2Vsinhϑ−x3VsinhϑctVsinhϑλ1Vsinhϑcoshϑ−x3Vsinhϑx2VsinhϑctVsinhϑ−λ3Vsinhϑλ2Vsinhϑ−x1Vsinhϑλ2Vsinhϑx3Vsinhϑcoshϑ−x1Vsinhϑλ3VsinhϑctVsinhϑ−λ1Vsinhϑ−x2Vsinhϑλ3Vsinhϑ−x2Vsinhϑx1Vsinhϑcoshϑ−λ2Vsinhϑλ1VsinhϑctVsinhϑ−x3Vsinhϑx1VsinhϑctVsinhϑλ3Vsinhϑ−λ2Vsinhϑcoshϑx3Vsinhϑ−x2Vsinhϑ−λ1Vsinhϑx2Vsinhϑ−λ3VsinhϑctVsinhϑλ1Vsinhϑ−x3Vsinhϑcoshϑx1Vsinhϑ−λ2Vsinhϑx3Vsinhϑλ2Vsinhϑ−λ1VsinhϑctVsinhϑx2Vsinhϑ−x1Vsinhϑcoshϑ−λ3VsinhϑctVsinhϑx1Vsinhϑx2Vsinhϑx3Vsinhϑ−λ1Vsinhϑ−λ2Vsinhϑ−λ3Vsinhϑcoshϑ,x3=−coshϑλ1Vsinhϑλ2Vsinhϑλ3Vsinhϑ−x1Vsinhϑ−x2Vsinhϑ−x3VsinhϑctVsinhϑλ1Vsinhϑcoshϑ−x3Vsinhϑx2VsinhϑctVsinhϑ−λ3Vsinhϑλ2Vsinhϑ−x1Vsinhϑλ2Vsinhϑx3Vsinhϑcoshϑ−x1Vsinhϑλ3VsinhϑctVsinhϑ−λ1Vsinhϑ−x2Vsinhϑλ3Vsinhϑ−x2Vsinhϑx1Vsinhϑcoshϑ−λ2Vsinhϑλ1VsinhϑctVsinhϑ−x3Vsinhϑx1VsinhϑctVsinhϑλ3Vsinhϑ−λ2Vsinhϑcoshϑx3Vsinhϑ−x2Vsinhϑ−λ1Vsinhϑx2Vsinhϑ−λ3VsinhϑctVsinhϑλ1Vsinhϑ−x3Vsinhϑcoshϑx1Vsinhϑ−λ2Vsinhϑx3Vsinhϑλ2Vsinhϑ−λ1VsinhϑctVsinhϑx2Vsinhϑ−x1Vsinhϑcoshϑ−λ3VsinhϑctVsinhϑx1Vsinhϑx2Vsinhϑx3Vsinhϑ−λ1Vsinhϑ−λ2Vsinhϑ−λ3Vsinhϑcoshϑ,whereϑ=Υ/n.(ii)
Ifn is an odd number, the only 1− solution is(74)x0=sinhϑλ1Vcoshϑλ2Vcoshϑλ3Vcoshϑ−x1Vcoshϑ−x2Vcoshϑ−x3VcoshϑctVcoshϑλ1Vcoshϑsinhϑ−x3Vcoshϑx2VcoshϑctVcoshϑ−λ3Vcoshϑλ2Vcoshϑ−x1Vcoshϑλ2Vcoshϑx3Vcoshϑsinhϑ−x1Vcoshϑλ3VcoshϑctVcoshϑ−λ1Vcoshϑ−x2Vcoshϑλ3Vcoshϑ−x2Vcoshϑx1Vcoshϑsinhϑ−λ2Vcoshϑλ1VcoshϑctVcoshϑ−x3Vcoshϑx1VcoshϑctVcoshϑλ3Vcoshϑ−λ2Vcoshϑsinhϑx3Vcoshϑ−x2Vcoshϑ−λ1Vcoshϑx2Vcoshϑ−λ3VcoshϑctVcoshϑλ1Vcoshϑ−x3Vcoshϑsinhϑx1Vcoshϑ−λ2Vcoshϑx3Vcoshϑλ2Vcoshϑ−λ1VcoshϑctVcoshϑx2Vcoshϑ−x1Vcoshϑsinhϑ−λ3VcoshϑctVcoshϑx1Vcoshϑx2Vcoshϑx3Vcoshϑ−λ1Vcoshϑ−λ2Vcoshϑ−λ3Vcoshϑsinhϑ.Proof.
The proof can be obtained like Theorem16.Theorem 18.
Given the matrixsR for a unit split octonion with an inner product having a positive value and a unit spacelike 7− vector U. Then, the equation xn=sR has n-solutions in the following form:(75)xp=cosϑλ1Vsinϑλ2Vsinϑλ3Vsinϑ−x1Vsinϑ−x2Vsinϑ−x3VsinϑctVsinϑλ1Vsinϑcosϑ−x3Vsinϑx2VsinϑctVsinϑ−λ3Vsinϑλ2Vsinϑ−x1Vsinϑλ2Vsinϑx3Vsinϑcosϑ−x1Vsinϑλ3VsinϑctVsinϑ−λ1Vsinϑ−x2Vsinϑλ3Vsinϑ−x2Vsinϑx1Vsinϑcosϑ−λ2Vsinϑλ1VsinϑctVsinϑ−x3Vsinϑx1VsinϑctVsinϑλ3Vsinϑ−λ2Vsinϑcosϑx3Vsinϑ−x2Vsinϑ−λ1Vsinϑx2Vsinϑ−λ3VsinϑctVsinϑλ1Vsinϑ−x3Vsinϑcosϑx1Vsinϑ−λ2Vsinϑx3Vsinϑλ2Vsinϑ−λ1VsinϑctVsinϑx2Vsinϑ−x1Vsinϑcosϑ−λ3VsinϑctVsinϑx1Vsinϑx2Vsinϑx3Vsinϑ−λ1Vsinϑ−λ2Vsinϑ−λ3Vsinϑcosϑ,where ϑ=Υ+2pπ/n,p=0,1,2,…,n−1.Proof.
