TITLE: Caculus of $j^j$, $i^j$ and $j^i$ (where $i^2=-1$, $j^2=1$, $j \neq -1$ and $j\neq 1$), and others like $Ln(i+j)$
QUESTION [3 upvotes]: First of all, I am do math only for fun, and I am only amateur, so excuse me because of my lack on knowledge in this field.
I was curious about some answers and I want to know if my judgment was ok or wrong.
First I wanted to find out the answer of:
$$i^j$$ (where $i^2=-1, j^2=1, j≠-1, j≠1$), j is split complex number
$$j^j$$
$$j^i$$
Maybe I was wrong to combine complex number with split numbers. I don’t know.
Please let me know If my calculus is ok or there are some mistakes in it.
Thank you.
First:
$$\ln(j)$$
I use the formula for logarithm of split complex numbers:
$$\ln (a + bj) = ½ (\ln ((a + b)(a - b)) + ½ j( \ln (a + b)/(a-b))$$
I know that $(a + b)(a - b) < 0$, we will get a complex number in our answer and that that the split complex logarithm is not closed.
Of course for all formulas I take into consideration only the principal branch.
So:
$$Ln(j)= ½((\ln (0+1)(0-1))+1/2 j (\ln (0+1)/(0-1))$$
$$=1/2 \ln (-1) + ½ j \ln (-1)$$
$$= ½ (i \pi ) +1/2 j ( i \pi)$$
BUT, when I wanted to verify I am correct I discover that, the answer $ ½ (i \pi ) +1/2 j ( i \pi)$ is actually $\ln(-j)$.
I also discover that are at least two correct answer for $\ln(j)$ and for $\ln(-j)$.
Ex. for $\ln(j)$ :
$$Ln(j)=i\pi/2 – j i\pi/2$$ and $$Ln(j)=-i\pi/2 + j i\pi/2$$
Proof:
$$e^{i\pi/2–ji\pi/2}$$
$$=(e^{i\pi/2})*(\cosh(-i\pi/2)+j\sinh(-i\pi/2))$$
$$=(e^{i\pi/2})(0 -ij)$$
$$=(-ij)e^{i\pi/2})$$
$$=(-ij)(\cos(\pi/2)+i\sin(\pi/2)$$
$$=-iji=j$$
I hope is nothing wrong in my calculus. 
So $$j^j=e^{j\ln(j)}$$
$$=$$
1) $$e^{(i\pi/2–ji\pi/2)j}=e^{ji\pi/2–i\pi/2}=-j$$
$$=e^{(-i\pi/2)}(cosh(i\pi/2)+jsinh(I\pi/2)) =e^{(-i\pi/2)}(0-ij)=-ij(cos(-\pi/2)+isin(-\pi/2))=-ij(0-1i)=-j$$
2) $$e^{(-i\pi/2+ji\pi/2)j}= e^{-ji\pi/2+i\pi/2}=-j$$
because $$Ln(j)=i\pi/2 – j i\pi/2$$ and $$Ln(j)=-i\pi/2 + j i\pi/2$$
$$=e^{(i\pi/2)}(cosh(-I\pi/2)+jsinh(-i\pi/2))= e^{(i\pi/2)}(0+ij) =ij(cos(\pi/2)+isin(\pi/2))=ij(0+1i)=-j$$
So... $$j^j=-j$$....
Am I wright? 
$$Ln(i)= i\pi/2$$
$$j^i=e^{(ilnj)}$$
1) $$e^{(i\pi/2 – j i\pi/2)i}=e^{-\pi/2 + j pi/2}= (e^{-\pi/2})(\cosh (\pi/2)+ j \sinh (\pi/2))$$
$$=0.521606...+j0.478393...$$
2) $$e^{(-i\pi/2 + j i\pi/2)i}= e^{i\pi/2 - i\pi/2} = (e^{\pi/2})(\cosh (-\pi/2)+ j \sinh (-\pi/2))$$
$$=12.070346...-j11.070346$$
Am I wright?
$$i^j$$
$$=e^{ji\pi/2}$$
$=\cosh(i\pi/2)+j(\sinh(i\pi/2)$
$=0+j*i$
So ... $i^j=ij$....
Am I right?
Proof:
$i^j=ij$
$$jlni=ln(ij)$$
$$ln(ij)=1/2(ln (0+i)(0-i)) +1/2j ((ln (0+i)/(0-i))$$
$$ln(ij)=1/2ln1 +1/2j ln (-1)$$
$$ln(ij)= (ipi/2)j$$
$$jlni=ln(ij)$$
$$j(ipi/2)=(ipi/2)j$$
Is it possible to find $ln(i+j)$? Cand I find $\ln(i+j+\varepsilon)$, where $\varepsilon^2=0$, with  $\varepsilon \neq 0$?
Thank you.

REPLY [0 votes]: Split-complex numbers can be represented as 2x2 matrices: $a+bj=\left(
\begin{array}{cc}
 a & b \\
 b & a \\
\end{array}
\right)$.
So, $j^j=\left(
\begin{array}{cc}
 0 & 1 \\
 1 & 0 \\
\end{array}
\right)^ \left(
\begin{array}{cc}
 0 & 1 \\
 1 & 0 \\
\end{array}
\right)=\left(
\begin{array}{cc}
 0 & 1 \\
 1 & 0 \\
\end{array}
\right)=j$
Verify it with Wolfram Alpha.
It seems you possibly made a mistake with sign.
The combination of complex and split-complex numbers is called tessarines, they can be represented as 2x2 matrices $a+bj=\left(
\begin{array}{cc}
 a & b \\
 b & a \\
\end{array}
\right)$ with complex elements $a$ and $b$.
So, in tessarines, you are correct that $i^j=ij$.
The $j^i$ is $\left(\frac{1}{2}-\frac{e^{-\pi }}{2}\right) j+\frac{e^{-\pi }}{2}+\frac{1}{2}$
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