How do I solve (a+bi+cj+dk)^(f+gi+hj+nk)?

Question

I purposely skipped using e as a factor in the title because e is Eulers Number.
I have found $$e^{(a+bi+cj+dk)}$$ and $$(a+bi+cj+dk)^n$$ but no way to combine them together. My current theory is adding on to the formula for imaginary numbers to get: $$n=e^{(w_2*\ln{⁡(\sqrt{w_1^2+x_1^2})}+x_2*\arctan⁡(\frac{x_1}{w_1})-y_2*\ln{⁡(\sqrt{y_1^2+z_1^2})}-z_2*\arctan{⁡(\frac{z_1}{y_1})})}$$ where (n,w,x,y,z) is (a,a,b,c,d),(b,b,a,d,c),(c,c,d,a,b),(d,d,c,b,a) Could someone please give me the true formula? It would be very well appreciated. Thank you so very much!

Answer 1

If you want to define $p^q$ for quaternions $p$ and $q$, you ought to first start at $w^z$ for complex numbers $w$ and $z$. Plugging something like $i^i$ or $\log i$ into a calculator tells you they can be defined, but it doesn't tell you how it's defined, so you stopped short of doing more research you could've been doing.

For complex numbers $w$ and $z$, the value of $w^z$ is defined to be $\exp(z\ln w)$, assuming you have a way to define $\ln w$. Famously, $\exp$ is not one-to-one on $\mathbb{C}$ (since $\exp 0=\exp2\pi i$ for instance), so it doesn't have an inverse which is holomorphic on all of $\mathbb{C}^\times=\mathbb{C}\setminus\{0\}$. That means we have to make some compromises in order to even define $\ln w$ at all. The way this is done in practice is with branch cuts. The "standard branch cut" is considered to be the nonpositive real axis $(-\infty,0]$, in which case $\ln$ is defined on $\mathbb{C}^\times$ but is only holomorphic on $\mathbb{C}\setminus(-\infty,0]$. If you graph $\ln w$ as $w$ crosses $(-\infty,0]$, you'll see a jump discontinuity where the jump is $\pm2\pi i$. Specifically, if $w=r\exp(i\theta)$ with $r>0$ and $-\pi<\theta\le\pi$ then $\ln w:=\ln r+i\theta$. Many argue, though, that the branch cut is arbitrary, and even that $\ln w$ isn't, or shouldn't, be defined.

This works for quaternions too. All quaternions have a polar form $p=r\exp(\theta\mathbf{u})$ for some unit vector $\mathbf{u}$, and we can arrange for $-\pi<\theta\le\pi$. There is some redundancy, since $\theta\mathbf{u}=(-\theta)(-\mathbf{u})$, but the resulting expression $\ln p:=\ln r+\theta\mathbf{u}$ is well-defined regardless. Yet, it still would be nice to have unique polar forms, so I do recommend restricting to convex angles $0\le\theta\le\pi$, in which case $\theta$ is unique, and $\mathbf{u}$ is unique unless $\theta=0$ or $\pi$ in which case it is arbitrary.

For real values $s$, you can define $q^s:=\exp(s\ln q)$ using the previous discussion to make sense of $\ln q$. But for quaternion bases you come to a decision point: should you define $p^q$ to be $\exp(q\ln p)$ or $\exp\big((\ln p)q\big)$. Or you could even pick things even between these two, like $\exp(q^s(\ln p)q^{1-s})$ for real values $s$.

And even if you do make an arbitrary choice to define $p^q$ ... why? Rotational dynamics don't demand this - in the broader context of Lie theory it makes sense to express everything with the natural exponential function. The function isn't holomorphic on any domain - indeed, the only quaternion holomorphic functions are affine functions $ax+b$ or $xa+b$ (depending on whether you're considering the left or right derivative). Not even power functions like $x^2$ are holomorphic, let alone power series or whatnot.