The proof is similar to Theorem16.Theorem 19.
Given the matrixsR for a split octonion with an inner product having zero value and a unit 7− vector U. Then, the equation xn=sR has the following solutions as follows:(i)
Ifn is an even number, there exist 2− solutions in the following forms:(76)x0=Ω1λ1wλ2wλ3w−x1w−x2w−x3wctwλ1w1−x3wx2wctw−λ3wλ2w−x1wλ2wx3w1−x1wλ3wctw−λ1w−x2wλ3w−x2wx1w1−λ2wλ1wctw−x3wx1wctwλ3w−λ2w1x3w−x2w−λ1wx2w−λ3wctwλ1w−x31x1w−λ2wx3wλ2w−λ1wctwx2w−x1w1−λ3wctwx1wx2wx3w−λ1w−λ2w−λ3w1,x1=−Ω1λ1wλ2wλ3w−x1w−x2w−x3wctwλ1w1−x3wx2wctw−λ3wλ2w−x1wλ2wx3w1−x1wλ3wctw−λ1w−x2wλ3w−x2wx1w1−λ2wλ1wctw−x3wx1wctwλ3w−λ2w1x3w−x2w−λ1wx2w−λ3wctwλ1w−x31x1w−λ2wx3wλ2w−λ1wctwx2w−x1w1−λ3wctwx1wx2wx3w−λ1w−λ2w−λ3w1,whereΩ=2wn/2.(ii)
Ifn is an odd number, there exists only 1− solution(77)x0=Ω1λ1wλ2wλ3w−x1w−x2w−x3wctwλ1w1−x3wx2wctw−λ3wλ2w−x1wλ2wx3w1−x1wλ3wctw−λ1w−x2wλ3w−x2wx1w1−λ2wλ1wctw−x3wx1wctwλ3w−λ2w1x3w−x2w−λ1wx2w−λ3wctwλ1w−x31x1w−λ2wx3wλ2w−λ1wctwx2w−x1w1−λ3wctwx1wx2wx3w−λ1w−λ2w−λ3w1,whereΩ=2wn/2.Proof.
The proof can be presented by the similar way to Theorem16.Theorem 20.
LetsR be a right matrix representation for a split octonion with an inner product having zero value and a unit 7− vector U. Then, the equation xn=sR has the following solutions as follows:(i)
Ifn is an even number and sgnw>0 , there exist 2− solutions in the following forms:(78)x0=wn1λ1nλ2nλ3n−x1n−x2n−x3nctnλ1n1−x3nx2nctn−λ3nλ2n−x1nλ2nx3n1−x1nλ3nctn−λ1n−x2nλ3n−x2nx1n1−λ2nλ1nctn−x3nx1nctnλ3n−λ2n1x3n−x2n−λ1nx2n−λ3nctnλ1n−x31x1n−λ2nx3nλ2n−λ1nctnx2n−x1n1−λ3nctnx1nx2nx3n−λ1n−λ2n−λ3n1,x1=−wn1λ1nλ2nλ3n−x1n−x2n−x3nctnλ1n1−x3nx2nctn−λ3nλ2n−x1nλ2nx3n1−x1nλ3nctn−λ1n−x2nλ3n−x2nx1n1−λ2nλ1nctn−x3nx1nctnλ3n−λ2n1x3n−x2n−λ1nx2n−λ3nctnλ1n−x31x1n−λ2nx3nλ2n−λ1nctnx2n−x1n1−λ3nctnx1nx2nx3n−λ1n−λ2n−λ3n1.But there is no solution forsgnw<0.(ii)
Ifn is an odd number, there exists only 1− solution in the following form:(79)x0=wn1λ1nλ2nλ3n−x1n−x2n−x3nctnλ1n1−x3nx2nctn−λ3nλ2n−x1nλ2nx3n1−x1nλ3nctn−λ1n−x2nλ3n−x2nx1n1−λ2nλ1nctn−x3nx1nctnλ3n−λ2n1x3n−x2n−λ1nx2n−λ3nctnλ1n−x31x1n−λ2nx3nλ2n−λ1nctnx2n−x1n1−λ3nctnx1nx2nx3n−λ1n−λ2n−λ3n1Proof.
The proof can be given similarly to Theorem16.


