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The complex split numbers are defined as $w\in\mathbb{D}:=\{x+jy\,|\;j^2=1,j\neq\pm1\;\; x,y\in \mathbb{R} \}$ now say i want to integrate the function $f(w)=\frac1{w-2}$ and I want to integrate it around the unit hyperbola $\gamma=e^{jt}\;|t\in[-\infty,\infty]$ is there a theorem similar to the Residue one (even if $\mathbb{D}$ is not a field because $0=(1-j)(1+j)$) that lets me evaluate. Also I'm already seeing a bit of ambiguities here. Using the series expansion of $e^x$, analogously to the Euler formula one finds $e^{jt}=\cosh t+j\sinh t$ which is the same as the curve: $x^2-y^2=1$ but then how does create a contour that encloses the other hyperbola $y^2-x^2=1\cong \sinh t+j \cosh t$? Is $e^{jt}=\cosh t+ j\sinh t=j \cosh t+ \sinh t?$ Getting back to our integral: $$ \int_\gamma \frac{dw}{w-2}$$ since $\gamma$ encloses the pole at $w=2$. Working it out by substituting $\gamma(t)=e^{jt} \;|t\in [-T,T] $ with $T\to \infty$ one finds $$\int_{-T}^{T}\frac{1}{e^{jt}-2}\cdot je^{jt}dt=\ln\left( \frac{e^{jT}-2}{e^{-jT}-2}\right)$$ But I'm quite unsure on how to proceed with the limit.

$\textbf{Edit}$ Say instead I integrate the function $g(w)=\frac1w$ with the same contour $\gamma$, then the pole $w=0$ is outside the hyperbola, working it out one finds: $$ \int_\gamma\frac{dw}{w}=\ln \left(\frac{e^{jT}}{e^{-jT}}= \right)=2jT$$ With $T\to \infty$ then $$\int_\gamma\frac{dw}{w}=j\infty$$ I suppose something similar happens with the previous integral. It makes sense since $\gamma$ is a contour that extends to infinity, but does this mean that it doesn’t matter wether the contour encloses the pole or not?

Roccooi
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  • I've thought about this before. I think I decided there's probably not a nice analog to the residue theorem, but this could be wrong. There is an analog to Cauchy's theorem, however. See here. – Geoffrey Sangston Jul 11 '25 at 13:13
  • Very interesting, thanks! – Roccooi Jul 11 '25 at 13:25
  • Also does the equation $w^2=1$ have 4 solutions $w_1=1; w_2=-1; w_3=j;w_4=-j$? or just 2 $w_1=j ; w_2=-j $? – Roccooi Jul 11 '25 at 13:30
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    Well, $1$ is a real number here, so $1^2=1$ – jjagmath Jul 11 '25 at 19:38
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    @Roccooi 4 solutions. All equations of order $n$ with non-zero-divisor coefficients have $n^2$ solutions (if we also allow complex unity). https://math.stackexchange.com/questions/4458423/what-is-sqrt-1-or-sqrt-j-in-the-hyperbolic-split-complex-numbers/4461481#4461481 – Anixx Jul 11 '25 at 19:51

1 Answers1

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We have $$\frac{1}{e^{jt}-2}\cdot je^{jt} = \frac{j (1-2 \cosh (t))-2 \sinh (t)}{5-4 \cosh (t)}$$ so $$\int_{-\infty}^\infty \frac{1}{e^{jt}-2}\cdot je^{jt}~dt=-2\int_{-\infty}^\infty \frac{\sinh(t)}{5-4\cosh(t)}~dt+j \int_{-\infty}^\infty \frac{1-2\cosh(t)}{5-4\cosh(t)}~dt$$

but neither of those integral converge, so the original integral doesn't converge.

jjagmath
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