I am trying to calculate the value of $\ln j$ where $j^2=1, j\ne\pm1$($j$ is split complex).
This is how I did it:
given $e^{j\theta}=\cosh\theta+j\sinh\theta$ I can set $\cosh\theta=0\implies \theta = i\pi n - \frac{i \pi}2, n \in \Bbb Z,i$ is the imaginary number, for convenience sake i'll take $n=1$, using this i can calculate $\sinh\theta$ and get $e^{ji\frac\pi2}=0+ji=ji$(which also implies that $\cos\left(j\frac\pi2\right)=0$ and $\sin\left(j\frac\pi2\right)=j)$
now divide it by $i$(or multiply by $-i$) and i get $\frac{e^{ji\frac\pi2}}i=j$
so: $$\ln j=\ln\left(\frac{e^{ji\frac\pi2}}i\right)=\ln\left(e^{ji\frac\pi2}\right)-\ln i=ji\frac\pi2-i\frac\pi2$$
Am I right by doing this?
Split-complex numbers are not closed under logarithm, so one has to assume tessarines (a combination of complex and split-complex numbers). They can be represented as 2x2 matrices of the form $a+bj=\left( \begin{array}{cc} a & b \\ b & a \\ \end{array} \right)$ with complex elements $a$ and $b$.
So, we can verify your answer with Wolfram Alpha. The result is $\ln j=i\frac\pi2-ji\frac\pi2$, it differs from your result in sign, but it depends on the branch choice, as $1/j=j$.