## 7. An Application

Example 1.
Let us consider a split octonion as(80)s=12+2J1+12J2+12J3+12j1+12j2+12j3.
SinceIs=−1 and V2=3/2, s is a split octonion with the inner product having a negative value and timelike vector part. So the split octonion s is written in the polar form(81)s=sinhΥ+UcoshΥ,where(82)U=2J1+1/2J2+1/2J3+1/2j1+1/2j2+1/2j33/2and sinhΥ=1/2, coshΥ=3/2. Then, from Theorem 1, the 7th− power of the split octonion s is obtained as(83)s7=sinh7Υ+Ucosh7Υ,=50.2046+57.9828J1+28.9914J2+28.9914J3+28.9914j1+28.9914j2+28.9914j3.And by Theorem 6, the 7th− root of the split octonion s is calculated as(84)s7=sinhΥ7+UcoshΥ7,=0.0942+1.1598J1+0.5799J2+0.5799J3+0.5799j1+0.5799j2+0.5799j3.
The right representation of the split octonions denoted by sR can be written as follows:(85)sR=1/221/21/2−1/2−1/2−1/2021/2−1/21/20−1/21/2−1/21/21/21/2−1/21/20−2−1/21/2−1/21/21/2−1/220−1/21/201/2−1/21/21/2−1/2−21/2−1/202−1/21/21/2−1/21/21/2−201/2−1/21/2−1/201/21/21/2−2−1/2−1/21/2,or in the polar form(86)sinhΥ23/2coshΥ1/23/2coshΥ1/23/2coshΥ−1/23/2coshΥ−1/23/2coshΥ−1/23/2coshΥ023/2coshΥsinhΥ−1/23/2coshΥ1/23/2coshΥ0−1/23/2coshΥ1/23/2coshΥ−1/23/2coshΥ1/23/2coshΥ1/23/2coshΥsinhΥ−1/23/2coshΥ1/23/2coshΥ0−23/2coshΥ−1/23/2coshΥ1/23/2coshΥ−1/23/2coshΥ1/23/2coshΥsinhΥ−1/23/2coshΥ23/2coshΥ0−1/23/2coshΥ1/23/2coshΥ01/23/2coshΥ−1/23/2coshΥsinhΥ1/23/2coshΥ−1/23/2coshΥ−23/2coshΥ1/23/2coshΥ−1/23/2coshΥ023/2coshΥ−1/23/2coshΥsinhΥ1/23/2coshΥ−1/23/2coshΥ1/23/2coshΥ1/23/2coshΥ−23/2coshΥ01/23/2coshΥ−1/23/2coshΥsinhΥ−1/23/2coshΥ01/23/2coshΥ1/23/2coshΥ1/23/2coshΥ−23/2coshΥ−1/23/2coshΥ−1/23/2coshΥsinhΥ.
In addition, the7th− power of the matrix sR by Theorem 11 is(87)50.204657.982828.991428.9914−28.9914−28.9914−28.99140.000057.982850.2046−28.991428.99140.0000−28.991428.9914−28.991428.991428.991450.2046−28.991428.99140.0000−57.9828−28.991428.9914−28.991428.991450.2046−28.991457.98280.0000−28.991428.99140.000028.9914−28.991450.204628.9914−28.9914−57.982828.9914−28.99140.000057.9828−28.991450.204628.9914−28.991428.991428.9914−57.98280.000028.9914−28.991450.2046−28.99140.000028.991428.991428.9914−57.9828−28.9914−28.991450.2046.
However, if we want to find the7th− root of sR from Theorem 16, we find(88)0.09421.15980.57990.5799−0.5799−0.5799−0.579901.15980.0942−0.57990.57990−0.57990.5799−0.57990.57990.57990.0942−0.57990.57990−1.1598−0.57990.5799−0.57990.57990.0942−0.57991.15980−0.57990.579900.5799−0.57990.09420.5799−0.57991.15980.5799−0.579901.1598−0.57990.09420.5799−0.57990.57990.5799−1.159800.5799−0.57990.0942−0.579900.57990.57990.5799−1.1598−0.5799−0.57990.0942.
The accuracy of each of the above calculations forΥ=sinh−11/2 can be checked with the following Matlab codes.


## 8. Conclusions

In this article, we present polar forms for the split octonions with the inner product being zero and the split octonions with the lightlike vector part. By using the polar forms of the split octonions, we give De Moivre’s formulas for split octonions and matrices related with split octonions. Thus, we obtain thenth− powers of split octonions and matrices related with split octonions. In addition to this, we calculate the nth− roots of the split octonions and the matrix representations of split octonions. Finally, we give an example to illustrate the obtained results.



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