Maybe you just think $p^q$ is such a basic operation for real numbers that it makes sense to trek for a sensible interpretation of it for other number systems out of a sense of completeness. But it just turns out that it's mostly ugly and unnatural. (I would argue writing $p^t$ for real $t$ as shorthand is debatably a positive choice in certain contexts, but besides that, a negative choice.)

Answer 2

A quaternion $q$ with zero real part has negative square, so with e.g. Taylor series you can show $e^q=\cos|q|+q\operatorname{sinc}|q|$ ($\operatorname{sinc}$ is explained here), in analogy with $e^{i\theta}=\cos\theta+i\sin\theta$ for imaginary $i\theta\in\Bbb C$. Now you know how to evaluate $e^z$ for any quaternion $z$, we can use $z=|z|e^q$ with $\Re q=0$ to obtain $z^w=|z|^we^{qw}=e^{w(\ln|z|+q)}$. The hard part is determining which $q$ satisfy $\Re q=0,\,e^q=z/|z|$, which I leave you to ponder; but there will be plenty because, as with complex exponentiation, quaternionic exponentiation is multi-valued.

Answer 3

As others have already pointed out that this expression would be ambiguous, I will try to provide a formula for obtaining at least one value.

We start with this Mathematica code for matrix algebra:

$PrePrint =.; FormatValues@Matrix = {};
Hold[MakeBoxes[Matrix[a_], StandardForm] ^:= 
    Block[{Internal`$ConvertForms = {}}, 
     TemplateBox[{MakeBoxes[MatrixForm@a][[1, 1, 3]]}, "Matrix", 
      DisplayFunction -> (RowBox@{style@"(", "\[NoBreak]", #, 
           "\[NoBreak]", style@")"} &)]]] /. 
  style[string_] -> StyleBox[string, FontColor -> Red] // ReleaseHold
Matrix /: Plus[Matrix[A_?SquareMatrixQ], Matrix[B_?SquareMatrixQ]] := 
 Matrix[KroneckerProduct[A, 
    IdentityMatrix[LCM[Length[A], Length[B]]/Length[A]]] + 
   KroneckerProduct[B, 
    IdentityMatrix[LCM[Length[A], Length[B]]/Length[B]]]]
Matrix /: Dot[Matrix[A_?SquareMatrixQ], Matrix[B_?SquareMatrixQ]] := 
 Matrix[KroneckerProduct[A, 
    IdentityMatrix[LCM[Length[A], Length[B]]/Length[A]]] . 
   KroneckerProduct[B, 
    IdentityMatrix[LCM[Length[A], Length[B]]/Length[B]]]]
Matrix /: Matrix[x_?MatrixQ] := 
  First[First[x]] /; x == First[First[x]] IdentityMatrix[Length[x]];
Matrix /: x_?NumericQ + Matrix[y_] := 
  Matrix[x IdentityMatrix[Length[y]] + y];
Matrix /: x_?NumericQ Matrix[y_] := Matrix[x y];
Matrix /: Matrix[x_]*Matrix[y_] := 
  Matrix[x] . Matrix[y] /; 
   Matrix[x] . Matrix[y] == Matrix[y] . Matrix[x];
Matrix /: Power[Matrix[x_?MatrixQ], 0] := 1;
Matrix /: Re[Matrix[x_]] := Matrix[Re[x]];
Matrix /: Power[Matrix[x_?MatrixQ], y_?NumericQ] := 
  Matrix[MatrixPower[x, y]];
Matrix /: Power[Matrix[x_?MatrixQ], Matrix[y_?MatrixQ]] := 
  Exp[Log[Matrix[x]] . Matrix[y]];
Matrix /: Log[Matrix[x_?MatrixQ], Matrix[y_?MatrixQ]] := 
 Log[Matrix[x]]^-1 . Log[Matrix[y]]
Matrix /: Scalar[Matrix[x_?MatrixQ]] := Tr[x]/Length[x]
Scalar[x_?NumericQ] := Re[x]
$Post2 = 
  FullSimplify[# /. 
      f_[args1___?NumericQ, Matrix[mat_], args2___?NumericQ] :> 
       Matrix[MatrixFunction[f[args1, #, args2] &, mat]] /. 
     NonCommutativeMultiply -> Dot] &;
$Post = Nest[$Post2, #, 3] /. Dot -> NonCommutativeMultiply &;

