6

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?

J. Zimmerman
  • 842
Holo
  • 10,246
  • 1
    Is this linked with https://math.stackexchange.com/questions/2521299/caculus-of-jj-ij-and-ji-where-i2-1-j2-1-j-neq-1-and-j-n ? You should define carefully a log in this context, as for complex logarithm – ahstat Nov 16 '17 at 09:20
  • @ahstat that question made me trying to do this. And what do you mean by define carefully log? – Holo Nov 16 '17 at 09:25
  • 1
    For complex numbers, $e^{2i\pi}=e^0$, but $2i\pi \neq 0$. You can probably read about "complex logarithm" and adapt it in your context – ahstat Nov 16 '17 at 09:32
  • Using Mathematica and matrix representation of tessarines I get the same answer but with opposite sign. Maybe, branch choice issue. – Anixx Mar 10 '21 at 11:57
  • All nonzero split-complex numbers are of exactly one of the four forms $\pm e^{jx},,\pm je^{jx}$ with $x\in\Bbb R$. – J.G. Mar 10 '21 at 12:38

2 Answers2

1

Your reasoning is correct. Check it as follows: Let $\, t := (-1+j)\pi i/2.\, $ Then let $$ u := \exp(t) = \exp(j\pi i/2 - \pi i/2) = \exp(j\pi i/2)/\exp(\pi i/2). $$ Now $\, \exp(\pi i/2) = i\,$ and $$\exp(j\pi i/2) = \cosh(\pi i/2) +j\sinh(\pi i/2) = 0 + ji = ji $$ which implies $\, u = ji/i = j\,$ and $\, \log(j) = t\,$ which we wanted to prove.

Somos
  • 37,457
  • 3
  • 35
  • 85
0

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 $1/j=j$.

Anixx
  • 10,161