On the exponentiation of three-dimensional numbers.

Question

There exists a general formula to calculate $({a+bi})^{c+di}$. However, it there a formula to calculate exponentiation with three-dimensional numbers, such as $({a+bi+cj})^{d+ei+fj}$ ?

Answer 1

Yes. But what exactly 3-dimensional algebra are you talking about? There are several. In the linked post are descriptions of 3 3-dimensional commutative number systems.

If we define

$1=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$

$j=\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\ \end{array} \right)$

$k=\left( \begin{array}{ccc} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{array} \right)$

then

$(a_1+b_1j+c_1k)^{(a_2+b_2j+c_2k)}=a_2 \left(b_1 k+c_1 j\right)+a_1 \left(a_2+b_2 k+c_2 j\right)+b_2 \left(b_1 j+c_1\right)+c_2 \left(b_1+c_1 k\right)$

If we define

$1=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$

$j=\left( \begin{array}{ccc} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$

$k=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \\ \end{array} \right)$

then

$(a_1+b_1j+c_1k)^{(a_2+b_2j+c_2k)}=\frac{1}{2} (1-j) \left(\left(a_1-b_1+c_1\right){}^{a_2-b_2+c_2}-\left(a_1+b_1+c_1\right){}^{a_2+b_2+c_2}\right)+\frac{1}{2} (1-k) \left(\left(a_1+b_1-c_1\right){}^{a_2+b_2-c_2}-\left(a_1+b_1+c_1\right){}^{a_2+b_2+c_2}\right)+\left(a_1+b_1+c_1\right){}^{a_2+b_2+c_2}$

If we define

$1=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$

$j=\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{array} \right)$

$k=\left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right)$

then the formula is

$(a_1+b_1j+c_1k)^{(a_2+b_2j+c_2k)}=\frac{1}{2} a_2^{a_1-2} \left(a_2^2 \left(\log \left(a_2\right) \left(b_1^2 k \log \left(a_2\right)+2 b_1 j+2 c_1 k\right)+2\right)+2 a_2 \left(a_1 \left(b_2 \left(b_1 k \log \left(a_2\right)+j\right)+c_2 k\right)+b_1 b_2 k\right)+\left(a_1-1\right) a_1 b_2^2 k\right)$