It allows to insert matrices into normal expressions. It also defines matrix exponentiation as $A^B=\exp(\ln A \cdot B)$. As the matrices are generally not commutative, one can define it in another way: $A^B=\exp(B \cdot\ln A)$, but I consider the former definition to be more natural.

Now, we define basis, isomorphic to the quaternions:

i := Matrix[( {
     {I, 0},
     {0, -I}
    } )];
j := Matrix[( {
     {0, 1},
     {-1, 0}
    } )];
k := Matrix[( {
     {0, I},
     {I, 0}
    } ) ];

Declare the coefficients to be numbers:

NumericQ[a] := True
NumericQ[b] := True
NumericQ[c] := True
NumericQ[d] := True
NumericQ[e] := True
NumericQ[f] := True
NumericQ[g] := True
NumericQ[h] := True

Evaluate the expression:

(a + b i + c j + d k)^(e + f i + g j + h k)

Wait a few hours and get the formula:

$\left( \begin{array}{cc} \frac{\left(a-\sqrt{-b^2-c^2-d^2}\right)^{\frac{b f+c g+d h+e \sqrt{-b^2-c^2-d^2}}{2 \sqrt{-b^2-c^2-d^2}}} \left(a+\sqrt{-b^2-c^2-d^2}\right)^{-\frac{b f+c g+d h+e \left(-\sqrt{-b^2-c^2-d^2}\right)}{2 \sqrt{-b^2-c^2-d^2}}} e^{-\frac{\sqrt{\left(\left(b f+c g+d h+e \sqrt{-b^2-c^2-d^2}\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right)-\left(b f+c g+d h+e \left(-\sqrt{-b^2-c^2-d^2}\right)\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right)\right)^2+4 \left(b^2+c^2+d^2\right) \left(e^2+f^2+g^2+h^2\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right)}}{2 \sqrt{-b^2-c^2-d^2}}} \left(-i \sqrt{-b^2-c^2-d^2} f \log \left(a-\sqrt{-b^2-c^2-d^2}\right)-i d g \log \left(a-\sqrt{-b^2-c^2-d^2}\right)+b e i \log \left(a-\sqrt{-b^2-c^2-d^2}\right)+c h i \log \left(a-\sqrt{-b^2-c^2-d^2}\right)-i b e \log \left(a+\sqrt{-b^2-c^2-d^2}\right)-i \sqrt{-b^2-c^2-d^2} f \log \left(a+\sqrt{-b^2-c^2-d^2}\right)-i c h \log \left(a+\sqrt{-b^2-c^2-d^2}\right)+d g i \log \left(a+\sqrt{-b^2-c^2-d^2}\right)+\sqrt{\left(\left(b f+c g+d h+e \sqrt{-b^2-c^2-d^2}\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right)-\left(b f+c g+d h+e \left(-\sqrt{-b^2-c^2-d^2}\right)\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right)\right)^2+4 \left(b^2+c^2+d^2\right) \left(e^2+f^2+g^2+h^2\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right)}+e^{\frac{\sqrt{\left(\left(b f+c g+d h+e \sqrt{-b^2-c^2-d^2}\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right)-\left(b f+c g+d h+e \left(-\sqrt{-b^2-c^2-d^2}\right)\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right)\right)^2+4 \left(b^2+c^2+d^2\right) \left(e^2+f^2+g^2+h^2\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right)}}{\sqrt{-b^2-c^2-d^2}}} \left(\left(b e-\sqrt{-b^2-c^2-d^2} f-d g+c h\right) (-i) \log \left(a-\sqrt{-b^2-c^2-d^2}\right)+\left(b e-d g+c h+f \sqrt{-b^2-c^2-d^2}\right) i \log \left(a+\sqrt{-b^2-c^2-d^2}\right)+\sqrt{\left(\left(b f+c g+d h+e \sqrt{-b^2-c^2-d^2}\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right)-\left(b f+c g+d h+e \left(-\sqrt{-b^2-c^2-d^2}\right)\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right)\right)^2+4 \left(b^2+c^2+d^2\right) \left(e^2+f^2+g^2+h^2\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right)}\right)\right)}{2 \sqrt{\left(\left(b f+c g+d h+e \sqrt{-b^2-c^2-d^2}\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right)-\left(b f+c g+d h+e \left(-\sqrt{-b^2-c^2-d^2}\right)\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right)\right)^2+4 \left(b^2+c^2+d^2\right) \left(e^2+f^2+g^2+h^2\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right)}} & \frac{\left(a-\sqrt{-b^2-c^2-d^2}\right)^{\frac{b f+c g+d h+e \sqrt{-b^2-c^2-d^2}}{2 \sqrt{-b^2-c^2-d^2}}} \left(a+\sqrt{-b^2-c^2-d^2}\right)^{-\frac{b f+c g+d h+e \left(-\sqrt{-b^2-c^2-d^2}\right)}{2 \sqrt{-b^2-c^2-d^2}}} e^{-\frac{\sqrt{\left(-\left(\left(b^2+c^2+d^2\right) e^2\right)+2 (b f+c g+d h) \sqrt{-b^2-c^2-d^2} e+(b f+c g+d h)^2\right) \log ^2\left(a-\sqrt{-b^2-c^2-d^2}\right)+2 \left(\left(e^2+f^2+2 \left(g^2+h^2\right)\right) b^2-2 f (c g+d h) b-2 c d g h+d^2 \left(e^2+h^2+2 \left(f^2+g^2\right)\right)+c^2 \left(e^2+2 f^2+g^2+2 h^2\right)\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right)+\left(-\left(\left(b^2+c^2+d^2\right) e^2\right)-2 \sqrt{-b^2-c^2-d^2} (b f+c g+d h) e+(b f+c g+d h)^2\right) \log ^2\left(a+\sqrt{-b^2-c^2-d^2}\right)}}{2 \sqrt{-b^2-c^2-d^2}}} \left(-1+e^{\frac{\sqrt{\left(-\left(\left(b^2+c^2+d^2\right) e^2\right)+2 (b f+c g+d h) \sqrt{-b^2-c^2-d^2} e+(b f+c g+d h)^2\right) \log ^2\left(a-\sqrt{-b^2-c^2-d^2}\right)+2 \left(\left(e^2+f^2+2 \left(g^2+h^2\right)\right) b^2-2 f (c g+d h) b-2 c d g h+d^2 \left(e^2+h^2+2 \left(f^2+g^2\right)\right)+c^2 \left(e^2+2 f^2+g^2+2 h^2\right)\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right)+\left(-\left(\left(b^2+c^2+d^2\right) e^2\right)-2 \sqrt{-b^2-c^2-d^2} (b f+c g+d h) e+(b f+c g+d h)^2\right) \log ^2\left(a+\sqrt{-b^2-c^2-d^2}\right)}}{\sqrt{-b^2-c^2-d^2}}}\right) \left(\left(\left(-i b+\sqrt{-b^2-c^2-d^2}\right) (g+h i)-(c+d i) (e-i f)\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right)+\left((c+d i) (e-i f)+(g+h i) \left(i b+\sqrt{-b^2-c^2-d^2}\right)\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right)\right)}{2 \sqrt{\left(-\left(\left(b^2+c^2+d^2\right) e^2\right)+2 (b f+c g+d h) \sqrt{-b^2-c^2-d^2} e+(b f+c g+d h)^2\right) \log ^2\left(a-\sqrt{-b^2-c^2-d^2}\right)+2 \left(\left(e^2+f^2+2 \left(g^2+h^2\right)\right) b^2-2 f (c g+d h) b-2 c d g h+d^2 \left(e^2+h^2+2 \left(f^2+g^2\right)\right)+c^2 \left(e^2+2 f^2+g^2+2 h^2\right)\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right)+\left(-\left(\left(b^2+c^2+d^2\right) e^2\right)-2 \sqrt{-b^2-c^2-d^2} (b f+c g+d h) e+(b f+c g+d h)^2\right) \log ^2\left(a+\sqrt{-b^2-c^2-d^2}\right)}} \\ \frac{\left(a-\sqrt{-b^2-c^2-d^2}\right)^{\frac{b f+c g+d h+e \sqrt{-b^2-c^2-d^2}}{2 \sqrt{-b^2-c^2-d^2}}} \left(a+\sqrt{-b^2-c^2-d^2}\right)^{-\frac{b f+c g+d h+e \left(-\sqrt{-b^2-c^2-d^2}\right)}{2 \sqrt{-b^2-c^2-d^2}}} e^{-\frac{\sqrt{\left(-\left(\left(b^2+c^2+d^2\right) e^2\right)+2 (b f+c g+d h) \sqrt{-b^2-c^2-d^2} e+(b f+c g+d h)^2\right) \log ^2\left(a-\sqrt{-b^2-c^2-d^2}\right)+2 \left(\left(e^2+f^2+2 \left(g^2+h^2\right)\right) b^2-2 f (c g+d h) b-2 c d g h+d^2 \left(e^2+h^2+2 \left(f^2+g^2\right)\right)+c^2 \left(e^2+2 f^2+g^2+2 h^2\right)\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right)+\left(-\left(\left(b^2+c^2+d^2\right) e^2\right)-2 \sqrt{-b^2-c^2-d^2} (b f+c g+d h) e+(b f+c g+d h)^2\right) \log ^2\left(a+\sqrt{-b^2-c^2-d^2}\right)}}{2 \sqrt{-b^2-c^2-d^2}}} \left(-1+e^{\frac{\sqrt{\left(-\left(\left(b^2+c^2+d^2\right) e^2\right)+2 (b f+c g+d h) \sqrt{-b^2-c^2-d^2} e+(b f+c g+d h)^2\right) \log ^2\left(a-\sqrt{-b^2-c^2-d^2}\right)+2 \left(\left(e^2+f^2+2 \left(g^2+h^2\right)\right) b^2-2 f (c g+d h) b-2 c d g h+d^2 \left(e^2+h^2+2 \left(f^2+g^2\right)\right)+c^2 \left(e^2+2 f^2+g^2+2 h^2\right)\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right)+\left(-\left(\left(b^2+c^2+d^2\right) e^2\right)-2 \sqrt{-b^2-c^2-d^2} (b f+c g+d h) e+(b f+c g+d h)^2\right) \log ^2\left(a+\sqrt{-b^2-c^2-d^2}\right)}}{\sqrt{-b^2-c^2-d^2}}}\right) \left(\left((c-i d) (e+f i)-\left(i b+\sqrt{-b^2-c^2-d^2}\right) (g-i h)\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right)-\left((c-i d) (e+f i)+(g-i h) \left(-i b+\sqrt{-b^2-c^2-d^2}\right)\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right)\right)}{2 \sqrt{\left(-\left(\left(b^2+c^2+d^2\right) e^2\right)+2 (b f+c g+d h) \sqrt{-b^2-c^2-d^2} e+(b f+c g+d h)^2\right) \log ^2\left(a-\sqrt{-b^2-c^2-d^2}\right)+2 \left(\left(e^2+f^2+2 \left(g^2+h^2\right)\right) b^2-2 f (c g+d h) b-2 c d g h+d^2 \left(e^2+h^2+2 \left(f^2+g^2\right)\right)+c^2 \left(e^2+2 f^2+g^2+2 h^2\right)\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right)+\left(-\left(\left(b^2+c^2+d^2\right) e^2\right)-2 \sqrt{-b^2-c^2-d^2} (b f+c g+d h) e+(b f+c g+d h)^2\right) \log ^2\left(a+\sqrt{-b^2-c^2-d^2}\right)}} & \frac{\left(a-\sqrt{-b^2-c^2-d^2}\right)^{\frac{b f+c g+d h+e \sqrt{-b^2-c^2-d^2}}{2 \sqrt{-b^2-c^2-d^2}}} \left(a+\sqrt{-b^2-c^2-d^2}\right)^{-\frac{b f+c g+d h+e \left(-\sqrt{-b^2-c^2-d^2}\right)}{2 \sqrt{-b^2-c^2-d^2}}} e^{-\frac{\sqrt{\left(\left(b f+c g+d h+e \sqrt{-b^2-c^2-d^2}\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right)-\left(b f+c g+d h+e \left(-\sqrt{-b^2-c^2-d^2}\right)\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right)\right)^2+4 \left(b^2+c^2+d^2\right) \left(e^2+f^2+g^2+h^2\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right)}}{2 \sqrt{-b^2-c^2-d^2}}} \left(-i c h \log \left(a-\sqrt{-b^2-c^2-d^2}\right)+b e (-i) \log \left(a-\sqrt{-b^2-c^2-d^2}\right)+d g i \log \left(a-\sqrt{-b^2-c^2-d^2}\right)+f i \sqrt{-b^2-c^2-d^2} \log \left(a-\sqrt{-b^2-c^2-d^2}\right)-i d g \log \left(a+\sqrt{-b^2-c^2-d^2}\right)+b e i \log \left(a+\sqrt{-b^2-c^2-d^2}\right)+c h i \log \left(a+\sqrt{-b^2-c^2-d^2}\right)+f i \log \left(a+\sqrt{-b^2-c^2-d^2}\right) \sqrt{-b^2-c^2-d^2}+\sqrt{\left(\left(b f+c g+d h+e \sqrt{-b^2-c^2-d^2}\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right)-\left(b f+c g+d h+e \left(-\sqrt{-b^2-c^2-d^2}\right)\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right)\right)^2+4 \left(b^2+c^2+d^2\right) \left(e^2+f^2+g^2+h^2\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right)}+e^{\frac{\sqrt{\left(\left(b f+c g+d h+e \sqrt{-b^2-c^2-d^2}\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right)-\left(b f+c g+d h+e \left(-\sqrt{-b^2-c^2-d^2}\right)\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right)\right)^2+4 \left(b^2+c^2+d^2\right) \left(e^2+f^2+g^2+h^2\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right)}}{\sqrt{-b^2-c^2-d^2}}} \left(i \left(\left(b e-\sqrt{-b^2-c^2-d^2} f-d g+c h\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right)-\left(b e-d g+c h+f \sqrt{-b^2-c^2-d^2}\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right)\right)+\sqrt{\left(\left(b f+c g+d h+e \sqrt{-b^2-c^2-d^2}\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right)-\left(b f+c g+d h+e \left(-\sqrt{-b^2-c^2-d^2}\right)\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right)\right)^2+4 \left(b^2+c^2+d^2\right) \left(e^2+f^2+g^2+h^2\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right)}\right)\right)}{2 \sqrt{\left(\left(b f+c g+d h+e \sqrt{-b^2-c^2-d^2}\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right)-\left(b f+c g+d h+e \left(-\sqrt{-b^2-c^2-d^2}\right)\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right)\right)^2+4 \left(b^2+c^2+d^2\right) \left(e^2+f^2+g^2+h^2\right) \log \left(a-\sqrt{-b^2-c^2-d^2}\right) \log \left(a+\sqrt{-b^2-c^2-d^2}\right)}} \\ \end{array} \right)$

It should be noted that the less coefficients you include, the simpler is the result. For instance, $i^j=k$.